18.6. THE MATRIX EXPONENTIAL, DIFFERENTIAL EQUATIONS ∗ 391 each determining a solution as described above. Letting xk (t) denote the solution which comes from the chain (v1, v2, · · · , vk) and the above formula involving a sum, it follows that xk (0) = vk. Thus if you consider the solutions coming from the chains v1, (v1, v2) , · · · , (v1, v2, · · · , vm) and consider the vectors obtained by letting t = 0, this results in the ordered list of vectors v1, v2, · · · , vm This is a linearly independent set of vectors. Suppose for some l ≤ m ∑l ckvk = 0 k=1 where not all the ck = 0 and l is as small as possible for this to occur. Then since v1 ≠ 0, it must be that l ≥ 2. Also, cl ̸= 0. Do A − λI to both sides. This gives ∑l ∑l−1 0 = ckvk−1 = ck+1vk k=2 k=1 and so l was not as small as possible after all. Thus the set must be linearly independent after all. Note that for C (λk) , a chain based on an eigenvector corresponding to λk, A : span (C (λk)) → span (C (λk)) Letting Ak be the linear transformation which comes from the restriction of A to span (C (λk)) , what is the matrix of Ak with respect to the ordered basis (v1, v2, · · · , vm) = C (λk)? (A − λkI) vj = vj−1, j > 1 while (A − λkI) v1 = 0. Then formally, the matrix of Ak is given by M where ( )( ) λkv1 v1 + λkv2 · · · vm−1 + λkvm = v1 v2 · · · vm M It follows that M is of the form 1 , λk 1 λk . . . ... λk a matrix which has all zeros except for λk down the main diagonal and 1 down the super diagonal. This is called a Jordan block corresponding to the eigenvalue λk. It can be proved that there are chains {C (λk)}rk=1 of such vectors associated with each eigenvalue λk such that the totality of these vectors form a basis for Cn. Then you simply use these solutions as just described to obtain a matrix Φ (t) whose columns are each solutions to the differential equation x′ = Ax and since the vectors just mentioned form a basis, this yields a fundamental matrix. With respect to this basis, the matrix A becomes similar to one in Jordan Canonical form and the existence of this basis is equivalent to obtaining the existence of the Jordan form. This very interesting theorem is discussed in an appendix if you are interested.
392 CHAPTER 18. LINEAR TRANSFORMATIONS Example 18.6.8 Find a fundamental matrix for the system 2 1 −1 x′ = −1 0 1 x −1 −2 3 In this case the eigenvectors and eigenvalues are 0 −1 1 ↔ 1, 1 ↔ 2 11 The characteristic polynomial is (λ − 1) (λ − 2)2 and so 2 is an eigenvalue of algebraic multiplicity 2. Corresponding to this eigenvalue, there is the above eigenvector and a generalized eigenvector v2 which comes from solving the system −1 (A − 2I) v2 = 1 1 This leads to the augmented matrix 0 1 −1 −1 −1 −2 1 1 −1 −2 1 1 1 a solution to this is −1 . Therefore, there are three solutions to this differential equation 0 0 −1 −1 1 1 et, 1 e2t, t 1 e2t + −1 e2t 11 1 0 Then a fundamental matrix is −e2t 0 e2t e2t − te2t e2t te2t − e2t Φ (t) = et et te2t You can check and see that this works. Other situations are similar. If you can believe that there will be enough of these vectors to obtain a basis formed by chains, then this gives a way to obtain a formula for the fundamental matrix. However, to verify that this is the case, you will need to deal with the Jordan canonical form. From the above discussion involving a power series, the existence of a fundamental matrix is not an issue. This is about finding it in closed form using well known functions and it is only this which involves the necessity of considering the Jordan canonical form. 18.7 Exercises 1. Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R2 through an angle of π/3.
18.7. EXERCISES 393 2. Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R2 through an angle of π/4. 3. Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R2 through an angle of π/12. Hint: Note that π/12 = π/3 − π/4. 4. Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R2 through an angle of 2π/3 and then reflects across the x axis. 5. Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R2 through an angle of π/3 and then reflects across the y axis. 6. Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R2 through an angle of 5π/12. Hint: Note that 5π/12 = 2π/3 − π/4. 7. Let V be an inner product space and u ≠ 0. Show that the function Tu defined by Tu (v) ≡ v − proju (v) is also a linear transformation. Here proju (v) ≡ ⟨v, u⟩ |u|2 u Now show directly from the axioms of the inner product that ⟨Tuv, u⟩ = 0 8. Let V be a finite dimensional inner product space, the field of scalars equal to either R or C. Verify that f given by f v ≡ ⟨v, z⟩ is in L (V, F). Next suppose f is an arbitrary element of L (V, F). Show the following. (a) If f = 0, the zero mapping, then f v = ⟨v, 0⟩ for all v ∈ V . (b) If f ̸= 0 then there exists z ≠ 0 satisfying ⟨u, z⟩ = 0 for all u ∈ ker (f ) . (c) Explain why f (y) z − f (z) y ∈ ker (f ). (d) Use part b. to show that there exists w such that f (y) = ⟨y, w⟩ for all y ∈ V . (e) Show there is at most one such w. You have now proved the Riesz representation theorem which states that every f ∈ L (V, F) is of the form f (y) = ⟨y, w⟩ for a unique w ∈ V. 9. ↑Let A ∈ L (V, W ) where V, W are two finite dimensional inner product spaces, both having field of scalars equal to F which is either R or C. Let f ∈ L (V, F) be given by f (y) ≡ ⟨Ay, z⟩ where ⟨⟩ now refers to the inner product in W. Use the above problem to verify that there exists a unique w ∈ V such that f (y) = ⟨y, w⟩ , the inner product here being the one on V . Let A∗z ≡ w. Show that A∗ ∈ L (W, V ) and by construction, ⟨Ay, z⟩ = ⟨y,A∗z⟩ . In the case that V = Fn and W = Fm and A consists of multiplication on the left by an m × n matrix, give a description of A∗. 10. Let A be the linear transformation defined on the vector space of smooth functions (Those which have all derivatives) given by Af = D2 + 2D + 1. Find ker (A). Hint: First solve (D + 1) z = 0. Then solve (D + 1) y = z.
394 CHAPTER 18. LINEAR TRANSFORMATIONS 11. Let A be the linear transformation defined on the vector space of smooth functions (Those which have all derivatives) given by Af = D2 + 5D + 4. Find ker (A). Note that you could first find ker (D + 4) where D is the differentiation operator and then consider ker (D + 1) (D + 4) = ker (A) and consider Sylvester’s theorem. 12. Suppose Ax = b has a solution where A is a linear transformation. Explain why the solution is unique precisely when Ax = 0 has only the trivial (zero) solution. 13. Verify the linear transformation determined by the matrix () 102 014 maps R3 onto R2 but the linear transformation determined by this matrix is not one to one. 14. Let L be the linear transformation taking polynomials of degree at most three to polynomials of degree at most three given by D2 + 2D + 1 where Dbasisisth{e1,dxi,ffxe2r,exnt3i}at.ioFninodptehraetmora. tFriixndditrhecetmlyaatnridx of this linear transformation relative to the then find the matrix with respect to the differential operator D + 1 and multiply this matrix by itself. You should get the same thing. Why? 15. Let L be the linear transformation taking polynomials of degree at most three to polynomials of degree at most three given by D2 + 5D + 4 where D is the{differentiat}ion operator. Find the matrix of this linear transformation relative to the bases 1, x, x2, x3 . Find the matrix directly and then find the matrices with respect to the differential operators D + 1, D + 4 and multiply these two matrices. You should get the same thing. Why? 16. Show that if L ∈ L (V, W ) (linear transformation) where V and W are vector spaces, then if Lyp = f for some yp ∈ V, then the general solution of Ly = f is of the form ker (L) + yp. 17. Let L ∈ L (V, W ) where V, W are vector spaces, finite or infinite dimensional, and define x ∼ y if x − y ∈ ker (L) . Show that ∼ is an equivalence relation. (x ∼ x, if x ∼ y, then y ∼ x, and x ∼ y and y ∼ z implies x ∼ z.) Next define addition and scalar multiplication on the space of equivalence classes as follows. [x] ≡ {y : y ∼ x} . [x] + [y] ≡ [x + y] α [x] = [αx] Show that these are well defined definitions and that the set of equivalence classes is a vector space with respect to these operations. The zero is [ker L] . Denote the resulting vector space by V / ker (L) . Now suppose L is onto W. Define a mapping A : V / ker (K) → W as follows. A [x] ≡ Lx Show that A is well defined, one to one and onto. 18. If V is a finite dimensional vector space and L ∈ L (V, V ) , show that the minimal polynomial for L equals the minimal polynomial of A where A is the n × n matrix of L with respect to some basis. 19. Let A be an n × n matrix. Describe a fairly simple method based on row operations for computing the minimal polynomial of A. Recall, that this is a monic polynomial p (λ) such that p (A) = 0 and it has smallest degree of all such monic polynomials. Hint: Consider I, A2, · · · . Regard each as a vector in Fn2 and consider taking the row reduced echelon form or something like this. You might also use the Cayley Hamilton theorem to note that you can stop the above sequence at An.
18.7. EXERCISES 395 20. Let A be an n × n matrix which is non defective. That is, there exists a basis of eigenvectors. Show that if p (λ) is the minimal polynomial, then p (λ) has no repeated roots. Hint: First show that the minimal polynomial of A is the same as the minimal polynomial of the diagonal matrix D = D (λ1) ... D (λr) Where D (λ) Sishoawdtiahgaotntahlemmaitnriimx ahlapvionlygnλomdoiawlnist∏heir=m1a(iλn diagonal and in the above, the λi are distinct. − λi) . 21. Show that if A is an n × n matrix and the minimal polynomial has no repeated roots, then A is non defective and there exists a basis of eigenvectors. Thus, from the above problem, a matrix may be diagonalized if and only if its minimal polynomial has no repeated roots. (It turns out this condition is something which is relatively easy to determine. You look at the polynomial and its derivative and ask whether these are relatively prime. The answer to this question can be determined using routine algorithms as discussed above in the section on polynomials and fields. Thus it is possible to determine whether an n × n matrix is defective.) Hint: You might want to use Theorem 18.3.1. 22. Recall the linearization of the Lotka Volterra equations used to model the interaction between predators and prey. It was shown earlier that if x, y are the deviations from the equilibrium point, then x′ = −bxy − b c y d a y′ = dxy + dx b If one is interested only in small values of x, y, that is, in behavior near the equilibrium point, these are approximately equal to the system x′ = −b c y d a y′ = dx b Written more simply, for α, β > 0, x′ = −αy y′ = βx Find the solution to the initial value problem ( )( )( ) ( )( ) x′ 0 −α y′ = β 0 x x (0) = x0 , y (0) y0 y Also have a computer graph the vector fields (−xy − y, xy + x) and (−y, x) which result from setting all the parameters equal to 1. 23. This and the next problems will demonstrate how to solve any second order linear differential equation with constant coefficients. Suppose you want to find all smooth solutions y to y′′ + (a + b) y′ + aby = 0 where a, b are complex numbers, a ≠ b. Then in terms of the differentiation operator D, this can be written as (D + a) (D + b) y = 0
396 CHAPTER 18. LINEAR TRANSFORMATIONS Thus you want to have ker ((D + a) (D + b)) . Thus the two operators (D + a) and (D + b) commute. Show this carefully. It just comes down to standard calculus manipulations. Also show that ker (D + a) is of the form Ce−at where C is a constant. Use this to verify that (D + a) is one to one on ker (D + b) and (D + b) is one to one on ker (D + a). Then use Lemma 18.3.4 to verify that the solution to the above equation is of the form ker ((D + a) (D + b)) = C1e−at + C2e−bt Here we write e−at to signify the function t → e−at. This shows how to solve any equation y′′ +αy′ +βy = 0 where α, β are real or complex numbers. This is because you can always write such an equation in the form discussed above by simply factoring the quadratic polynomial r2 + αr + β using the quadratic formula. Of course you might need to use DeMoivre’s theorem to take square roots. This is called the general solution to the homogeneous equation. Once you have found the general solution, you can then determine the constants C1, C2 in such a way that the solution to a given initial condition where y (0) , y′ (0) are given may be obtained. For example, if I wanted y (0) = 1, y′ (0) = 0, this would be easy to find. I would just need to solve the equations y (0) = C1 + C2 = 1 y′ (0) = (−a) C1 + (−b) C2 = 0 then solve these equations as in the first few chapters of the book. 24. What if the second order equation is of the form (D + a)2 y = 0? Show that () ker (D + a)2 = C1e−at + C2te−at () To show this, verify directly that the right side is in ker (D + a)2 . Next suppose y ∈ () ker (D + a)2 . Thus (D + a) ((D + a) y) = 0 and so (D + a) y = Ce−at because you know ker (D + a) . Now simply solve the above equation as follows. You have y′ + ay = Ce−at Multiply both sides by eat and verify that this yields d (eaty) = C dt Now finish the details to verify that everything in ker (D + a)2 is of the desired form. 25. The case of real coefficients is very important and in fact is the case of most interest. Show that if α, β are real, then if y is a solution to y′′ + αy′ + βy = 0, then so is y¯, the complex conjugate. (( )) 26. If you have the solution to ker D2 + (a + b) D + ab = C1eat + C2ebt and a, b are complex but a + b, ab are real, show that in tfhacets,atmheeysoalruetcioonmfpolrexkecro(n(jDug2a+tes(aso+ab)=Dα++aibβ),)b = α − iβ. Then explain why you get exactly by simply writing in the form e−αt (B1 (cos (βt)) + B2 sin (βt)) The advantage in doing this is that everything is expressed in terms of real functions and typically you are looking for real solutions. 27. Find the solutions to the following initial value problems. Write in terms of real valued functions.
18.7. EXERCISES 397 (a) y′′ + 4y′ + 3y = 0, y (0) = 1, y′ (0) = 1 (d) y′′ + 5y′ + 4y = 0, y (0) = 0, y′ (0) = 1 (b) y′′ + 2y′ + 2y = 0, y (0) = 0, y′ (0) = 1 (e) y′′ + 4y = 0, y (0) = 1, y′ (0) = 0 (c) y′′ + 4y′ + 4y = 0, y (0) = 1, y′ (0) = 1 (f) y′′ − 2y′ + 5y = 0, y (0) = 2, y′ (0) = 1 28. If you have an initial value problem of the form y′′ + b (t) y′ + c (t) y = 0, y (t0) = y0, y′ (t0) = y1 Explain why it can be written in the form ( )′ ( )( ) ( )( ) x1 = 0 1 x1 , x1 (t0) = y0 x2 −c (t) −b (t) x2 x2 (t0) y1 Easy if you take x1 = y and x′1 = x2. Then you just plug in to the equation and get what is desired. 29. Suppose you have two solutions y, yˆ to the “homogeneous” equation y′′ + b (t) y′ + c (t) y = 0 (18.17) The Wronskian of these is defined as ) yˆ (t) ( y (t) yˆ′ (t) W (y, yˆ) (t) ≡ W (t) ≡ det y′ (t) Show that W ′ (t) + b (t) W (t) = 0 Now let B′ (t) = b (t) . Multiply by eB(t) and verify that ( )′ eB(t)W (t) = 0 Hence W (t) = Ce−B(t) This is Abel’s formula. Note that this shows that the Wronskian either vanishes for all t or for no t. 30. Suppose you have found two solutions y, yˆ to (18.17). Consider their Wronskian. Show that the Wronskian is nonzero if and only if the ratio of these solutions is not a constant. Hint: This is real easy. Just use the quotient rule and observe that in the quotient rule the numberator is a multiple of the Wronskian. 31. ↑Show that if you have two solutions y, yˆ of (18.17), whose Wronskian is nonzero, then for any choice of y0, y1, there exist constants C1, C2 such that there is exactly one solution to the initial value problem y′′ + b (t) y′ + c (t) y = 0, y (t0) = y0, y′ (t0) = y1 which is of the form C1y + C2yˆ. When this happens, we say that we have the general solution in the form C1y + C2yˆ. Explain why all solutions to the equation must be of this form. 32. ↑Suppose you have found a single solution to the equation y′′ + b (t) y′ + c (t) y = 0
398 CHAPTER 18. LINEAR TRANSFORMATIONS How can you go about finding the general solution to y′′ + b (t) y′ + c (t) y = 0? Hint: You know that all you have to do is to find another solution yˆ such that it is not a scalar multiple of y. Also, you know Abel’s formula. Letting B′ (t) = b (t) , yˆ (t) y (t) = Ce−B(t) yˆ′ (t) y′ (t) and so you have a differential equation for yˆ −y (t) yˆ′ (t) + yˆ (t) y′ (t) = e−B(t) You don’t care to find all examples. All you need is one which is not a multiple of y. That is why it suffices to let C = 0. Then yˆ′ (t) − y′ (t) = − e−B(t) yˆ (t) y (t) y (t) Now solve this equation for yˆ. First multiply by 1/y. Verify that d (yˆ/y) = − e−B(t) dt y2 (t) Now from this describe how to find yˆ which is not a constant multiple of y. 33. If you have the general solution for y′′ + b (t) y′ + c (t) y = 0, two solutions y, yˆ having nonzero Wronskian, show that it is always possible to obtain all solutions to y′′ + b (t) y′ + c (t) y = f (t) and that such a solution will be of the form A (t) y (t) + B (t) yˆ (t). This is called variation of parameters. Hint: You want it to work. Plug in and require it to do so. ≡0 c (t) (y = Ay + Byˆ) , b (t) y′ = A′y + B′yˆ + Ay′ + Byˆ′ 1 (y′′ = A′y′ + B′yˆ′ + Ay′′ + Byˆ′′) Then show we get a solution if A′y + B′yˆ = 0, A′y′ + B′yˆ′ = f . Now get a solution to this using that the Wronskian is not zero. Then if z is any solution and y˜ is the one you just found, z − y˜ solves y′′ + b (t) y′ + c (t) y = 0 and so is of the form C1y + C2yˆ. 34. Explain how to write the higher order differential equation y(k) + ak−1y(k−1) + · · · + a1y′ + a0y = f (t) as a first order system of the form x′ = Ax + F Thus the theory of this chapter includes all ordinary differential equations with constant coefficients.
Appendix A The Jordan Canonical Form* Recall Corollary 18.4.4. For convenience, this corollary is stated below. Corollary A.0.1 Let A be an n×n matrix. Then A is similar to an upper triangular, block diagonal matrix of the form 0 T ≡ T1 ··· ... ... ... 0 · · · Tr where Tk is an upper triangular matrix having only λk on the main diagonal. The diagonal blocks can be arranged in any order desired. If Tk is an mk × mk matrix, then mk = dim ker (A − λkI)rk . where the minimal polynomial of A is ∏p (λ − λk)rk k=1 The Jordan Canonical form involves a further reduction in which the upper triangular matrices, Tk assume a particularly revealing and simple form. Definition A.0.2 Jk (α) is a Jordan block if it is a k × k matrix of the form α1 0 Jk (α) = 1 0 ... ... ... ... ... 0 ··· 0 α In words, there is an unbroken string of ones down the super diagonal and the number, α filling every space on the main diagonal with zeros everywhere else. A matrix is strictly upper triangular if it is of the form 0∗∗ ... . . . ∗ , 0 ··· 0 where there are zeroes on the main diagonal and below the main diagonal. 399
400 APPENDIX A. THE JORDAN CANONICAL FORM* The Jordan canonical form involves each of the upper triangular matrices in the conclusion of Corollary 18.4.4 being a block diagonal matrix with the blocks being Jordan blocks in which the size of the blocks decreases from the upper left to the lower right. The idea is to show that every square matrix is similar to a unique such matrix which is in Jordan canonical form. It is assumed here that the field of scalars is C but everything which will be done below works just fine if the minimal polynomial can be completely factored into linear factors in the field of scalars. Note that in the conclusion of Corollary 18.4.4 each of the triangular matrices is of the form αI + N where N is a strictly upper triangular matrix. The existence of the Jordan canonical form follows quickly from the following lemma. Lemma A.0.3 Let N be an n × n matrix which is strictly upper triangular. Then there exists an invertible matrix S such that Jr2 (0) S−1N S = Jr1 (0) 0 ... 0 Jrs (0) where r1 ≥ r2 ≥ ·· · ≥ rs ≥ 1 and ∑s ri = n. i=1 Proof: First note the only eigenvalue of N is 0. Let v1 be an eigenvector. Then {v1, v2, · · · , vr} is called a chain if N vk+1 = vk for all k = 1, 2, · · · , r and v1 is an eigenvector so N v1 = 0. It will be called a maximal chain if there is no solution v, to the equation, N v = vr. Claim 1: The vectors in any chain are linearly independent and for {v1, v2, · · · , vr} a chain based on v1, N : span (v1, v2, · · · , vr) → span (v1, v2, · · · , vr) . (1.1) Also if {v1, v2, · · · , vr} is a chain, then r ≤ n. Proof: First note that (1.1) is obvious because ∑r ∑r N civi = civi−1. i=1 i=2 It only remains to verify the vectors of a chain are independent. Suppose then ∑r ckvk = 0. k=1 Do N r−1 to it to conclude cr = 0. Next do N r−2 to it to conclude cr−1 = 0 and continue this way. Now it is obvious r ≤ n because the chain is independent. This proves the claim. Consider the set of all chains based on eigenvectors{. Since all h}ave total length no larger than n it follows there exists one which has maximal length, v11, · · · , vr11 ≡ B1. If span (B1) contains all eigenvectors of N, the{n stop. Othe}rwise, consider all chains based on eigenvectors not in span (B1) and pick one, B2 ≡ v12, · · · , vr22 which is as long as possible. Thus r2 ≤ r1. If span (B1, B2) contains all eigenvectors of N, sto{p. Otherwis}e, consider all chains based on eigenvectors not in span (B1, B2) and pick one, B3 ≡ v13, · · · , vr33 such that r3 is as large as possible. Continue this way. Thus rk ≥ rk+1. Claim 2: The above process terminates with a finite list of chains {B1, · · · , Bs} because for any k, {B1, · · · , Bk} is linearly independent.
401 Proof of Claim 2: The claim is true if k = 1. This follows from Claim 1. Suppose it is true for k − 1, k ≥ 2. Then {B1, · · · , Bk−1} is linearly independent. Suppose ∑p cqwq = 0, cq ̸= 0 q=1 where the wq come from {B1, · · · , Bk−1, Bk} . By induction, some of these wq must come from Bk. Let vik be the one for which i is as large as possible. Then do N i−1 to both sides to obtain v1k, the eigenvector upon which the chain Bk is based, is a linear combination of {B1, · · · , Bk−1} contrary to the construction. Since {B1, · · · , Bk} is linearly independent, the process terminates. This proves the claim. Claim 3: Suppose N w = 0. (w is an eigenvector) Then there exist scalars, ci such that ∑s w = civ1i . i=1 Recall that v1i is the eigenvector in the ith chain on which this chain is based. Proof of Claim 3: From the construction, w ∈ span (B1, · · · , Bs) since otherwise, it could serve as a base for another chain. Therefore, ∑s ∑ri cki vki . w= i=1 k=1 Now apply N to both sides. ∑s ∑ri cki vki −1 0= i=1 k=2 and so by Claim 2, cik = 0 if k ≥ 2. Therefore, ∑s w = c1i v1i i=1 and this proves the claim. If N w = 0, then w ∈ span (B1, · · · , Bs) . In fact, it was a particular linear combination involving the bases of the chains. What if N kw = 0? Does it still follow that w ∈ span (B1, · · · , Bs)? Claim 4: If N kw = 0, then w ∈ span (B1, · · · , Bs) . Proof of Claim 4: Suppose this true that if N k−1w, k ≥ 2, it follows that w ∈ span (B1, · · · , Bs) . Then suppose N kw = 0. If N k−1w = 0, then by induction, w ∈ span (B1, · · · , Bs). The other case is that N k−1w ≠ 0. Then you have the chain N k−1w, · · · , w where N k−1w is an eigenvector. The chain has length no longer than the lengths of any of the Bi by construction. Then by Claim 3, N k−1w = ∑s ∑s ciN k−1vki civ1i = i=1 i=1 And so, () which implies by induction that ∑s N k−1 w− civki = 0 i=1 ∑s w− civki ∈ span (B1, · · · , Bs) . i=1
402 APPENDIX A. THE JORDAN CANONICAL FORM* This proves Claim 4. Since every w satisfies N nw = 0, this shows that span (B1, · · · , Bs) = Fn. By Claim 2, {B1, · · · , Bs} is independent. Therefore, this is a basis for Fn. Now consider the block matrix ( ) S = B1 · · · Bs where () Bk = v1k · · · vrkk . Thus S−1 = C1 ... Cs From the construction, N vjk ∈ span (Bk) , and so CiN vjk = 0. It follows that CiBi = Iri×ri and CiN Bj = 0 if i ≠ j. Let uT1 Ck = ... . urTk Then noting that Bk is n × rk and Ck is rk × n, u1T CkN Bk = ... ( N v1k ··· ) N vrkk uTrk uT1 = ... ( 0 v1k ··· ) vrkk −1 urTk which equals an rk × rk matrix of the form 0 1 ··· 0 Jrk (0) = ... ... ... ... ... ... 1 0 ··· ··· 0 That is, it has ones down the super diagonal and zeros everywhere else. It follows S−1N S = C1 N ( B1 ··· ) ... Bs Cs 0 = Jr1 (0) Jr2 (0) . . . 0 Jrs (0)
403 as claimed. Now let the upper triangular matrices, Tk be given in the conclusion of Corollary 18.4.4. Thus, as noted earlier, Tk = λkIrk×rk + Nk where Nk is a strictly upper triangular matrix of the sort just discussed in Lemma A.0.3. Therefore, Sk−1NkSk is of Now Sk−1λkIrk×rk Sk there exists Sk such that is of the form the form given in Lemma A.0.3. = λkIrk×rk and so Sk−1TkSk Ji2 (λk) Ji1 (λk) 0 ... 0 Jis (λk) where i1 ≥ i2 ≥ ··· ≥ is and ∑s ij = rk . This proves the following corollary. j=1 Corollary A.0.4 Suppose A is an upper triangular n × n matrix having α in every position on the main diagonal. Then there exists an invertible matrix S such that Jk2 (α) S−1AS = Jk1 (α) 0 ... 0 Jkr (α) where k1 ≥ k2 ≥ · · · ≥ kr ≥ 1 and ∑r ki = n. i=1 The next theorem is gives the existence of the Jordan canonical form. Theorem A.0.5 Let A be an n × n matrix having eigenvalues λ1, · · · , λr where the multiplicity of λi as a zero of the characteristic polynomial equals mi. Then there exists an invertible matrix S such that S−1AS = J (λ1) . . . 0 (1.2) 0 J (λr) where J (λk) is an mk × mk matrix of the form Jk2 (λk) (1.3) Jk1 (λk) 0 ... 0 Jkr (λk) where k1 ≥ k2 ≥ · · · ≥ kr ≥ 1 and ∑r ki = mk . i=1 Proof: From Corollary 18.4.4, there exists S such that S−1AS is of the form ··· ... 0 T ≡ T1 ... ... 0 · · · Tr
404 APPENDIX A. THE JORDAN CANONICAL FORM* where Tk is an upper triangular mk × mk matrix having only λk on the main diagonal. By Corollary A.0.4 There exist matrices, Sk such that Sk−1TkSk = J (λk) where J (λk) is described in (1.3). Now let M be the block diagonal matrix given by M = S1 . . . 0 . 0 Sr It follows that M −1S−1ASM = M −1T M and this is of the desired form. What about the uniqueness of the Jordan canonical form? Obviously if you change the order of the eigenvalues, you get a different Jordan canonical form but it turns out that if the order of the eigenvalues is the same, then the Jordan canonical form is unique. In fact, it is the same for any two similar matrices. Theorem A.0.6 Let A and B be two similar matrices. Let JA and JB be Jordan forms of A and B respectively, made up of the blocks JA (λi) and JB (λi) respectively. Then JA and JB are identical except possibly for the order of the J (λi) where the λi are defined above. Proof: First note that for λi an eigenvalue, the matrices JA (λi) and JB (λi) are both of size mi × mi because the two matrices A and B, being similar, have exactly the same characteristic equation and the size of a block equals the algebraic multiplicity of the eigenvalue as a zero of the characteristic equation. It is only necessary to worry about the number and size of the Jordan blocks making up JA (λi) and JB (λi) . Let the eigenvalues of A and B be {λ1, · · · , λr} . Consider the two sequences of numbers {rank (A − λI)m} and {rank (B − λI)m}. Since A and B are similar, these two sequences coincide. (Why?) Also, for the same reason, {rank (JA − λI)m} coincides with {rank (JB − λI)m} . Now pick λk an eigenvalue and consider {rank (JA − λkI)m} and {rank (JB − λkI)m} . Then JA (λ1 − λk) 0 JA − λkI = ... JA (0) ... 0 JA (λr − λk) and a similar formula holds for JB − λkI. Here 0 JA (0) = Jk1 (0) Jk2 (0) ... 0 Jkr (0) and JB (0) = Jl1 (0) 0 Jl2 (0) ... 0 Jlp∑(0) ∑ and it suffices to verify that li = ki for all i. As noted above, ki = li. Now from the above formulas, rank (JA − λkI)m = ∑ mi + rank (JA (0)m) i≠ k ∑ = mi + rank (JB (0)m) i̸=k = rank (JB − λkI)m ,
405 which shows rank (JA (0)m) = rank (JB (0)m) for all m. However, Jl1 (0)m = 0 JB (0)m Jl2 (0)m ... 0 Jlp (0)m with a similar formula holding for JA (0)mand rank (JB (0)m) = ∑p rank (Jli (0)m) , similar for rank (JA (0)m) . In going from m to m + 1, i=1 () rank (Jli (0)m) − 1 = rank Jli (0)m+1 until m = li at which time there is no further change. Therefore, p = r since otherwise, there would exist a discrepancy right away in going from m = 1 to m = 2. Now suppose the sequence {li} is not equal to the sequence, {ki}. Then lr−b ̸= kr−b for some b a nonnegative integer taken to be a small as possible. Say lr−b > kr−b. Then, letting m = kr−b, ∑r ∑r rank (Jli (0)m) = rank (Jki (0)m) i=1 i=1 and in going to m + 1 a discrepancy must occur because the sum on the right will contribute less to the decrease in rank than the sum on the left.
406 APPENDIX A. THE JORDAN CANONICAL FORM*
Appendix B Directions For Computer Algebra Systems B.1 Finding Inverses B.2 Finding Row Reduced Echelon Form B.3 Finding P LU Factorizations B.4 Finding QR Factorizations B.5 Finding The Singular Value Decomposition First here is how you do it using Scientific Notebook. It does it very easily. Next, here is the rather elaborate procedure to get the singular value decomposition using maple. I am using maple 12. Maybe there is an easier way with a more up to date version. I am not sure. B.6 Use Of Matrix Calculator On Web There is a really nice service on the web which will do all of these things very easily. It is www.bluebit.gr/matrix-calculator/ or click on following link or google matrix calculator. You enter a matrix row by row, placing a space between each number. When you come to the end of a row, you press enter on the keyboard to enter the next row. When you have entered the matrix, you select what you want it to do. 407
408 APPENDIX B. DIRECTIONS FOR COMPUTER ALGEBRA SYSTEMS
Bibliography [1] Apostol T. Calculus Volume II Second edition, Wiley 1969. [2] Baker, Roger, Linear Algebra, Rinton Press 2001. [3] Davis H. and Snider A., Vector Analysis Wm. C. Brown 1995. [4] Edwards C.H. Advanced Calculus of several Variables, Dover 1994. [5] Chahal J.S., Historical Perspective of Mathematics 2000 B.C. - 2000 A.D. Kendrick Press, Inc. (2007) [6] Golub, G. and Van Loan, C.,Matrix Computations, Johns Hopkins University Press, 1996. [7] Greenberg M.D. Advanced Engineering Mathematics Prentice Hall 1998 Second edition. [8] Gurtin M. An introduction to continuum mechanics, Academic press 1981. [9] Hardy G. A Course Of Pure Mathematics, Tenth edition, Cambridge University Press 1992. [10] Horn R. and Johnson C. matrix Analysis, Cambridge University Press, 1985. [11] Jacobsen N. Basic Algebra Freeman 1974. [12] Karlin S. and Taylor H. A First Course in Stochastic Processes, Academic Press, 1975. [13] Kuttler K.Linear Algebra On web page. Linear Algebra [14] Nobel B. and Daniel J. Applied Linear Algebra, Prentice Hall, 1977. [15] Rudin W. Principles of Mathematical Analysis, McGraw Hill, 1976. [16] Salas S. and Hille E., Calculus One and Several Variables, Wiley 1990. [17] Strang Gilbert, Linear Algebra and its Applications, Harcourt Brace Jovanovich 1980. [18] Wilkinson, J.H., The Algebraic Eigenvalue Problem, Clarendon Press Oxford 1965. [19] Yosida K., Functional Analysis, Springer Verlag, 1978. 409
410 BIBLIOGRAPHY
Appendix C Answers To Selected Exercises C.1 Exercises 11 De Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for sin (5x) and 5 Let z = 5 + i9. Find z−1. one for cos (5x). (5 + i9)−1 = 5 − 9 i sin (5x) = 5 cos4 x sin x − 10 cos2 x sin3 x + 106 106 sin5 x 6 Let z = 2 + i7 and let w = 3 − i8. Find cos (5x) = cos5 x−10 cos3 x sin2 x+5 cos x sin4 x zw, z + w, z2, and w/z. 13 Factor x3 + 8 as a product of linear factors. 62 + 5i, 5 − i, −45 + 28i, and − 50 − 37 i. √√ 53 53 x3 + 8 = 0, Solution is: i 3 + 1, 1 − i 3, −2 8 Graph the complex cube roots of 8 in the and so this polynomial equals complex plane. Do the same for the four fourth roots of 16. ( ( √ )) ( ( √ )) (x + 2) x − i 3 + 1 x − 1 − i 3 The cube roots a√re the soluti√ons to z3 +8 = 0, Solution is: i 3 + 1, 1 − i 3, −2 () 14 Write x3+27 in the form (x + 3) x2 + ax + b The fourth roots are the solutions to z4 + 16 = 0, Solution is: where x2 + ax + b cannot be factored any more using only real numbers. √√√ (1 − i) 2, − (1 + i) 2, − (1 − i) 2, (1 + i) () √ x3 + 27 = (x + 3) x2 − 3x + 9 2. When you graph these, you will have 16 Factor x4+16 as the product of two quadratic three equally spaced points on the circle of polynomials each of which cannot be fac- radius 2 for the cube roots and you will tored further without using complex num- have four equally spaced points on the cir- bers. cle of radius 2 for the fourth roots. Here ( √ )( √ ) are pictures which should result. x4+16 = x2 − 2 2x + 4 x2 + 2 2x + 4 . You can use the information in the preced- ing problem. Note that (x − z) (x − z) has real coefficients. 17 If z, w are complex numbers prove zw = zw and then show by induction that z1 · · · zm = z1 · · · zm. Also verify that ∑m ∑m zk = zk. 9 If z is a complex number, show there exists k=1 k=1 ω a complex number with |ω| = 1 and ωz = In words this says the conjugate of a prod- |z| . uct equals the product of the conjugates and the conjugate of a sum equals the sum If z = 0, let ω = 1. If z ̸= 0, let ω = z of the conjugates. |z| (a + ib) (c + id) = ac − bd + i (ad + bc) = 11 You already know formulas for cos (x + y) (ac − bd) − i (ad + bc) and sin (x + y) and these were used to prove 411
412 APPENDIX C. ANSWERS TO SELECTED EXERCISES (a − ib) (c − id) = ac−bd−i (ad + bc) which 21 De Moivre’s theorem is really a grand thing. is the same thing. Thus it holds for a prod- I plan to use it now for rational exponents, uct of two complex numbers. Now suppose not just integers. you have that it is true for the product of n complex numbers. Then z1 · · · zn+1 = 1 = 1(1/4) = (cos 2π + i sin 2π)1/4 z1 · · · zn zn+1 and now, by induction this = cos (π/2) + i sin (π/2) = i. equals Therefore, squaring both sides it follows 1 = z1 · · · zn zn+1 −1 as in the previous problem. What does this tell you about De Moivre’s theorem? As to sums, this is even easier. Is there a profound difference between rais- ing numbers to integer powers and raising ∑n ∑n ∑n numbers to non integer powers? (xj + iyj) = xj + i yj It doesn’t work. This is because there are j=1 j=1 j=1 four fourth roots of 1. ∑n ∑n ∑n 22 Here is another question: If n is an integer, = xj − i yj = xj − iyj is it always true that (cos θ − i sin θ)n = cos (nθ) − i sin (nθ)? Explain. j=1 j=1 j=1 Yes, this is true. ∑n = (xj + iyj). (cos θ − i sin θ)n = (cos (−θ) + i sin (−θ))n j=1 = cos (−nθ) + i sin (−nθ) = cos (nθ) − i sin (nθ) 18 Suppose p (x) = anxn + an−1xn−1 + · · · + a1x + a0 where all the ak are real numbers. 23 Suppose you have any polynomial in cos θ Suppose also that p (z) = 0 for some z ∈ C. Show it follows that p (z) = 0 also. and sin θ. By this I mean an expression of You just use the above problem. If p (z) = ∑themfor∑m n aαβ cosα θ sinβ θ where aαβ ∈ 0, then you have p (z) = 0 = α=0 β=0 anzn + an−1zn−1 + · · · + a1z + a0 ∑C.mC+ann this baγlwcoays sγθb+e ∑wrτni=+tt−me(nn+inmt)hceτ form = anzn + an−1zn−1 + · · · + a1z + a0 sin τ θ? = an zn + an−1 zn−1 + · · · + a1 z + a0 γ=−(n+m) = anzn + an−1zn−1 + · · · + a1z + a0 Explain. = p (z) Yes it can. It follows from the identities for 19 Show that 1 + i, 2 + i are the only two zeros the sine and cosine of the sum and differ- to ence of angles that p (x) = x2 − (3 + 2i) x + (1 + 3i) sin a sin b = 1 (cos (a − b) − cos (a + b)) 2 so the zeros do not necessarily come in con- jugate pairs if the coefficients are not real. cos a cos b = 1 (cos (a + b) + cos (a − b)) 2 (x − (1 + i)) (x − (2 + i)) = x2−(3 + 2i) x+ 1 + 3i sin a cos b = 1 (sin (a + b) + sin (a − b)) 2 20 I claim that 1 = −1. Here is why. Now cos θ = 1 cos θ + 0 sin θ and sin θ = √√ √ √ 0 cos θ + 1 sin θ. Suppose that whenever k ≤ −1 −1 (−1)2 1 n, −1 = i2 = = = = 1. ∑k This is clearly a remarkable result but is cosk (θ) = aj cos (jθ) + bj sin (jθ) there something wrong with it? If so, what is wrong? j=−k √Something is wrong. There is no single for some numbers aj, bj. Then −1. cosn+1 (θ) =
C.2. EXERCISES 23 413 ∑n (a) x−212i1√++2221x√+2 1+ i√= 0, Solution is :√x = aj cos (θ) cos (jθ) + bj cos (θ) sin (jθ) i 2, 2 + − 1 x = −1 − 1 j=−n 2 2 Now use the above identities to write all (b) 4x2 + 4ix − 5 = 0, Solution is : x = products as sums of sines and cosines of (j − 1) θ, jθ, (j + 1) θ. Then adjusting the 1− 1 i, x = −1 − 1 i constants, it follows 2 2 cosn+1 (θ) = (c) 4x2 + (4 + 4i) x + 1 + 2i = 0, Solution n∑+1 is : x = − 1 , x = − 1 −i aj′ cos (θ) cos (jθ) + bj′ cos (θ) sin (jθ) 2 2 j=−n+1 (d) x2 − 4ix − 5 = 0, Solution is : x = You can do something similar with sinn (θ) −1 + 2i, x = 1 + 2i and with products of the form (e) 3x2 + 61(161√+−1619i√)+x1(9+16++3(i 1616=√−01,619√)Soi1l9u)tiio, n is : cosα θ sinβ θ. x= x=− 24 Suppose p (x) = anxn + an−1xn−1 + · · · + a1x+a0 is a polynomial and it has n zeros, − 1 − 6 z1, z2, · · · , zn 27 Prove the fundamental theorem of algebra listed according to multiplicity. (z is a root for quadratic polynomials having coefficients of multiplicity m if the polynomial f (x) = in C. (x − z)m divides p (x) but (x − z) f (x) does not.) Show that This is pretty easy because you can simply write the quadratic formula. Finding the square roots of complex numbers is easy from the above presentation. Hence, ev- ery quadratic polynomial has two roots in C. Note that the two square roots in the quadratic formula are on opposite sides of the unit circle so one is −1 times the other. p (x) = an (x − z1) (x − z2) · · · (x − zn) . C.2 Exercises 23 p (x) = (x − z1) q (x) + r (x) where r (x) is () a nonzero constant or equal to 0. However, 1 50 + 3√00 i+ 3√00 j 22 r (z1) = 0 and so r (x) = 0. Now do to q (x) 2 θ = 9.56 degrees. what was done to p (x) and continue until the degree of the resulting q (x) equals 0. 3 Will need 68. 966 gallons of gas. Therefore, Then you have the above factorization. it will not make it. ( √) 25 Give the solutions to the following quadratic equations having real coefficients. 4 At 155 75 3 + 150 () (a) x2 − 2x + 2 = 0, Solution is: 1 + i, 1 − i = 155.0 279. 9 (b) 3x2 + x√+ 3 = 0, Solution is: 1 √ − 5 It takes 2 hours 6 i 35 √ 1 , − 1 i 35 − 1 6 1 13 miles. He ends up 1/3 mile down 6 6 6 6 (c) x2 −6x+13 = 0, Solution is: 3+2i, 3− stream. ( √) 2i 2 5 √ 7 3 , 3 In the second case, he could not (d) x2 +√4x + 9 = 0, Solution is: i 5 − do it. 2, −i 5 − 2 (e) 4x2 + 4x + 5 = 0, Solution is: − 1 + 8 (−3, 2, −5) . 2 − 1 − i, 2 i 9 (3, 0, 0). 26 Give the solutions to the following quadratic ( √ √ √ ) equations having complex coefficients. Note 52 3 −5 2 how the solutions do not come in conjugate 10 + 5 11 pairs as they do when the equation has real 2 2 √ 11 T = 50 26. coefficients.
414 APPENDIX C. ANSWERS TO SELECTED EXERCISES C.3 Exercises 33 7 113 1 This formula says that 9 It means that if you place them so that they u · v = |u| |v| cos θ where θ is the included all have their tails at the same point, the angle between the two vectors. Thus three will lie in the same plane. |u · v| = |u| |v| |cos θ| ≤ |u| |v| 12 a × b × c is meaningless. and equality holds if and only if θ = 0 or 13 (u × v) ×w = (u · w) v− (v · w) u, π. This means that the two vectors either u× (v × w) = (w · u) v− (v · u) w point in the same direction or opposite di- rections. Hence one is a multiple of the 14 u· (z × w) v − v· (z × w) u other. = [u, z, w] v− [v, z, w] u 3 −0.197 39 = cos θ, Thus θ = 1. 769 5 radi- 15 [v, w, z] [u, v, w] ans. 16 0 4 −0.444 44 = cos θ, θ = 2. 031 3 radians. 18 u · u = c, Therefore, u′ · u + u · u′ = 0 so 5 uu··uv(u = −5 (1, 2, 3) ) u′ · u = 0. 14 = − 5 − 5 − 15 14 7 14 C.5 Exercises 59 7 uu··uv(u = (1,2,−2,1)·(1,2,3,0) (1, 2, 3, 0) 1 [ 10 , y = 1] 1+4+9 ) x= 13 13 1 1 3 = − 14 − 7 − 14 0 3 [x = 1, y = 0] 8 It makes no sense. 5 [ 3 , y = 1] x= 5 5 9 projD (F) ≡ F·D D 6 No solution exists. |D| |D| = (|F| cos θ) D = (|F| cos θ) u 8 It appears that the solution exists but is |D| not unique. 11 40 cos ( 20 ) 100 = 3758. 8 180 π 13 20 ( π ) 300 = 4242. 6 9 It appears that there is a unique solution. cos 4 15 (−4, 3, −4) · (0, 1, 0) × 10 = 30 11 There might be a solution. If so, there are infinitely many. ( )√ 17 (2, 3, −4) · 0, √1 , √1 20 = −10 2 12 No. Consider x+y+z = 2 and x+y+z = 1. 22 19 (1, 2, 3, 4) · (2, 0, 1, 3) = 17 13 These can have a solution. For example, 21 |a + b|2 + |a − b|2 = x + y = 1, 2x + 2y = 2, 3x + 3y = 3 even |a|2 + |b|2 + 2a · b + |a|2 + |b|2 − 2a · b has an infinite set of solutions. = 2 |a|2 + 2 |b|2 14 h = 4 15 Any h will work. C.4 Exercises 41 16 Any h will work. 1 If a ̸= 0, then the condition says that 17 If h ̸= 2 there will be a unique solution for any k. If h = 2 and k ̸= 4, there are no |a × u| = |a| sin θ = 0 solutions. If h = 2 and k = 4, then there for all angles θ. Hence a = 0 after all. are infinitely many solutions. 3 1 √374 19 There is no solution. 2 [ 3 − 2 − 1 1] √ 20 w= 2 y 1, x = 3 2 y, z = 3 58 3
C.6. EXERCISES 82 415 22 z = t, y = 3 + t ( 1 ) , 4 4 7 x = 1 − 1 t where t ∈ R. Not possible, 9 , 2 2 23 z = t, y = 4t, x = 2 − 4t. () −2 24 x5 = t, x4 = 1 − 6t, x3 = −1 + 7t, −7 1 5 , x2 = 4t, x1 = 3 − 9t () Not possible, 2 , 25 x5 = t, x3 = s. Then the other variables are −5 given by x4 = − 1 − 3 t, () 2 2 13 x2 = 3 − t 1 , x1 = 5 + 1 t − 2s. , 10 2 2 2 2 39 26 [x = 1 − 2t, z = 1, y = t] 7 27 [x = 2 − 4t, y = −8t, z = t] 4 5 , Not ( −14 13 ) possible, 2, 25 [x = −1, y = 2, z = −1] −5 29 [x = 2, y = 4, z = 5] Not possible, () 30 [x = 1, y = 2, z = −5] −1 −3 11, 4 12 31 [x = −1, y = −5, z = 4] () xy 32 [x = 2t + 1, y = 4t, z = t] 7. −x −y 33 [x = 1, y = 5, z = 3] 0 −1 −2 34 [x = 4, y = −4, z = −2] 8 0 −1 −2 , 1 35 These are not legitimate row operations. 01 2 They do not preserve the solution set of the system. 9 [k = 4] 36 {g = 60, I = 90, b = 200, s = 50} 10 There is no possible choice of k which will make these matrices commute. 37 [w = 15, x = 15, y = 20, z = 10] . C.6 Exercises 82 11 To get −A, just replace every entry of A with its additive inverse. The 0 matrix is () the one which has all zeros in it. −3 −6 −9 12 A= A+AT + A−AT . 1. , 2 2 −6 −3 −21 13 If A = −AT , then aii = −aii and so each () aii = 0. 8 −5 3 , ) i jk −11 5 −4 4 ( 14 Ω × u : ω1 ω2 ω3 −3 3 , Not possi- 7 u1 u2 u3 Not possible, 6 −1 = iω2u3 − iω3u2 − jω1u3 + jω3u1 + kω1u2 − kω2u1. ble, Not possible. () In terms of matrices, this is −3 −9 −3 ω2u3 − ω3u2 3, −ω1u3 + ω3u1 . The matrix is of the −6 −6 3 ω1u2 − ω2u1 () form 5 −18 5 , −11 4 4 11 2 13 6 , −4 2
416 APPENDIX C. ANSWERS TO SELECTED EXERCISES −ω3 0 0 ω2 −2 0 3 −ω1 for suitable ωi since 34 0 ω3 1 − 2 3 3 −ω2 ω1 0 1 0 −1 it is skew symmetric. Now you note that 12 3 multiplication on the left by this matrix 35 2 1 4 , row echelon form: gives the same thing as the cross product. 15 −A = −A + (A + B) 4 5 10 = (−A + A) + B = 0 + B = B 1 0 5 0 1 16 0′ = 0′ + 0 = 0. 3 . There is no inverse. 2 3 00 0 17 0A = (0 + 0) A = 0A+0A. Now add − (0A) 1 1 1 to both sides. Then 0 = 0A. −1 2 1 2 2 2 18 A + (−1) A = (1 + (−1)) A = 0A = 0. 36 3 − 1 − 5 −1 0 2 2 Therefore, from the uniqueness of the addi- 0 1 tive inverse, it follows that −A = (−1) A. −2 − 3 1 9 4 4 4 () 19 (αA + βB)T 1 −1 2 0 ij A = 1 0 2 0 = (αA + βB)ji = αAji + βBji 37 0 0 3 0 = α (AT ) + (βB)Tij = (αAT + β B T ) 1 3 03 ij ij ∑ 20 (ImA)ij ≡ j δikAkj = Aij . x1 + 3x2 + 2x3 24 Explain why if AB = AC and A−1 exists, 38 Write 2x3 + x1 in the form then B = C. 6x3 Because you can multiply on the left by x4 + 3x2 + x1 A−1. )−1 ( ) = x1 ( 3 1 A 2 1 7 − 7 x2 3 1 x3 28 7 2 −1 7 ( 1 )−1 ( − 3 ) x4 0 = 5 1 29 31 5 where A is an appropriate matrix. 5 0 ( 2 )−1 ( 1 ) 1320 10 30 3 A = 1 0 2 0 3 = 0 0 6 0 01 − 2 3 ( )−1 1301 21 31 does not exist. The row re- 42 1 1 1 0 ( ) 1 1 2 1 . 39 A = −1 0 1 0 1 2 0 duced echelon form of this matrix is 0 0 () 1 003 d − b ad−bc 32 ad−bc , assuming ad−bc ̸= 1 202 − c a ad−bc ad−bc 1 1 2 0 . 2 1 −3 2 0 of course. 42 Thus the solution is −2 4 −5 33 0 1 −2 1 212 1 −2 3
C.7. EXERCISES 101 417 5 −4 x y = 6 −6 z 7 −32 w 8 63 12 0 2 a 1 1 2 0 b 9 211 2 1 −3 2 c 11 It does not change the determinant. This 12 1 2 d was just taking the transpose. 13 The determinant is unchanged. It was just a + 2b + 2d the first row added to the second. = a + b + 2c 15 In this case the two columns were switched 2a + b − 3c + 2d so the determinant of the second is −1 times the determinant of the first. a + 2b + c + 2d 43 Multiply on the left of both sides by A−1. 17 det (aA) = det (aIA) = det (aI) det (A) = an det (A) . The matrix which has a down 44 Multiply on both sides on the left by A−1. the main diagonal has determinant equal to Thus 0 = A−10 = an. (A−1 ) A−1 (Ax) = A x = Ix = x 45 A−1 = A−1I = A−1 (AB) 19 This is not true at all. Consider (A−1 ) () A = B = IB = B. A= 10 , 46 You just need to show that (A−1)T acts like the inverse of AT because from unique- 01 ness in the above problem, this will imply () it is the inverse. But from properties of the transpose, B= −1 0 . 0 −1 AT (A−1)T = (A−1A)T = IT = I 20 It must be 0 because (A−1)T AT = (AA−1)T = IT = I 0 = det (0) = det (Ak) = (det (A))k . (A−1)T ( )−1 21 det (A) = 1, or −1. AT Hence = and this last ma- 23 If A = S−1BS, then SAS−1 = B and so if A ∼ B, then B ∼ A. It is obvious that trix exists. A ∼ A because you can let S = I. Say A ∼ B and B ∼ C. Then A = P −1BP and 47 (AB) B−1A−1 = A ( −1) A−1 = AA−1 = B = Q−1CQ. Therefore, BB A = P −1Q−1CQP = (QP )−1 C (QP ) I (A−1 ) A B−1A−1 (AB) = B−1 B = B−1IB = B−1B = I 51 Ax · y = ∑ k (Ax)k yk and so A ∼ C. ∑∑ = i Akixiyk = 25 Say M = S−1N S. Then ∑ ∑k ATikxiyk = (x,AT ) y det ((λI − M) ) i k det λI − S −1 = (λS−1S N S ) S C.7 Exercises 101 = det ( − S−1N S ) 26 = det −1 (λI − N) S 32 46 = det ( −1) det (λI − N) det (S) S ( −1) = det (λI − N ) det S det (S) = det (λI − N )
418 APPENDIX C. ANSWERS TO SELECTED EXERCISES C.8 Exercises 149 1 0 −3 31 − 2 1 5 4 a. is not, b. is, and so is c. 3 3 3 2 − 1 − 2 1 0 3 3 3 100 3 0 − 1 3 5 0 1 0 0 2 0 2 1 0 1 0 0 0 1 33 3 − 9 2 2 0 0 1 −2 1 − 1 0 00 0 2 2 38 Show that if det (A) ̸= 0 for A an n × n ma- trix, it follows that if Ax = 0, then x = 0. 10 0 0 0 0 If det (A) ≠ 0, then A−1 exists and so you 1 1 could multiply on both sides on the left by 2 00 0 1 A−1 and obtain that x = 0. 7 It is because you cannot have more than 39 Suppose A, B are n × n matrices and that min (m, n) nonzero rows in the row reduced AB = I. Show that then BA = I. Hint: echelon form. Recall that the number of You might do something like this: First ex- pivot columns is the same as the number of plain why det (A) , det (B) are both nonzero. nonzero rows from the description of this Then (AB) A = A and then show row reduced echelon form. {( ) ( )} BA (BA − I) = 0. 11 9 , is a basis. 23 From this use what is given to conclude A (BA − I) = 0. Then use Problem 38. 1 1 10 2 , 3 is a basis. You have 1 = det (A) det (B). Hence both 0 1 A and B have inverses. Letting x be given, A (BA − I) x = (AB) Ax − Ax 12 Yes. This is a subspace. It is closed with = Ax − Ax = 0 respect to vector addition and scalar mul- tiplication. and so it follows from the above problem 14 Yes, this is a subspace. that 16 This is a subspace. (BA − I) x = 0. Since x is arbitrary, it follows that BA = I. 18 Not a subspace. 0 20 ∑k 0xk = 0. e−t e−t (cos t + sin t) 0 − (sin t) e−t i=1 40 0 22 If AB = I, then B must be one to one. 0 −e−t (cos t − sin t) (cos t) e−t Otherwise there exists x ≠ 0 such that 41 The columns of the matrix are Bx = 0. 1 e−t 0 But then you would have 2 − sin t , x = Ix = ABx = A0 = 0 1 cos t + 1 sin t 2 2 In particular, the columns of B are linearly independent. Therefore, B is also onto. 1 sin t − 1 cos t cos t Also, 2 2 (BA − I) Bx = B (AB) x−Bx = 0 1 e−t Since B is onto, it follows that BA−I maps 2 every vector to 0 and so this matrix is 0. Thus BA = I. 1 sin t − 1 cos t 2 2 − 1 cos t − 1 sin t 2 2
C.8. EXERCISES 149 419 24 These vectors are not linearly independent. They are linearly dependent. In fact −1 is 0 2 1 times the first added to 2 times the second 0 , 1 , 3 is the third. 1 1 0 001 26 These cannot be linearly independent be- 38 This is obvious. If x, y ∈ V ∩ W, then for scalars α, β, the linear combination αx+βy cause there are 4 of them. You can have at must be in both V and W since they are both subspaces. most three. However, it might be that they span R3. 1432 2 3 1 4 , row echelon form: 40 Let {x1, · · · , xk} be a basis for V ∩W. Then there is a basis for V and W which are re- 3306 spectively {x1, · · · , xk, yk+1, · · · , yp} , 1 0 −1 2 {x1, · · · , xk, zk+1, · · · , zq} 0 1 1 0 . The dimension of the 00 0 0 span of these vectors is 2 so they do not It follows that you must have span R3. 28 These vectors are not linearly independent k+p−k+q−k ≤ n and so they are not a basis. The remaining and so you must have question is whether they span. p+q−n≤k 1412 0 3 2 4 , row echelon form: 42 No. There must then be infinitely many so- lutions. If the system is Ax = b, then there 3300 are infinitely many solutions to Ax = 0 and so the solutions to Ax = b are a particu- lar solution to Ax = b added to the solu- 1000 tions to Ax = 0 of which there are infinitely 0 1 0 0 . The dimension of the span many. 0012 of these vectors is 3 and so they do span R3. 32 Yes it is. It is the span of the vectors 44 Yes. It has a unique solution. 23 46 a. Infinite solution set. b. This surely can’t −1 , 1 . happen. If you add in another column, the rank does not get smaller. c. You can’t 11 have the rank equal 4 if you only have two columns. d. In this case, there is no solu- Since these two vectors are a linearly inde- tion to the system of equations represented pendent set, the given subspace has dimen- by the augmented matrix. e. In this case, sion 2. there is a unique solution since the columns of A are independent.The columns are in- 34 It is a subspace and it equals the span of dependent. Therefore, A is one to one. the vectors which form the columns of the following matrix. 0210 0 1 3 0 , 50 Suppose ABx = 0. ∑Thken Bx ∈ ker (A) ∩ 1 1 0 1 B (Fp) and so Bx = Bzi showing that row echelon form: i=1 0010 ∑k x− zi ∈ ker (B) . 1001 i=1 0 1 0 0 . It follows that the di- Consider B (Fp) ∩ ker (A) and let a basis be 0 0 1 0 0000 {w1, · · · , wk} . mension of this subspace equals 3. A basis
420 APPENDIX C. ANSWERS TO SELECTED EXERCISES Then each wi is of the form Bzi = wi. () Therefore, {z1, · · · , zk} is linearly indepen- Suppose AT Ax = 0. Then AT Ax, x = dent and ABzi = 0. Now let {u1, · · · , ur} be a basis for ker (B) . If ABx = 0, then (Ax,Ax) and so Ax = 0. Therefore, Bx ∈ ker (A) ∩ B (Fp) and so (AT b, ) = (b,Ax) = (b, 0) = 0 ∑k x Bx = ciBzi AT b ∈ ( (( A)T ))⊥ i=1 ker AT It follows that which implies and so there exists a solution x to the equa- ∑k x− cizi ∈ ker (B) tion AT Ax = AT b i=1 by the Fredholm alternative. and so it is of the form 56 ∑k ∑r x− cizi = dj uj |b−Ay|2 i=1 j=1 = |b−Ax+Ax−Ay|2 It follows that if ABx = 0 so that = |b−Ax|2 + |Ax − Ay|2 x ∈ ker (AB) , +2 (b−Ax,A (x − y)) = |+b2−(AAxT|b2 −+A|AT xAx−, Ay|2 ) (x − y) = |b−Ax|2 + |Ax − Ay|2 then and so, Ax is closest to b out of all vectors x ∈ span (z1, · · · , zk, u1, · · · , ur) . Ay. Therefore, 57 The dimension of Fn2 is n2. Therefore, there exist scalars ck such that dim (ker (AB)) ∑n2 ≤ k + r = dim (B (Fp) ∩ ker (A)) ckAk = 0 + dim (ker (B)) k=0 ≤ dim (ker (A)) + dim (ker (B)) Let p (λ) be the monic polynomial having smallest degree such that p (A) = 0. If q (A) = 52 If det (A − λI) = 0 then (A − λI)−1 does 0 then from the Euclidean algorithm, not exist and so the columns are not inde- q (λ) = p (λ) l (λ) + r (λ) pendent which means that for some x ≠ 0, (A − λI) x = 0. where the degree of r (λ) is less than the degree of p (λ) or else r (λ) equals 0. How- 54 Since A1 is not one to one, it follows there ever, if it is not zero, you could plug in A exists x ≠ 0 such that A1x = 0. Hence Ax = 0 and obtain although x ̸= 0 so it follows that A is not 0 = q (A) = 0 + r (A) one to one. From another point of view, if A and this would contradict the definition of were one to one, then ker (A)⊥ = Rn and so p (λ) as being the polynomial having small- by the Fredholm alternative, AT would be est degree which sends A to 0. Hence q (λ) = onto Rn. However, AT has only m columns p (λ) l (λ) . so this cannot take place. 55 That ( A)T = AT A follows from the prop- C.9 Exercises 166 AT erties of the transpose. Therefore, ( ) ( ((AT A)T ))⊥ ( (AT A))⊥ 1 − 1 √ ker ker 2 3 = 3 √2 1 3 1 2 2
C.9. EXERCISES 166 421 ( 1 √ −2121√√22 ) 2 √2 4 1 Now, why does this matrix preserve dis- 2 2 tance? For short, call it Q and note that ( √ ) QT Q = I. Then 3 1 1 |x|2 (QT ) |Qx|2 2 x 5 2√ = Qx, = (Qx,Qx) = − 1 3 1 2 2 ( √ ) and so Q preserves distances. 6 21−√123 3 − 1 2 27 From the above problem this preserves dis- tances and QT = Q. Now do it to x. − 1 2 ( 1 √√ 1 √ 1 √√22√−314+√412√√32 ) 4 √2√3 4 √2 4 7 1 + 1 1 Q (x − y) 4 23 − 4 2 4 = x − y − x−y · ( √) 2 |x − y|2 8 −−12 √12 3 − 1 3 2 (x − y, x − y) 1 2 = y−x ( 1 √ )( 1 √ ) 2 3 2 3 21−√213 1 − − 9 − 212√3 − 1 − 1 Q (x + y) = x + y − x−y · 2 2 2 |x − y|2 ( √√ √ √√ √ ) 1 √2√3 − 1 √2 − 1 2√ 3 − 1 2 (x − y)T (x + y) 16 4 4 4√ 4√ 1 2 3 + 1 2 1 2 3 − 1 2 4 4 4 4 1 √ 21−√123 = x + y − 2 x−y ( − ) 2 3 0 |x − y|2 |x|2 |y|2 0 17 1 2 = x+y 0 0 −1 and so 15 3 15 19 1 5 25 Qx − Qy= y − x 35 Qx+Qy= x + y 3 15 9 21 Hence, adding these yields Qx = y and then subtracting them gives Qy = x. Tu (av+bw) 28 Linear transformations take 0 to 0. Also (av+bw · u) = av+bw− |u|2 u Ta (u + v) ̸= Tau + Tav. = av − a (v · u) u + bw−b (w · u) u |u|2 |u|2 −3tˆ3 −tˆ3 0 = aTu (v) + bTu (w) 32 + −1 , tˆ3 ∈R () tˆ3 0 a2 − b2 2ab 25 2ab b2 − a2 3tˆ −3 26 34 2tˆ + −1 , tˆ ∈ R ( )T ( ) tˆ 0 I I − 2uuT ) − 2uuT ( ( ) −4tˆ 0 = I − 2uuT I − 2uuT 36 −2tˆ + −1 , tˆ ∈ R. =1 tˆ 0 = I − 2uuT − 2uuT + 4uuT uuT −tˆ −1 =I 38 2tˆ + −1 tˆ 0
422 APPENDIX C. ANSWERS TO SELECTED EXERCISES 0 1 −2 −5 0 5 11 3 = 39 −tˆ , tˆ ∈ R 3 −2 −tˆ −6 −15 1 3 tˆ 00 1 −2 −5 0 1 1 0 0 1 1 3 02 −2 −tˆ + −1 3 01 00 01 −tˆ −1 40 1 −3 −4 −3 tˆ 0 10 10 10 = 5 −3 −tˆ −6 2 −5 −8 1 42 + tˆ 5 00 1 −3 −4 −3 tˆ 0 1 1 0 0 1 −2 1 −3 05 −3 1 00 0 1 1 −1 −1 1 , 1 3 −2 1 2 0 −8 44 A basis is 0 1 7 9 2 6 = −6 2 3 2 −7 49 1 0 00 3 −2 1 . −2 −1 s + t 3 1 0 0 0 −2 3 −1/2 −1/2 −2 1 1 0 0 0 1 1 0 0 1 1 −2 −2 1 000 −1 −3 −1 00 9 = 1 3 0 +w 14 3 9 0 + −1 7/2 4 12 16 0 0 0 0 1000 −1 −3 −1 10 −1 1 0 0 0 0 −1 −3 0 1 0 0 0 −3 52 If x, y ∈ ker (A) then −4 0 −4 1 000 A (ax+by) = aAx + bAy = a0 + b0 = 0 11 An LU factorization of the coefficient ma- trix is and so ker (A) is closed under linear com- binations. Hence it is a subspace. 121 0 1 3 230 C.10 Exercises 183 100 121 = 0 1 0 0 1 3 120 2 −1 1 001 1 2 1 3 = 123 120 100 2 1 0 0 −3 3 101 003
C.11. EXERCISES 207 423 First solve 0 20 1 √ −6166356√√√22√√√1111 − 1616√√√22 1 0u 11 √11 − · 1 0 v 3 √11 0 1w 11 1 11 1 2 11 2 2 11 33 3 √ 1−61 √1412√√1111 12116√1√√2√1111 2 −1 11 0 1 00 2 = 2 6 22 You would have QRx = b and so then you would have Rx = QT b. Now R is upper tri- which yields u = 1, v = 2, w = 6. Next angular and so the solution of this problem is fairly simple. solve 121 x 0 1 3 y C.11 Exercises 207 001 z 1 The minimum is −11/2 and it occurs when x1 = x3 = x6 = 0 and x2 = 7/2, x4 = 1 13/2, x6 = −11/2. = 2 The maximum is 7 and it occurs when x1 = 6 7, x2 = 0, x3 = 0, x4 = 3, x5 = 5, x6 = 0. This yields z = 6, y = −16, x = 27. 2 Maximize and minimize the following if pos- ( ) ( )( ) sible. All variables are nonnegative. 01 10 01 (a) The maximum is 7 when x1 = 7 and 14 = 00 x1, x3 = 0. 01 11 The minimum is −7 and it happens when x1 = 0, x2 = 7/2, x3 = 0. ( ) ( )( ) (b) The minimum is −21 and it occurs 01 = 10 01 when x1 = x2 = 0, x3 = 7. 01 The maximum is 7 and it occurs when 01 01 x1 = 7, x2 = 0, x3 = 0. (c) The minimum is 0 and it occurs when 121 x1 = x2 = 0, x3 = 1. The maximum is 14 and it happens 15 1 2 2 = when x1 = 7, x2 = x3 = 0. 211 (d) The maximum is 7 and it happens when x2 = 7/2, x3 = x1 = 0. The minimum is 0 when x1 = x2 = 100 100 0, x3 = 1. 0 0 1 2 1 0 · 4 Find solutions if possible. 010 101 (a) There is no solution to these inequal- ities with x1, x2 ≥ 0. 12 1 0 −3 −1 (b) A solution is x1 = 8/5, x2 = x3 = 0. (c) No solution to these inequalities for 00 1 which all the variables are nonnega- tive. 121 (d) There is a solution when x2 = 2, x3 = 17 1 2 2 = 0, x1 = 0. 2 4 1 (e) There is no solution. 321 1000 10 0 0 0 0 0 1 3 1 0 0 · 0 0 1 0 2 0 1 0 0100 1 0 −1 1 12 1 0 −4 −2 0 0 −1 00 0
424 APPENDIX C. ANSWERS TO SELECTED EXERCISES C.12 Exercises 236 This matrix is defective. In this case, there 3 If it did have λ ∈ R as an eigenvalue, then is only one eigenvalue, −1 of multiplicity 3 there would exist a vector x such that Ax = λx for λ a real number. Therefore, Ax and but the dimension of the eigenspace is only x would need to be parallel. However, this doesn’t happen because A rotates the vec- 2. tors. 5 Amx = λmx for any integer. In the case of 3 ↔ 0 This one is −1, 18 eigenvectors: 4 A−1λx = AA−1x = x 1 4 so A−1x = λ−1x. Thus the eigenvalues of A−1 are just λ−1 where λ is an eigenvalue 1 of A. defective. 7 Let x be the eigenvector. Then Amx = λmx,Amx = Ax = λx and so 3 ↔ 1 λm = λ 19 eigenvectors: 4 1 4 1 This is defective. 21 eigenvectors: 1 ↔ 4, −i −1 −i 1 1 Hence if λ ≠ 0, then λm−1 = 1 ↔ 2 − 2i, i ↔ 2 + 2i i and so |λ| = 1. 1 23 eigenvectors: 3 2 1 ↔ 4, 9 eigenvectors: 1 ↔ 1, 1 ↔ 2. −1 1 1 1 This is a defective matrix. −2 5 −i ↔ 2 − 2i, 11 eigenvectors: 1 , 0 ↔ −1, −i 0 1 1 i 2 ↔ 2 i ↔ 2 + 2i This matrix is not 1 1 1 defective. This matrix is not defective because, even though λ = 1 is a repeated eigenvalue, it 25 eigenvectors: has a 2 dimensional eigenspace. 1 ↔ −6, −1 1 1 0 ↔ 3, 0 , 13 eigenvectors: 0 − 1 2 1 −i ↔ 2 − 6i, −i 1 −1 ↔ 6 1 1 i ↔ 2 + 6i i This matrix is not defective. 1 eigenvectors: − 1 5 ↔ −1 3 , 6 15 1 0 This is not defective. 0 1
C.12. EXERCISES 236 425 27 The characteristic polynomial is of degree 39 Since the vectors are linearly independent, three and it has real coefficients. Therefore, the matrix S has an inverse. Denoting this there is a real root and two distinct complex inverse by roots. It follows that A cannot be defective w1T because it has three distinct eigenvalues. S−1 = ... {( )} wnT 29 eigenvectors: −i ↔ −i, it follows by definition that 1 {( )} wiT xj = δij . i ↔i 1 Therefore, 0 31 eigenvectors: 0 ↔ −1, S−1M S = S−1 (M x1, · · · , M xn) 1 w1T = ... (λ1x1, · · · , λnxn) 1 0 ↔ 1 0 , 1 wnT 0 0 = λ1 0 0 ↔ 1, 33 eigenvectors: 0 ... 1 0 λn 41 The diagonally dominant condition implies 1 ↔ 2, that none of the Gerschgorin disks contain 1 0. Therefore, 0 is not an eigenvalue. Hence 0 A is one to one, hence invertible. 43 First note that (AB)∗ = B∗A∗. Say M x = −1 ↔ 3 λx, x ̸= 0. Then 1 0 λ |x|2 = λx∗x = (λx)∗ x = (M x)∗ x = x∗M ∗x 35 In terms of percentages in the various loca- tions, 21. 429 = x∗M x = x∗λx = λ |x|2 21. 429 Hence λ = λ. 57. 143 46 Suppose A is skew symmetric. Then what about iA? 37 Obviously A cannot be onto because the range of A has dimension 1 and the dimen- (iA)∗ = −iA∗ = −iAT = iA sion of this space should be 3 if the matrix is onto. Therefore, A cannot be invertible. and so iA is self adjoint. Hence it has all Its row reduced echelon form cannot be I real eigenvalues. Therefore, the eigenval- since if it were, A would be onto. Aw = w ues of A are all of the form iλ where λ so it has an eigenvalue equal to 1. Now is real. Now what about the eigenvectors? suppose Ax = λx. Thus, from the Cauchy You need Schwarz inequality, Ax = iλx |x| = |x| |w| |w| |w|2 where λ ̸= 0 is real and A is real. Then A Re (x) = iλ Re (x) ≥ |(x, w)| |w| = |λ| |x| The left has all real entries and the right has |w|2 all pure imaginary entries. Hence Re (x) = 0 and so x has all imaginary entries. and so |λ| ≤ 1.
426 APPENDIX C. ANSWERS TO SELECTED EXERCISES C.13 Exercises 273 9 Yes. 1 ↔ c, 1 a. orthogonal and transformation, b. sym- 11 eigenvectors: 0 metric, c. skew symmetric. 0 4 ∥U x∥2 = (U x, U x) ( ) = U T U x, x = (I x, x) = ∥x∥2 0 ↔ −ib, −i Next suppose distance is preserved by U. 1 Then 0 ↔ ib (U (x + y) , U (x + y)) i 1 = ∥U x∥2 + ∥U y∥2 + 2 (U x, U y) ( ) = ∥x∥2 + ∥y∥2 + 2 U T U x, y 0 ↔ But since U preserves distances, it is also 12 eigenvectors: −i a − ib, the case that 1 (U (x + y) , U (x + y)) = ∥x∥2 + ∥y∥2 + 2 (x, y) 0 ↔ a + ib, i Hence ( ) 1 and so U y (x, y) = T U x, (( T ) ) 1 ↔ c U I y 0 U − x, = 0 0 Since y is arbitrary, it follows that U T U − 13 eigenvectors: I = 0. Thus U is orthogonal. 1 ↔ 6, 5 () √1 1 xyz · 3 1 −1 ↔ 12, a1 a4/2 a5/2 x 1 a4/2 a2 a6/2 y √1 0 2 a5/2 a6/2 a3 z −1 ↔ 18 7 If A is symmetric, then A = U T DU for √1 −1 6 2 some D a diagonal matrix in which all the diagonal entries are non zero. Hence A−1 = 15 U −1D−1U −T . Now λxT x = (Ax)T x U −1U −T = ( T U )−1 U (CD)T =DT CT = I−1 = I xT Ax = and so A−1 = QD−1QT , where Q is orthog- A is=real xT Ax onal. Is this thing on the right symmet- ric? Take its transpose. This is QD−1QT λ is eig=envalue xT λx = λxT x which is the same thing, so it appears that a symmetric matrix must have symmetric and so λ = λ. This shows that all eigenval- inverse. Now consider raising it to a power. ues are real. It follows all the eigenvectors are real. Why? Ak = U T DkU Because A is real. Ax = λx, Ax = λx, so x+x is an eigenvector. Hence it can be assumed all eigenvectors are real. and the right side is clearly symmetric.
C.13. EXERCISES 273 427 Now let x, y, µ and λ be given as above. 34 eigenvectors: √ −31316121−1613√√√√12√√130√326236√233↔↔2↔.1T,0h, e columns are these λ (x · y) = λx · y = Ax · y = x · Ay = x·µy = µ (x · y) = µ (x · y) and so (λ − µ) x · y = 0. Since λ ≠ µ, it follows x · y = 0. 17 vectors. λx · x = Ax · x A He=rmitian x·Ax 37 If A is given by the formula, then = x·λx AT = U T DT U = U T DU = A rule for comple=x inner product λx · x Next suppose A = AT . Then by the the- and so λ = λ. This shows that all eigenval- orems on symmetric matrices, there exists ues are real. Now let x, y, µ and λ be given an orthogonal matrix U such that as above. λ (x · y) = λx · y = Ax · y = U AU T = D x · Ay= x·µy for D diagonal. Hence A = UT DU rule for comple=x inner product µ (x · y) = µ (x · y) 39 There exists U unitary such that A = U ∗T U and so such that T is uppser triangular. Thus A (λ − µ) x · y = 0. and T are similar. Hence they have the Since λ ̸= µ, it follows x · y = 0. same determinant. Therefore, det (A) = () det (T ) , but det (T ) equals the product of 13 the entries on the main diagonal which are 19 Certainly not. 02 the eigenvalues of A. () 2 29 eigenvectors: −31−−12√21√06116√2√√2√2663↔↔126,, 43 3 1 3 44 y = −0.125x2 + 1. 425x + 0.925 46 Find an orthonormal basis for the spans of the following sets of vectors. √ (a) (3, −4, 0) ,(7,−1, 0) ,(1, 7, 1). 1 √3 ↔ 18. 3/5 4/5 0 3 −4/5 , 3/5 , 0 1 √3 0 01 3 3 1 (b) (3, 0, −4) , (11, 0, 2) , (1, 1, 7) 3 The matrix U has these as its columns. 3 4 0 5 5 1 , , 0 0 − 4 3 0 5 5
428 APPENDIX C. ANSWERS TO SELECTED EXERCISES (c) (3, 0, −4) , (5, 0, 10) , (−7, 1, 1) Then 3 , 4 , 0 trace ((AA(∗) ) 5 5 1 0 0 − 4 3 0 = trace U σ 0 V ∗· 5 5 00 47 ( )) 1 √ −13052√√√22 V σ 0 U∗ 6 √6 , 00 1 √6 (( )) 3 = trace U σ2 0 U ∗ 00 61−−171561315√√√333 1 2 2 () ∑ σ2 0 = σj2 , = trace 00 j 49 ∑∑ 53 trace (AB) = ∑i ∑k AikBki, 1 √ 5 5 , trace (BA) = i k BikAki. These give the same thing. Now √0 trace (A) = trace (S−1BS) −25171334035√√5√555√√√111444 ( −1) = trace BSS = trace (B) . C.14 Exercises 292 51 It satisfies the properties of an inner prod- uct. Note that 0.39 trace (AB∗) = ∑∑ 1 −0.09 0.53 Aik Bik ∑i ∑k 5. 319 1 × 10−2 = AikBik 2 7. 446 8 × 10−2 ki 0.712 77 = trace (BA∗) 0.143 94 so (A, B)F = (B, A)F 5 0.939 39 0.280 3 The product is obviously linear in the first argument. If (A, A)F = 0, then 0.205 21 ∑ ∑ AikAik = ∑ |Aik|2 = 0 6 0.117 26 −2. 605 9 × 10−2 ik i,k 7 It indicates that they are no good for doing 52 From the singular value decomposition, it. () U ∗AV = σ0 C.15 Exercises 317 , 1 The actual largest eigenvalue is8 with cor- 00 −1 () responding eigenvector 1 A = U σ 0 V∗ 1 00
C.16. EXERCISES 323 429 4 The largest eigenvalue is −16 and an eigen- 2 First note that either m or −m is in S so S 1 is a nonempty set of positive integers. By well ordering, there is a smallest element of vector is −2 . S, called p = x0m + y0n. Either p divides 1 m or it does not. If p does not divide m, then by the above problem, 5 Eigenvalue near −1.35 : λ = −1. 341, 1.0 m = pq + r −0.456 06 where 0 < r < p. Thus −0.476 32 m = (x0m + y0n) q + r Eigenvalue near 1.5: λ = 1. 679 0, and so, solving for r, 0.867 41 r = m (1 − x0) + (−y0q) n ∈ S. 5. 586 9 −3. 528 2 However, this is a contradiction because p was the smallest element of S. Thus p|m. Eigenvalue near 6.5: λ = 6. 662, Similarly p|n.Now suppose q divides both m and n. Then m = qx and n = qy for integers, x and y. Therefore, 4. 405 2 3. 213 6 6. 171 7 8 Eigenvalue near −1 : λ = −0.703 69, 3. 374 9 p = mx0 + ny0 = x0qx + y0qy −1. 265 3 = q (x0x + y0y) 0.155 75 showing q|p. Therefore, p = (m, n) . Eigenvalue near .25 : λ = 0.189 11, 3 Suppose r is the greatest common divisor −0.242 20 of p and m. Then if r ̸= 1, it must equal p because it must divide p. Hence there exist −0.522 91 integers x, y such that 1.0 Eigenvalue near 7.5 : λ = 7. 514 6, p = xp + ym 0.346 92 which requires that p must divide m which 1.0 is assumed not to happen. Hence r = 1 and so the two numbers are relatively prime. 0.606 92 4 The only substantive issue is why Zp is a 10 22 1√ field. Let [x] ∈ Zp where [x] ≠ [0]. Thus x 3 2 is not a multiple of p. Then from the above λ − ≤ problem, x and p are relatively prime. Hence 3 from another of the above problems, there exist integers a, b such that 12 From the bottom line, a lower bound is −10 From the second line, an upper bound is 12. C.16 Exercises 323 1 = ap + bx 1 The hint is a good suggestion. Pick the first Then thing in S. By the Archimedean property, S ̸= ∅. That is km > n for all k sufficiently [1 − bx] = [ap] = 0 large. Call this first thing q + 1. Thus n − (q + 1) m < 0 but n − qm ≥ 0. Then and it follows that n − qm < m [b] [x] = [1] so [b] = [x]−1. and so 0 ≤ r ≡ n − qm < m.
430 APPENDIX C. ANSWERS TO SELECTED EXERCISES C.17 Exercises 342 (a) These are linearly independent. 1 No. (1, 0, 0, 0) ∈ M but 10 (1, 0, 0, 0) ∈/ M. (b) These are also linearly independent. 3 If not, you could add in a vector not in their 21 This is obvious because when you add two span and obtain 6 vectors which are linearly of these you get one and when you multiply independent. This cannot occur thanks to one of these by{a s√cal}ar, you get another the exchange theorem. one. A basis is 1, 2 . By definition, the span of these gives the collection of √vectors. 10 For each x ∈ [a, b] , let fx (x) = 1 and Are they independent? Say a + b 2 = 0 fx (y) = 0 if y ̸= x. Then these vectors where √a, b are rational numbers. If a ≠ 0, are obviously linearly independent. then b 2 = −a which can’t happe√n since a is rational. If b ̸= 0, then −a = b 2 which 12 A field also has multiplication. However, again can’t happen because on the left is you can consider the elements of the field a rational number and on the right is an as vectors and then it satisfies all the vector irrational. Hence both a, b = 0 and so this space axioms. When you multiply a num- is a basis. ber (vector) in R by a scalar in Q you get something in R. All the axioms for a vec- 29 Consider the claim about ln σ. tor space are now obvious. For example, if α ∈ Q and x, y ∈ R, 1eln(σ) + (−1) σe0 = 0 α (x + y) = αx + αy The equation shown does hold from the def- inition of ln σ. However, if ln σ were alge- from the distributive law on R. braic, then eln σ, e0 would be linearly de- pendent with field of scalars equal to the 13 Simply let f (i) be the ith component of a algebraic numbers, contrary to the Linde- mann Weierstrass theorem. The other in- vector x ∈ Fn. Thus a typical thing in Fn stances are similar. In the case of cos σ, you could use the identity is (f (1) , · · · , f (n)). 1 eiσ + 1 e−iσ − e0 cos σ = 0 14 Say for some n, ∑n ck ek = 0, the zero 22 k=1 contradicting independence of function. Then pick i, eiσ, e−iσ, e0. ∑n 0= ckek (i) k=1 = ciei (i) = ci Since i was arbitrary, this shows these vec- C.18 Exercises 359 tors are linearly independent. 1 I will show one of these. Verify that Exam- 15 Say ples 17.1.1 - 17.1.4 are each inner product spaces. ∑n ckyk = 0 First consider Example 17.1.1. All of the axioms of the inner product are obvious ex- k=1 cept one, the one which says that if ⟨f, f ⟩ = 0 then f = 0. This one depends on continu- Then taking derivatives you have ity of the functions. Suppose then that it is not true. In other words, ⟨f, f ⟩ = 0 and ∑n ckyk(j) = 0, j = 0, 1, 2 · · · , n − 1 yet f ≠ 0. Then for some x ∈ I, f (x) ̸= 0. By continuity, there exists δ > 0 such that k=1 if y ∈ I ∩ (x − δ, x + δ) ≡ Iδ, then This must hold when each equation is eval- |f (y) − f (x)| < |f (x)| /2 uated at x where you can pick the x at which the above determinant is nonzero. It follows that for y ∈ Iδ, Therefore, this is a system of n equations in n variables, the ci and the coefficient ma- trix is invertible. Therefore, each ci = 0. 19 Which are linearly independent? |f (y)| > |f (x)| − |f (x) /2| = |f (x)| /2.
C.18. EXERCISES 359 431 Hence be positive. However, eventually a repeat ∫ will ta(ke plac)e. Thus an = am m < n, and so am ak − 1 = 0 where k = n − m. Since ⟨f, f ⟩ ≥ |f (y)|2 p (x) dy ≥ am ≠ 0, it follows that ak = 1 for a suitable k. It follows that the sequence of powers Iδ of a must include each of {1, 2, · · · , p − 1} and all these would therefore, be positive. () However, 1 + (p − 1) = 0 contradicting the |f (x)|2 /2 (length of Iδ) (min (p)) > 0, assertion that Zp can be ordered. So what would you mean by saying ⟨z, z⟩ ≥ 0? The a contradiction. Note that min p > 0 be- Cauchy Schwarz inequality would not even cause p is a continuous function defined on a closed and bounded interval and so it apply. achieves its minimum by the extreme value theorem of calculus. 2 7 Let δ = r − |z − x| . Then if y ∈ B (z, δ) , ∫ |y − x| ≤ |y − z| + |z − x| < δ + |z − x| = r − |z − x| + |z − x| = r f (x) g (x)p (x) dx and so B (z, δ) ⊆ B (x,r). I (∫ )1/2 ≤ |f (x)|2 p (x) dx · I )1/2 8 ( ) (∫ |g (x)|2 p (x) dx 1 √ 1 √√ 3 √√ − 1 √√ 2 2 2 2 3x 4 2 5x2 4 25 I ∑n 9 Let y go with λ and z go with µ. f (xk) g (xk) z (p (x) y′)′ + (λq (x) + r (x)) yz = 0 k=0 y (p (x) z′)′ + (µq (x) + r (x)) zy = 0 ( )1/2 ( )1/2 ≤ ∑n |f (xk )|2 ∑n |g (xk )|2 Subtract. z (p (x) y′)′−y (p (x) z′)′+(λ − µ) q (x) yz = 0 k=0 k=0 ∑n ( )1/2 ( )1/2 Now integrate from a to b. First note that ∑n ∑n ukwk ≤ |uk |2 |wk |2 z (p (x) y′)′ − y (p (x) z′)′ = d (p (x) y′z − p (x) z′y) k=1 k=1 k=1 ∑∑ dx where u = k ukvk and w = k wkvk. and so what you get is ∑∞ ≤ (∑∞ )1/2 (∑∞ )1/2 p (b) y′ (b) z (b) − p (b) z′ (b) y (b) ak bk |ak |2 |bk |2 − (p (a) y′ (a) z (a) − p (a) z′ (a) y (a)) k=1 k=1 k=1 ∫b + (λ − µ) q (x) y (x) z (x) dx = 0 5 It might be the case that ⟨z, z⟩ = 0 and yet z ≠ 0. Just let z = (z1, · · · , zn) where ex- a actly p of the zi equal 1 but the remaining are equal to 0. Then ⟨z, z⟩ would reduce Look at the stuff on the top line. From the to 0 in the integers mod p. Another prob- assumptions on the boundary conditions, lem is the failure to have an order on Zp. Consider first Z2. Is 1 positive or negative? C1y (a) + C2y′ (a) = 0 If it is positive, then 1 + 1 would need to C1z (a) + C2z′ (a) = 0 be positive. But 1 + 1 = 0 in this case. and so If 1 is negative, then −1 is positive, but −1 is equal to 1. Thus 1 would be both y (a) z′ (a) − y′ (a) z (a) = 0 positive and negative. You can consider Similarly, the general case where p > 2 also. Sim- y (b) z′ (b) − y′ (b) z (b) = 0 ply take a ≠ 1. If a is positive, then con- sider a, a2, a3 · · · . These would all have to Hence, that stuff on the top line equals zero and so the orthogonality condition holds.
432 APPENDIX C. ANSWERS TO SELECTED EXERCISES 11 ∑5 2(−1)k+1 sin (kx) k k=1 15 |∑ni=1 xiyii| ≤ (∑n xi2i)1/2 (∑n yi2i)1/2 i=1 i=1 12 1 π − ∑2 4 cos ((2k + 1) x) 16 2 (2k+1)2 π k=0 ei(t/2)Dn (t) = 1 ∑n ei(k+(1/2))t 2π k=−n ei(−t/2)Dn (t) = 1 ∑n ei(k−(1/2))t 2π k=−n 1 n∑−1 ei(k+(1/2))t = 2π k=−(n+1) () Dn (t) ei(t/2) − e−i(t/2) = 1 () ei(n+(1/2))t − e−i(n+(1/2))t 2π (( ) ) 11 Dn (t) 2i sin (t/2) = 2i sin n+ t 2π 2 13 ∑5 4 (−1)k cos (kx) + π2 Dn (t) = 1 sin (sitn(n( 21+t)12 )) k2 3 2π k=1 You know that t → Dn (t) is periodic of period 2π. Therefore, if f (y) = 1, ∫π ∫π Snf (x) = Dn (x − y) dy = Dn (t) dt −π −π However, it follows directly from computa- tion that Snf (x) = 1. Just take the integral of the sum which defines Dn. 17 From Lemma 17.3.1 and Theorem 17.3.2 ⟨ ⟩ ∑n y − ⟨y, uk⟩ uk, w = 0 k=1
C.18. EXERCISES 359 433 for all w ∈ span ({ui}ni=1) . Therefore, without loss of generality, assume that f ∑n ∑n 2 has real values. Then the above limit re- |y|2 = y− ⟨y, uk⟩ uk + ⟨y, uk⟩ uk duces to having both the real and imagi- nary parts converge to 0. This implies the k=1 k=1 thing which was desired. Note also that if Now if ⟨u, v⟩ = 0, then you can see right α ∈ [−1, 1] , then away from the definition that ∫π |u + v|2 = |u|2 + |v|2 lim f (t) sin ((n + α) t) dt = lim n→∞ −π n→∞ Applying this to ∫π f (t) [sin (nt) cos α + cos (nt) sin α] dt = 0 ∑n u = y− ⟨y, uk⟩ uk, −π k=1 19 From the definition of Dn, ∑n ∫π v = ⟨y, uk⟩ uk, Snf (x) = f (x − y) Dn (y) dy k=1 −π the above equals Now observe that Dn is an even function. Therefore, the formula equals ∑n 2 ∑n 2 = y− ⟨y, uk⟩ uk + ⟨y, uk⟩ uk ∫π Snf (x) = f (x − y) Dn (y) dy k=1 k=1 0 ∑n 2 ∑n = y− ⟨y, uk⟩ uk + |⟨y, uk⟩|2 , ∫0 k=1 k=1 + f (x − y) Dn (y) dy the last step following because of similar ∫−ππ reasoning to the above and the assumption = f (x − y) Dn (y) dy that the ∑ukk∞=a1r|e⟨yo,rutkh⟩o|n2 orcmonavl.ergIets follows ∫ 0 the sum and so π limk→∞ ⟨y, uk⟩ = 0 because if a series con- + f (x + y) Dn (y) dy verges, then the kth term must converge to 0 0. ∫ π f (x + y) + f (x − y) = 2 2Dn (y) dy 18 Let f be any piecewise continuous function which is bounded on [−π, π] . Show, using 0 ∫π the above problem, that ∫π Now note that 0 2Dn (y) = 1 because lim f (t) sin (nt) dt n→∞ ∫−ππ ∫π = lim f (t) cos (nt) dt = 0 Dn (y) dy = 1 n→∞ −π −π and Dn is even. Therefore, Snf (x) − f (x+) + f (x−) = ∫π 2 Let the inner product space consist of piece- 2Dn (y) · wise continuous bounded functions with the 0 inner product defined by f (x + y) − f (x+) + f (x − y) − f (x−) ∫π dy ⟨f, g⟩ ≡ f (x) g (x)dx 2 −π From the formula for Dn (y) given earlier, this is dominated by an expression of the Then, from the abov{e problem} and the fact shown earlier that √1 eikx form 2π k∈Z form an C ∫ π f (x + y) − f (x+) + f (x − y) − f (x−) · orthonormal set of vectors in this inner prod- 0 sin (y/2) uct space, it follows that ⟨ einx⟩ sin ((n + 1/2) y)dy f, lim = 0 n→∞
434 APPENDIX C. ANSWERS TO SELECTED EXERCISES for a suitable constant C. The above is equal 21 Consider for t ∈ [0, 1] the following. to ∫π |y − (x + t (w − x))|2 C 0 y( y ) · where w ∈ K and x ∈ K. It equals f (t) = sin 2 |y − x|2 +t2 |w − x|2 −2t Re ⟨y − x, w − x⟩ f (x + y) − f (x+) + f (x − y) − f (x−) y Suppose x is the point of K which is closest to y. Then f ′ (0) ≥ 0. However, sin ((n + 1/2) y)dy f ′ (0) = −2 Re ⟨y − x, w − x⟩ . and the expression y equals a bounded sin(y/2) continuous function on [0, π] except at 0 Therefore, if x is closest to y, where it is undefined. This follows from elementary calculus. Therefore, changing Re ⟨y − x, w − x⟩ ≤ 0. the function at this single point does not Next suppose this condition holds. Then you have change the integral and so we can consider |y − (x + t (w − x))|2 ≥ this as a continuous bounded function de- |y − x|2 + t2 |w − x|2 ≥ |y − x|2 fined on [0, π] . Also, from the assumptions on f, y → f (x + y) − f (x+) + f (x − y) − f (x−) By convexity of K, a generic point of K is y of the form x + t (w − x) for w ∈ K. Hence x is the closest point. is equal to a piecewise continuous function on [0, π] except at the point 0. Therefore, 22 the above integral converges to 0 by the pre- vious problem. This shows that the Fourier |x + y|2 + |x − y|2 series generally tries to converge to the mid- = |x|2 + |y|2 + 2 Re ⟨x, y⟩ point of the jump. + |x|2 + |y|2 − 2 Re ⟨x, y⟩ 20 π ∑n (−1)k+1 = 2 |x|2 + 2 |y|2 = lim k=1 2k − 1 Of course the same reasoning yields 4 n→∞ You could also find the Fourier series for x2 1 ( + y|2 − |x − ) |x y|2 instead of x and get 4 1 ( ∑n π2 = |x|2 + |y|2 + 2 ⟨x, y⟩ lim 4 (−1)k cos (kx) + 3 x2 4 ( )) k2 = n→∞ k=1 − |x|2 + |y|2 − 2 ⟨x, y⟩ because the periodic extension of this func- = ⟨x, y⟩ tion is continuous. Let x = 0 ∑n 4 (−1)k + π2 =0 23 Let xk be a minimizing sequence. The con- lim k2 3 nection between xk and ck ∈ Fk is obvious n→∞ because the {uk} are orthonormal. That k=1 is, and so |xn∑− xm| = |cn − cm|Fp . π2 ∑n 4 (−1)k+1 where x ≡ j cjuj. Use the parallelogram 3 k2 identity. = lim n→∞ k=1 ≡ ∑∞ 4 (−1)k+1 y − xk − (y − xm) 2 k2 2 k=1 This is one of those calculus problems where + y − xk + (y − xm) 2 you show it converges absolutely by the com- 2 parison test with a p series. However, here is what it converges to. = 2 y − xk 2 y − xm +2 22
C.18. EXERCISES 359 435 Hence saying that ∥x∥ ≤ ∆ |x| . If this is not so, then there exists a sequence of vectors {xk} 1 |xm − xk |2 such that 4 ∥xk∥ > k |xk| = 1 |y − xk |2 2 dividing both sides by ∥xk∥ it can be as- sumed that 1 > k |xk| = xk . Hence xk → 1 y− xk + xm 2 0 in Fk. But from the triangle inequality, 2 2 + |y − xm|2 − ∑n ∥xk∥ ≤ xki ∥ui∥ ≤ 1 |y − xk |2 + 1 |y − xm|2 − λ2 2 2 i=1 Now the right hand side converges to 0 since Therefore, since limk→∞ xik = 0, this is a contradiction to each ∥xk∥ = 1. It follows {xk} is a minimizing sequence. Therefore, that there exists ∆ such that for all x, {xk} is a Cauchy sequence in U. H{ en}ce the sequence of component vectors ck is a ∥x∥ ≤ ∆ |x| Cauchy sequence in Fn and so it converges thanks to completeness of F. It follows that Now consider the other direction. If it is {xk} also must converge to some x. Then since K is closed, it follows that x ∈ K. not true, then there exists a sequence {xk} Hence such that λ = |x − y| . 24 1 |xk | > ∥xk ∥ k ⟨P x − P y, y − P y⟩ ≤ 0 Dividing both sides by |xk| , it can be as- ⟨P y − P x, x − P x⟩ ≤ 0 sumed that |xk| = xk = 1. Hence, by compactness of the closed unit ball in Fn, Thus there exists a further subsequence, still de- ⟨P x − P y, x − P x⟩ ≥ 0 noted by k such that xk → a ∈ Fn and it Hence also follows that |a|Fn = 1. Also the above inequality implies limk→∞ ∥xk∥ = 0. There- fore, ⟨P x − P y, x − P x⟩−⟨P x − P y, y − P y⟩ ≥ 0 ∑n ∑n and so aj uj = lim xjk uj = lim xk = 0 k→∞ k→∞ j=1 j=1 ⟨P x − P y, x − y − (P x − P y)⟩ ≥ 0 which is a contradiction to the uj being lin- early independent. Therefore, there exists |x − y| |P x − P y| ≥ δ > 0 such that for all x, ⟨P x − P y,P x − P y⟩ = |P x − P y|2 δ |x| ≤ ∥x∥ . 25 Let {uk}nk=1 be a basis for V and if x ∈ V, Now if you have any other norm on this finite dimensional vector space, say |||·||| , let xi be the components of x relative to then from what was just shown, there exist scalars δi and ∆i all positive, such that this basis. Thus the xi are defined accord- δ1 |x| ≤ ∥x∥ ≤ ∆1 |x| ing to ∑ δ2 |x| ≤ |||x||| ≤ ∆2 |x| xiui = x i Then decree that {ui} is an orthonormal It follows that basis. It follows |||x||| ≤ ∆2 ∥x∥ ≤ |x|2 = ∑ |xi|2 . δ1 i ∆2∆1 |x| ≤ ∆2∆1 |||x||| . δ1 δ1δ2 Now lleettt{inxgk }{xk} be a sequence of vectors of V denote the sequence of com- ponent vectors in Fn. One direction is easy,
436 APPENDIX C. ANSWERS TO SELECTED EXERCISES Hence (e) If w1, w2 both work, then for every y,0 = δ1 |||x||| ≤ ∥x∥ ≤ ∆1 |||x||| . ∆2 δ2 f (y) − f (y) = ⟨y, w1⟩ − ⟨y, w2⟩ = ⟨y, w1 − w2⟩ In other words, any two norms on a finite dimensional vector space are equivalent norms. Now let y = w1 −w2 and so w1 = w2. What this means is that every considera- tion which depends on analysis or topol- 9 It is required to show that A∗ is linear. ogy is exactly the same for any two norms. What might change are geometric proper- ⟨y,A∗ (αz + βw)⟩ ties of the norms. ≡ ⟨Ay,αz + βw⟩ C.19 Exercises 392 = α ⟨Ay, z⟩ + β ⟨Ay, w⟩ ≡ α ⟨y,A∗z⟩ + β ⟨y,A∗w⟩ ( 1 1 √) = ⟨y,αA∗z⟩ + ⟨y,βA∗w⟩ 2 = ⟨y,αA∗z + βA∗w⟩ 1 √2 − 3 1 3 1 2 2 3 ( 1 − 1 √ )( −2112√√22 −1212√√22 ) Since y is arbitrary, this shows that A∗ is = 2 3 √ √2 √2 1 3 1 linear. In case A is an m × n matrix as 2 (2 2 ) described, √√ −4141√√22−√314 √− 14√√2 1 √2√3 + 1 23 A∗ = (AT ) 4 − 4 1 23 1 11 The two operators D + 1 and D + 4 com- 4 4 mute and are each one to one on the ker- nel of the other. Also, it is obvious that ( −√21 1 √) ker (D + a) consists of functions of the form 2 Ce−at. Therefore, ker (D + 1) (D + 4) con- 5 3 sists of functions of the form 1 3 1 y = C1e−t + C2e−4t 2 2 8 Let f ∈ L (V, F). (a) If f = 0, the zero mapping, then f v = ⟨v, 0⟩ for all v ∈ V . ⟨v, 0⟩ = ⟨v, 0 + 0⟩ = ⟨v, 0⟩ + ⟨v, 0⟩ where C1, C2 are arbitrary constants. In oisth{eer−wt,oer−d4st,}a.basis for ker (D + 1) (D + 4) so ⟨v, 0⟩ = 0. (b) If f ≠ 0 then there exists z ̸= 0 satis- 14 1220 fying ⟨u, z⟩ = 0 for all u ∈ ker (f ) . ker (f ) is a subspace and so there ex- 0 1 4 6 ists z1 ∈/ ker (f ) . Then there exists a 0 0 1 6 closest point of ker (f ) to z1 called x. Then let z = z1 − x. Thus ⟨u, z⟩ = 0 0001 for all u ∈ ker (f ). 17 It is obvious that x ∼ x. If x ∼ y, then (c) y ∼ x is also clear. If x ∼ y and y ∼ z, then f (f (y) z − f (z) y) z−x=z−y+y−x = f (y) f (z) − f (z) f (y) = 0. (d) and by assumption, both z − y and y − x ∈ ker (L) which is a subspace. Therefore, z − x ∈ 0 = ⟨f (y) z − f (z) y, z⟩ ker (L) also and so ∼ is an equivalence re- = f (y) |z|2 − f (z) ⟨y, z⟩ lation. Are the operations well defined? and so ⟨⟩ If [x] = [x′] , [y] = [y′] , is it true that f (z) [x + y] = [y′ + x′]? Of course. x′ + y′ − (x + y) = (x′ − x) + (y′ − y) ∈ ker (L) be- f (y) = y, |z|2 z cause ker (L) is a subspace. Similar rea- so w = f (z) z appears to work. |z|2 soning applies to the case of scalar multi- plication. Now why is A well defined? If
C.19. EXERCISES 392 437 [x] = [x′] , is Lx = Lx′? Of course this is trix and then look for linear relationships. so. x − x′ ∈ ker (L) by assumption. There- fore, Lx = Lx′. It is clear also that A is linear. If A [x] = 0, then Lx = 0 and so 100 2 x ∈ ker (L) and so [x] = 0. Therefore, A is one to one. It is obviously onto L (V ) = W. 0 1 0 −5 0 0 1 4 19 An easy way to do this is to “unravel” the 0 0 0 0 powers of the matrix making vectors in Fn2 0 0 0 0 and then making these the columns of a 0 0 0 0 n2 × n matrix. Look for linear relation- 0 0 0 0 ships between the columns by obtaining the 0 0 0 0 row reduced echelon form and using Lemma 8.2.5. As an example, consider the follow- 000 0 ing matrix. From this and Lemma 8.2.5, you see that for A denoting the matrix, 110 A3 = 4A2 − 5A + 2I −1 0 −1 and so the minimal polynomial is 213 Lets find its minimal polynomial. We have λ3 − 4λ2 + 5λ − 2 the following powers No smaller degree polynomial can work ei- ther. Since it is of degree 3, this is also the 100 110 characteristic polynomial. Note how we got 0 1 0 , −1 0 −1 , this without expanding any determinants or solving any polynomial equations. If you 001 213 factor this polynomial, you get −4 λ3 − 4λ2 + 5λ − 2 = (λ − 2) (λ − 1)2 so this −1 −3 −1 −7 is an easy problem, but you see that this 01 −3 , −7 −6 procedure for finding the minimal polyno- −3 −2 mial will work even when you can’t factor the characteristic polynomial. 758 18 15 19 20 If two matrices are similar, then they must By the Cayley Hamilton theorem, I won’t have the same minimal polynomial. This need to consider any higher powers than is obvious from the fact that for p (λ) any this. Now I will unravel each and make polynomial and A = S−1BS, them the columns of a matrix. p (A) = S−1p (B) S 1 1 0 −3 So what is the minimal polynomial of the diagonal matrix shown? It is obviously 0 1 1 −1 0 0 −1 −4 ∏r 0 −1 −3 −7 (λ − λi) . 1 0 −2 −6 0 −1 −3 −7 i=1 0 2 7 18 0 1 5 15 Thus there are no repeated roots. 1 3 8 19 21 Show that if A is an n × n matrix and the minimal polynomial has no repeated roots, Next you can do row operations and obtain then A is non defective and there exists a the row reduced echelon form for this ma- basis of eigenvectors. Thus, from the above problem, a matrix may be diagonalized if and only if its minimal polynomial has no
438 APPENDIX C. ANSWERS TO SELECTED EXERCISES repeated roots. It turns out this condi- tion is something which is relatively easy to determine. Hint: You might want to use Theorem 18.3.1. If A has a minimal polynomial which has no repeated roots, say ∏m p (λ) = (λ − λi) , j=1 then from the material on decomposing into direct sums of generalized eigenspaces, you have Fn = ker (A − λ1I) ⊕ ker (A − λ2I) ⊕ · · · ⊕ ker (A − λmI) and by definition, the basis vectors for ker (A − λ2I) are all eigenvectors. Thus Fn has a basis of eigenvectors and is therefore diagonalizable or non defective.
Index ∩, 1 balancing, 55 ∪, 1 classical adjoint, 96 σ(A), 371 cofactor, 91, 92, 113 cofactor matrix, 92 Abel’s formula, 106 column rank, 130 Abelian group, 321 column space, 129 absolute value companion matrix, 305, 319 complex conjugate, 4 complex number, 5 complex eigenvalues, 225 adjoint, 258, 262 adjugate, 96, 115 shifted inverse power method, 304 algebraic multiplicity, 216 complex numbers, 4 algebraic number complex numbers minimal polynomial, 339 arithmetic, 4 algebraic numbers, 339 roots, 6 triangle inequality, 5 field, 340 component, 33 angle between vectors, 27 component of a force, 29 area components of a matrix, 68 components of a vector, 16 parallelogram, 36 composition of linear transformations, 383 area of a parallelogram, 35 condition number, 292 augmented matrix, 47 conformable, 71 axioms for a norm, 32 conjugate of a product, 11 back substitution, 46 consistent, 55 bases, 139 Coordinates, 13 basic feasible solution, 188 Cramer’s rule, 100, 115 cross product, 34 finding one, 201 area of parallelogram, 35 basic variables, 53, 188 coordinate description, 35 basis, 139, 324 distributive law, 37, 39 geometric description, 34 any two same size, 327 cross product column space, 141 coordinate description, 35 row space, 141 distributive law, 37 basis of eigenvectors geometric description, 34 diagonalizable, 220 parallelepiped, 38 bijective, 365 binomial theorem, 11 De Moivre’s theorem, 6 block matrix, 253 defective, 216 block multiplication, 253 defective eigenvalue, 216 box product, 38 derivative, 294 determinant, 109 Cartesian coordinates, 14 Cauchy Schwarz inequality, 25, 31 alternating property, 111 Cayley Hamilton theorem, 116, 241, 279 cofactor, 90 characteristic equation, 210 characteristic polynomial, 116 439 characteristic value, 210 chemical reactions
440 INDEX cofactor expansion, 113 Fourier coefficients, 355 expanding along row or column, 90 Fredholm alternative, 146, 264 expansion along row (column), 113 free variables, 53, 188 linear transformation, 381 Frobinius norm, 279 matrix inverse formula, 96, 114 fundamental matrix, 385 minor, 89 fundamental theorem of algebra, 8, 9 nonzero, 148 fundamental theorem of algebra product, 94, 112 product of eigenvalues, 278 plausibility argument, 10 row operations, 93 rigorous proof, 10 transpose, 110 zero, 148 Gauss Elimination, 55 determinant rank Gauss elimination, 47, 48 row rank, 147 Gauss Jordan method for inverses, 79 diagonal matrix, 220, 240 Gauss Seidel, 284 diagonalizable, 220, 240, 251, 381 Gauss Seidel method, 284 differential equations general solution, 165 first order systems, 345 dimension, 140 solution space, 164 definition, 140 generalized eigenspace, 371 dimension of vector space, 328 direct sum, 368 direct sum, 369 distance formula, 17 geometric multiplicity, 216 Dolittle’s method, 175 Gerschgorin’s theorem, 235 dot product, 25 Gram Schmidt process, 258, 350 properties, 25 Grammian matrix, 351 duality, 203 greatest common divisor, 331 dynamical system, 241 Hermitian, 260 echelon form, 48 homogeneous coordinates, 168 eigenspace, 212 homogeneous syster, 164 eigenvalue, 210, 371 homomorphism, 155 Householder matrix, 180 existence, 369 householder matrix, 167 eigenvalues, 116 eigenvector, 210 inconsistent, 52, 55 Einstein summation convention, 40 independent set elementary matrices, 119 elementary matrix extending to a basis, 142 independent set of vectors inverse, 123 properties, 123 extending to form a basis, 142 elementary operations, 45 injective, 365 empty set, 2 inner produc entries of a matrix, 68 equivalence class, 334, 379 strange examplet, 33 equivalence relation, 334, 379 inner product, 25, 30 exchange theorem, 138, 324 axioms, 347 field axioms, 4 Cauchy Schwarz inequality, 348 field extension, 334 inner product properties, 31 dimension, 336 integers mod a prime, 323 finite, 336 intersection, 1 field extensions, 337 intervals Field of scalars, 321 notation, 1 force, 22 inverse left inverse, 115 right inverse, 115 inverses and determinants, 98, 114 invertible, 77 irreducible, 331
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