MATH 110: LINEAR ALGEBRA PRACTICE FINAL SOLUTIONS

MATH 110: LINEAR ALGEBRA PRACTICE FINAL SOLUTIONS Question 1. (1) Write f(x) =a m x m + a m 1 x m 1 + + a 0.ThenF = f(t )=a m T m + a m 1 T m 1 + +a 0 I.SinceTis upper triangular, (T k ) ii = Tii k for all positive k and i {1,...,n}. Hence F ii = a m Tii m + a m 1 T m 1 ii + + a 0 = f(t ii ). (2) TF = Tf(T )=T(a m T m +a m 1 T m 1 + +a 0 I)=a m T m+1 +a m 1 T m + +a 0 T = (a m T m + a m 1 T m 1 + + a 0 I)T = f(t )T = FT. (3) Since T is upper triangular, so is each power of T, and hence so is F = f(t ). Therefore, T ij =0andF ij = 0whenever i>j. Hence and (FT) i,i+1 = (TF) i,i+1 = i+1 F ik T k,i+1 = F ik T k,i+1 = F ii T i,i+1 + F i,i+1 T i+1,i+1 k=1 k=i i+1 T ik F k,i+1 = T ik F k,i+1 = T ii F i,i+1 + T i,i+1 F i+1,i+1. k=1 k=i Then (FT) i,i+1 =(TF) i,i+1 implies that F i,i+1 (T ii T i+1,i+1 )=T i,i+1 F i+1,i+1 F ii T i,i+1. But, by hypothesis, T ii T i+1,i+1 0,so F i,i+1 =(T i,i+1 F i+1,i+1 F ii T i,i+1 )/(T ii T i+1,i+1 ). (4) As before, T ij =0andF ij = 0whenever i>j.sowehave and (FT) i,i+k = (TF) i,i+k = F ij T j,i+k = F ij T j,i+k = F i,i+k T i+k,i+k + j=1 T ij F j,i+k = T ij F j,i+k = T ii F i,i+k + j=1 Equating the two expressions, we obtain F i,i+k (T ii T i+k,i+k )= i+k 1 Again, by hypothesis, T ii T i+k,i+k 0,so i+k 1 F i,i+k =( F ij T j,i+k +1 1 F ij T j,i+k +1 +1 i+k 1 T ij F j,i+k. T ij F j,i+k )/(T ii T i+k,i+k ). F ij T j,i+k T ij F j,i+k.

2 MATH 110: PRACTICE FINAL SOLUTIONS (5) The above calculations work equally well for power series that converge at all eigenvalues of T, in place of polynomials. So, noting that cos(x) = ( 1) n x 2n n=0,wecan (2n)! apply the above considerations. Let us set ( ) π/4 7 T = 0 π/4 and F =cos(t ). By the above work, F 11 =cos(π/4) = 2/2, F 22 =cos( π/4) = 2/2, F21 =0,andF 12 =(T 12 F 22 F 11 T 12 )/(T 11 T 22 )= 7 2 2 2 2 7 π =0. So 4 + π 4 ( ) 2/2 0 F =. 0 2/2 Question 2. The first two parts of this problem only ask you to show that AB and BA have (almost) the same eigenvalues, but don t actually demand you show their characteristic polynomials (almost) agree. Showing the latter condition is stronger than the former, and the hints given actually lead to this stronger result. However there is a simpler method to show the weaker condition, and that is included after the full solution. (1) Without loss of generality, we may assume that A is nonsingular and hence invertible. So A 1 (AB)A = BA. SinceAB and BA are similar, they have the same eigenvalues. (2) Let us compute the characteristic polynomials of AB and BA. We ll be working in the field F (x), consisting of fractions of polynomials in x with coefficients in F. ([ ][ ]) (xi n AB) 0 xi n AB A I n A xi n =1 A I n A xi n A xi n = x n x A xi n n x 0 I n A xi n 0 x 1 I n ([ ][ ][ ]) x x 0 I n A xi n 0 x 1 I n A I n x x 1 A I n A I n A I n ([ ][ ]) x xin BA B A I n A I n 0 I n (xi n BA). Hence AB and BA have the same characteristic polynomial. (3) Again, we ll be working over F (x). ([ ][ ]) (xi m AB) 0 xi m AB A I m A xi m =1 A I m A xi m A xi m

]) 0 I m ([ ][ = x m n x 0 I m MATH 110: PRACTICE FINAL SOLUTIONS 3 ([ = x m n x n ([ = x m n x A xi m 0 x 1 I m ][ ]) A xi m 0 x 1 = x I m n x m A I m = x m n x 1=x A I m n x m A I m A I m ([ ][ ]) = x m n x = x A I m A I m n xin BA B m 0 I n A xi m ]) x m = x m n (xi n BA). Hence (xi m AB) =x m n (xi n BA). Here is a simpler method which just demonstrates that AB and BA have the same eigenvalues, except possibly for 0in the m n case. Suppose λ, v is an eigenpair for AB. ThenBA(Bv) = B(ABv) = Bλv = λbv. Aslongas Bv 0,wehavethatλ, Bv is an eigenpair for BA. But if λ 0,thenABv = λv 0,so Bv 0. This shows that without any restrictions on m and n, AB and BA have the same non-zero eigenvalues (the eigenvalues go back from BA to AB by symmetry). Now suppose m = n and 0is an eigenvalue for AB. Then either A or B is singular, so 0is also an eigenvalue for BA. Now suppose m>n. Then as B is n m, B has non-trivial nullspace, so 0must be an eigenvalue for AB. 0may or may not be an eigenvalue for BA. You may want to think of an example where it is not. Question 3. (1) Let a i denote the ith column of A and q i the ith column of Q. Using the algorithm for computing QR decompositions given in class, we obtain R 11 = a 1 =2and q 1 = a 1 /R 11 = 1/2 (1, 1, 1, 1) t. Also, R 12 = a 2,q 1 = 4, R 13 = a 3,q 1 = 6, R 22 = a 2 q 1 R 12 = (3, 3, 3, 3) t =6,andq 2 =(3, 3, 3, 3) t /R 22 =1/2 (1, 1, 1, 1) t. Continuing in this fashion, one obtains Q = 1 2 1 1 1 1 1 1 1 1 1 1 1 1 and R = 2 4 6 0 6 4 0 0 6 (2) By the computation done in class, x = R 1 Q t y =( 26, 23, 21) t. Question 4. (1) Multiplying X by P on the left interchanges rows i and n +1 i of X (for each i {1, 2,...,n}). So PX =(X n+1 i,j ). Multiplying PX by P on the right interchanges columns j and n+1 j of PX (for each j {1, 2,...,n}). So PXP =(X n+1 i,n+1 j ). (2) Let J be a Jordan block, with λ as its eigenvalue. For each i {1, 2,...,n}, J ii = λ, and hence λ = J n+1 i,n+1 i = (PXP) ii, by the above formula. Similarly, each J i,i+1 = 1, so 1 = J n+1 i,n+2 i = (PXP) i,i 1. Itisalsoeasytoseethatsince J ij = 0whenever j / {i, i +1}, (PXP) ij = 0whenever j / {i, i 1}. Hence, PJP 1 = PJP = J t..

4 MATH 110: PRACTICE FINAL SOLUTIONS Now consider a matrix in Jordan canonical form: J 1 0... 0 0 J J = 2... 0., 0 0... J k where the J i are Jordan blocks. Also, let P 1 0... 0 0 P P = 2... 0., 0 0... P k where for each i, P i has the same size as J i,andeachp i is of the form P of (1). Then P 1 J 1 P 1 0... 0 J t 1 0... 0 0 P P J P = 2 J 2 P 2... 0. = 0 J 2 t... 0. =(J ) t. 0 0... P k J k P k 0 0... Jk t Finally, let A be an n by n complex matrix, and suppose that A = SJS 1 is its Jordan decomposition. Then A t =(S 1 ) t J t S t =(S t ) 1 J t S t. But, by the above, J t is similar to J, and hence A t is similar to J. By the uniqueness of the Jordan canonical form, J is the Jordan canonical form of A t. Question 5. Let D be a diagonal matrix such that D ii = R ii. Also, let L = R D 1 (D is invertible, since A, and hence R, is invertible), and let U = DR. ThenLU = R D 1 DR = R R = R (Q Q)R =(QR) (QR) =A A,sinceQ is unital. Question 6. First note that because {u 1,u 2,v 1,v 2,v 3 } is a basis for F 5, these vectors must be distinct and the sets {u 1,u 2 } and {v 1,v 2,v 3 } are independent. Now clearly u i E c for each i, so span(u 1,u 2 ) E c,andspan(u 1,u 2 ) is 2-dimensional. So dim(e c ) 2. Likewise, span(v 1,v 2,v 3 ) is a 3-dimensional subspace of E d.nowwehave 5=dim(F 5 ) dim(e c + E d )=dim(e c )+dim(e d ) dim(e c E d ). But as c d, E c E d = {0}, so the last term is 0. Thus dim(e c ) = 2 and dim(e d )=3,so span(u 1,u 2 )=E c and likewise for d. Alternatively, as A s characteristic polynomial has degree 5, and dim(e λ ) mult(λ), the multiplicity of λ, it is easy to see that the multiplicities of c and d are 2 and 3 respectively. Thus E c cannot have dimension > 2, so must coincide with span(u 1,u 2 ), and likewise for d. Question 7. In the following I ll also define u j = v j for j>r, for simplicity of notation. The given basis consists of eigenvectors for A, and is orthnormal. Therfore A is normal, so A A = AA.LetA = QDQ be the decomposition of A given by this basis (so the columns of Q are, in order, {u 1,...,u n } and D = Q AQ). Then D =diag(λ 1,...,λ n ), where λ i = c for i r and λ i = d for i>r.forq is the change of co-ordinates matrix from the given eigenbasis to the standard basis. Or just consider D ij = e t ide j = e i Q AQe j =(Qe i ) A(Qe j )= = u i Au j = u i λ j u j = λ j δ ij.

MATH 110: PRACTICE FINAL SOLUTIONS 5 Here I have used that the u i s form an orthnormal set. (7.1) Using A = QDQ, A u i = QD Q u i = QD e i = Qλ i e i = λ i Qe i = λ i u i. So A u i = c u i for i r and A u i = d u i for i>r. (7.2) As β = {u 1,...,u n } forms a basis, CS(A) =span(l A (β)) = span(au 1,...,Au n )=span(cu 1,...,cu r,du r+1,...,du n ). Likewise, CS(A )=span(l A (β)) = span(c u 1,...,c u r,d u r+1,...,d u n ). If c 0andd 0, it is clear that these spans coincide with span(u 1,...,u n )=C n. If c 0=d, then both spans coincide with span(u 1,...,u n ). Generalizing the result of problem 6, this is E c.ifc =0 d, wegete d instead. We can t have c = d as we assumed they were distinct. So we are done. Note that having only 2 distinct eigenvalues wasn t important here; the same arguments generalize to give similar results (such as CS(A) =CS(A )foranormal) for A having more eigenvalues. Question 8. (8.1) True. A is skew-symmetric (A = A t )andreal,soa A = A t A = A 2 = AA, so A is normal. This is equivalent to A being diagonalizable by a unitary matrix. (Note we are talking about complex matrices here, as the question was about unitary matrices. There actually isn t a factorization A = QDQ t with Q real orthogonal and D real diagonal.) (8.2) False. A being unitarily diagonalizable is also equivalent to the existence of an orthnormal basis of eigenvectors for A. ButA is upper triangular with eigenvalues 1, 2, 3. Each of the eigenspaces must have dimension 1, and computing them, one easily sees that if v 1 E 1 and v 2 E 2, both non-zero, then v 1 and v 2 are not orthogonal. (8.3) False. The matrices [ 1 1 0 1 ], [ 1 0 0 2 do not commute. Just try it. (8.4) True. Such matrices may be simultaneously diagonalized, so they commute, by a homework problem. To see this, let J = QDQ 1. Q s columns are eigenvectors for J, so must also be eigenvectors for K by assumption. Therefore K = QCQ 1 for some diagonal matrix C (with the corresponding eigenvalues on its diagonal). Therefore JK = QDQ 1 QCQ 1 = QDCQ 1 = QCDQ 1 = KJ. (8.5) False. This requires the first column of Q 1 to be an eigenvector for A. For in the Q 1 -basis β (given by Q 1 s columns), [L A ] β = T, and the first standard basis vector is an eigenvector for an upper triangular matrix. Alternatively, letting Q s first column be q, q 0andAQ 1 = TQ,so Aq = A(Q 1 e 1 )=Q 1 Te 1 = Q 1 T 11 e 1 = T 11 Q 1 e 1 = T 11 q. ]

6 MATH 110: PRACTICE FINAL SOLUTIONS So A must have a real eigenvector. But this is false for the 90 o rotation matrix [ ] 0 1. 1 0 (8.6) False. Let u =[11111] t.notethatu t M = u t because M is a probability matrix. Now suppose Mx = y. Then 8=u t y = u t Mx = u t x =7. Question 9. Suppose all of f s roots are distinct, and are λ 1,...,λ n.letλ=diag(λ 1,...,λ n ). I will show that any complex matrix whose characteristic poly is f is similar to Λ. This is sufficient, because if M and N are both similar to Λ, then they are similar to one another. So let M be a complex matrix whose characteristic poly is f. ThenM is diagonalizable as it has n distinct eigenvalues. Thus Q 1 MQ = Θ, where Θ = diag(θ 1,...,θ n ). But the θ i s are the roots of f, as are the λ i s. So Θ and Λ have the same diagonal entries, but in possibly different orders. Thus they are similar through a permutation matrix P, and M was similar to Θ, so M is similar to Λ also. If you re interested, I ll now explicitly construct P. I ll define a permutation matrix P such that P ΘP t =Λ. Foreachi, there is a unique j such that λ i = θ j,asm s eigenvalues are f s roots. Let π : {1,...,n} {1,...,n} be the permutation (bijective function) that sends i to j (so λ i = θ π(i) ). We want the i th row of P to extract λ i from Θ, which is in the π(i) th position in Θ, so set P i,j = δ π(i),j. Because π is a permutation, it is easy to check that P is a permutation matrix. Let P i be the i th row of P.AsPP t = I, P i P t j = δ ij. Wehave (P ΘP t ) ij = P i ΘP t j = θ π(i)p i P t j = θ π(i) δ ij = λ i δ ij =Λ ij. Therefore P ΘP t =Λasrequired. Now we do the other direction. Suppose f s roots, counted according to multiplicity, are λ 1,...,λ n. By reordering if needed, suppose that r > 1, and λ 1 = λ i iff i r (so the multiplicity of λ = λ 1 is r>1). Let D =diag(λ 1,...,λ n ), let J = J(λ 1,r)bether r Jordan block with eigenvalue λ 1 and let C =diag(j, λ r+1,...,λ n ). Then both D and C have characteristic poly f, but are not similar. This follows from the uniqueness of the Jordan canonical form: any two Jordan matrices are similar iff for each i and λ, they have the same number of i i Jordan blocks with eigenvalue λ (but they may appear in different orders). D and C do not satisfy this, so are not similar. We may also see it directly as follows. Writing C as a 2 2 block matrix with first block r r, [ ][ ] J(0,r) 0 v Eλ C v1 (C λi)v =0 =0. 0 C 22 λi v 2 This is equivalent to requiring J(0,r)v 1 =0and(C 22 λi)v 2 =0. ButC 22 has diagonal entries distinct from λ (by choice of r), so C 22 λi is upper triangular invertible. So we must have v 2 = 0. Clearly the rank of J(0,r)isr 1, and its nullity is 1, which means N(J(0,r)) and Eλ C have dimension 1. But we assumed 1 <r,sodim(eλ C ) < mult(λ), so that C is not diagonalizable. As D is in fact diagonal, they cannot be similar.

MATH 110: PRACTICE FINAL SOLUTIONS 7 Question 10. Note that if C is an n n matrix with columns c i, e t ice j = e t ic j = C ij,and e t i = e i (e i is the standard basis vector, which is real). So Be j,e i = e i Be j = e t ibe j = B ij, and e j,ae i =(Ae i ) e j = e i A e j = e t i A e j = A ij. Applying the assumption, B ij = A ij for each i, j, sob = A. Question 11. LetW = {v V x, v =0}. We show that W satisfies the three conditions required of a subspace. By one of the first theorems on inner products, 0 W. If u,v W, x, u + v = x, u + x, v =0+0=0,sou + v W. If u W and c F, x, cu = c x, u = c0 =0,socu W.