MTH 102: Linear Algebra Department of Mathematics and Statistics Indian Institute of Technology - Kanpur Problem Set 6 Problems marked (T) are for discussions in Tutorial sessions. 1. Find the eigenvalues and corresponding eigenvectors of matrices 1 2 2 (a) (b) 2 2 2 4 1 3 6 6 Solution: (a) (1 λ) 2 4 = 0 (λ 3)(λ +1) = 0 λ 1 = 3, λ 2 = 1. Also, v 1 = [ 1/2 1 ] T, v 2 = [ 1/2 1 ] T. (b) λ 1 = 0, λ 2 = 2, λ 3 = 3 and v 1 = [ 0 ] T, v 2 = [ 2 1 0 ] T, v 3 = [ 1 0 1 ] T. 2. Construct a basisof R 3 consistingof eigenvectors of the following matrices 0 0 2 1 (a) 0 2 0 (b) 1 2 0 3 1 Solution: (a) Eigenvalues are λ 1 = 1, λ 2 = 2, λ 3 = 4 and eigenvectors are v 1 = [ 1 0 1/2 ] T, v 2 = [ 0 1 0 ] T, v 3 = [ 1 0 2 ] T. Since eigenvectors corresponding to different eigenvalues are linearly independent, {v 1,v 2,v 3 } is a basis of R 3. (b) Similar to (a). 3. (T) This question deals with the following symmetric matrix A: 1 0 1 A = 0 1-1 1-1 0 One eigenvalue is λ = 1 with the line of eigenvectors x = (c,c,0). (a) That line is the null space of what matrix constructed from A? Solution: The eigenvectors of λ = 1 makes the null space of A I. (b) Find the other two eigenvalues of A and two corresponding eigenvectors. Solution: A has trace 2 and determinant 2. So the two eigenvalues after λ 1 = 1 will add to 1 and multiply to 2. Those are λ 2 = 2 and λ 3 = 1. Corresponding eigenvectors are : v 2 = -1, v 3 = -1 1-2
2 (c) The diagonalization A = SΛS 1 has a specially nice form because A = A t. Write all entries in the three matrices in the nice symmetric diagonalization of A. Solution: Every symmetricmatrix hastheniceforma = QΛQ t withanorthogonal matrix Q. The columns of Q are orthonormal eigenvectors. 1/ 2 1/ 3 1/ 6 Q = 1/ 2 1/ 3 1/ 1 0 0 6 0 1/ 3 2/ Λ = 0 2 0 6 0 0-1 4. LetAbeann ninvertiblematrix. Showthateigenvalues ofa 1 arereciprocaloftheeigenvalues of A, moreover, A and A 1 have the same eigenvectors. Solution: Ax = λx x = λa 1 x A 1 x = 1 λx (Note that λ 0 as A is invertible implies that det(a) 0). 5. Let A bean n nmatrix andαbeascalar. Find theeigenvalues of A αi interms of eigenvalues of A. Further show that A and A αi have the same eigenvectors. Solution: If λ is an eigenvalue of A αi with eigenvector v, then Av = (A αi)v +αv = (λ+α)v. Thus, A and A αi have sameeigenvectors and eigenvalues of A αi is µ αif µ is an eigenvalue of A. 6. (T) Let A be an n n matrix. Show that A t and A have the same eigenvalues. Do they have the same eigenvectors? Solution: Follows directly from det(a λi) = det((a λi) t ) = det(a t λi). Eigenvectors are not same. Here is a counter example : 0 0 A =. 1 0 7. Let A be an n n matrix. Show that: (a) If A is idempotent (A 2 = A) then eigenvalues of A are either 0 or 1. Solution: Let Av = λv. Then λv = Av = A 2 v = λ 2 v λ(λ 1)v = 0. Result follows. (b) If A is nilpotent (A m = 0 for some m 1) then all eigenvalues of A are 0. Solution: Let Av = λv. Then A m v = λ m v. Now, A m = 0 λ m = 0 λ = 0. (c) If A = A then, the eigenvalues are all real. Solution: Let (λ,x) be an eigenpair. Then λx x = x (λx) = x (Ax) = (x Ax) = x A x = x Ax = λx x = λx x. Hence, the required result follows.
3 (d) If A = A then, the eigenvalues are either zero or purely imaginary. Solution: Proceed as in the above problem. (e) Let A be a unitary matrix (AA = I = A A). Then, the eigenvalues of A have absolute value 1. It follows that if A is real orthogonal then the eigenvalues of A have absolute value 1. Give an example to show that the conclusion may be false if we allow complex orthogonal. Solution: Let (λ,x) be an eigenpair of A. Then x 2 = x x = x (A A)x = (x A )(Ax) = (Ax) (Ax) = (λx) (λx) = x λλx = λ 2 x 2. [ So λ 2 2 i = 1. For counter example, take A = ]. i 2 8. (T) Suppose that A 15 5 5 = 0. Show that there exists a unitary matrix U such that U AU is upper triangular with diagonal entries 0. Solution: There exists U unitary such that U AU = T, upper triangular with diag(t) = {λ 1,...,λ 5 }. Hence T 15 has diagonal entries λ 15 1,...,λ15 5. As 0 = U A 15 U = T 15 we see that λ 15 i = 0. So, λ i = 0 for all i. 9. (T) Suppose that A 29 17 17 = 0. Show that A17 = 0. Solution: There exists U unitary such that U AU = T, upper triangular with diag(t) = {λ 1,...,λ 17 }. As A 29 = 0, it follows that λ i = 0. So, A = UTU, A 2 = UT 2 U,A 3 = UT 3 U and so on. Also, verify that as T is upper triangular with zeroes on the diagonal, we must have T 17 = 0. So, the result follows. Alternate: As each eigenvalue of A is 0, the characteristic polynomial, namely p A (x) = x 17. So, by Cayley Hamilton theorem, A 17 = 0. 10. The matrix A = is NOT diagonalizable. 0 1 11. The matrix A = is diagonalizable. 0 2 12. Show that Hermitian, Skew-Hermitian and unitary matrices are normal. 13. Suppose that A = A. Show that ranka = number of nonzero eigenvalues of A. Is this true for each square matrix? Is this true for each square symmetric complex matrix? Solution: By spectral theorem, there exists U, unitary such that U AU = D, diagonal. Since U is invertible, we see that ranka = ranku AU = rankd = number of nonzero entries of D = eigenvalues of A. 0 1 It is not true for general square matrices, consider A =. Here ranka = 1, whereas both 0 0 eigenvalues are 0. It is not true for a general complex symmetric matrix, consider A = whereas both eigenvalues are 0 (as deta = 0,trA = 0). 1 i. Here ranka = 1, i 1
14. Show that A = matrix. 2-1 0-1 2 0 2 2 3 is diagonalizable. Find a matrix S such that S 1 AS is a diagonal Solution: det(a λi) = (1 λ)(3 λ) 2. Therefore, eigen-values are 1 and 3. The eigen spaces (nullspaceofa λi), aregivenbye 1 = {x : Ax = x} = {(x 1,x 2,x 3 ) : x 2 = x 1, x 3 = 2x 1, x 1 R} = LS({(1,1, 2)}) and E 3 = {(x 1, x 1,x 3 ) : x 1, x 3 R} = LS({(1, 1,0), (0,0,1)}). Clearly, {(1, 1, 2), (1, 1, 0), (0, 0, 1)} are linearly independent and hence A is diagonalizable. 7-5 15 15. Let A = 6-4 15 0 0 1 calculate A 6. S = 0 1-1 0-2 0 1 Find a matrix S such that S 1 AS is a diagonal matrix and hence Solution: det(a λi) = (λ 1) 2 (λ 2). Therefore, eigen-values are 1and 2. E 1 = {(x 1,x 2,x 3 ) : 6x 1 5x 2 + 15x 3 = 0} = LS({(1,0, 6/15), (0,1,1/3)}). E 2 = {(x 1,x 1,0) : x 1 R} = LS({(1,1,0)}). For 1 0 1 S = 0, we have Therefore 16. Consider the 3 3 matrix - 6 15 S 1 AS = A 6 = S 1 A = Determine the entries a,b,c,d,e,f so that: 1 3 0 1 0 0 0 1 0 0 0 2 1 0 0 0 1 0 0 0 2 6 a b c 1 d e 0 1 f the top left 1 1 block is a matrix with eigenvalue 2; S. the top left 2 2 block is a matrix with eigenvalue 3 and -3; the top left 3 3 block is a matrix with eigenvalue 0, 1 and -2. 4
5 Solution: Let A i denote the top left i i block of A. The matrix A 1 is the matrix [a]. Since a is the only eigenvalue of this matrix, we conclude that a = 2. 2 b We now move onto determining the entries of the matrix A 2 : A 2 =. 1 d Since the sum of the eigenvalues of A 2 is 0 by hypothesis, and it is also equal to the trace of A 2, we obtain that 2+d = 0 or d = 2. Moreover the product of the eigenvalues of A 2 is 9 by hypothesis, and it is qual to the determinant of A 2. Thus we have 9 = 2d b = 4 b 2 5 and we deduce that b = 5 and therefore A 2 =. 1-2 Finally, consider A = A 3. Again, the sum of the eigenvalues of A is 1 and it is also equal to the trace of A. We deduce that f = 1. We still need to determine the entries c and e of A and we have 2 5 c A = 1-2 e 0 1-1 The characteristic polynomial of this matrix is λ 3 λ 2 +(e+9)λ+c 2e+9. We know that the roots of this polynomial must be 0, 1 and 2. Setting λ = 0 and λ = 1, we obtain which is equivalent to Thus c = 7 and e = 9 and we conclude 17. NOT for mid-sem or end-sem c 2e+9 = 0 1 1+(e+9)+c 2e+9 = 0 c 2e = 9 c e = 16. A = 2 5-7 1-2 9 0 1-1 (a) Find the eigenvalues and eigenvectors (depending on c) of 0.3 c A =. 0.7 1 c For which value of c is the matrix A not diagonallizable (so A = SΛS 1 is impossible)?
6 Solution: Eigen values are λ = 1 and λ = 0.3 c. The eigenvector for λ = 1 is in the null space of -0.7 c A I = 0.7 -c so x 1 = [ c 0.7 Similarly, the eigenvector for λ = 0.3 c is in the null space of c c A (0.3 c)i = 0.7 0.7 so x 2 = [ 1-1 A is not diagonalizable when its eigen values are equal : 1 = 0.3 c or c = 0.7. (b) What is the largest range of values of c (real number) so that A n approaches a limiting matrix A as n? Solution: ]. ]. [ A n = SΛ n S 0 = S 0 (0.3 c) n ] S 1. This approaches a limit if 0.3 c < 1. We could write that out as 0.7 < c < 1.3. (c) What is the limit of A n (still depending on c)? You could work from A = SΛS 1 to find A n. Solution: The eigen vectors are in S. As n, the smaller eigen value λ n 2 leaving A c 0 = /(c+0.7) 0.7 1 0 0 0.7 c c c = /(c+0.7). 0.7 0.7 goes to zero,