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Extra Credit Writing Assignment 4: Solutions
- Problem 20 (a), page 277.
Solution. For an arbitrary vector x we have the relations [x]D = PD←C [x]C , [x]C = PC←B[x]B,
(1) [x]D = PD←B[x]B,
with the change-of-coordinates matrices
PD←C =
[
[d 1 ]C , [d 2 ]C
]
PC←B =
[
[c 1 ]B, [c 2 ]B
]
PD←B =
[
[d 1 ]B, [d 2 ]B
]
Therefore, [x]D = PD←C [x]C = PD←C PC←B[x]B.
Comparing this with equality (1), we obtain that
PD←B = PD←C PC←B.
- Problem 26, page 309.
Solution. If λ is an eigenvalue of A and x is a corresponding eigenvector, i.e., Ax = λx, then A^2 x = λ^2 x, i.e., λ^2 is an eigenvalue of A^2 , with the same eigenvector x. If A^2 = 0 then necessarily λ^2 = 0 because x 6 = 0. Therefore, λ = 0 is the only possible eigenvalue of A.
- Problem 27, page 309.
Solution. The characteristic polynomials of A and AT^ coincide. Indeed, det(AT^ − λI) = det(A − λI)T^ = det(A − λI)
because the determinant of the transpose matrix is the same as the determi- nant of a matrix. Therefore, the eigenvalues of A and AT^ , i.e., the solutions of the characteristic equations det(A − λI) = 0 and det(AT^ − λI) coincide. In other words, λ is an eigenvalue of A if and only if λ is an eigenvalue of AT^.
- Let A and B be two n × n matrices such that AB = BA. (a) Show that if x is an eigenvector of A then either Bx = 0 or Bx is an eigenvector of A. (b) Show that if A has n distinct eigenvalues then B has the same eigenvec- tors as A. Moreover, both A and B can be diagonalized, and if A = P DP −^1 1
2
with an invertible matrix P and a diagonal matrix D, then B = P F P −^1 with the same P and a diagonal matrix F.
Solution. (a) If x is an eigenvector of A corresponding to an eigenvalue λ then ABx = BAx = λBx.
Thus, either Bx = 0 or Bx is another eigenvector of A corresponding to the eigenvalue λ. (b) If A has n distinct eigenvalues λ 1 ,... , λn then the corresponding eigen- vectors x 1 ,... , xn are linearly independent. Therefore, A is diagonalizable (see Theorem 6 in section 5.3), i.e., A = P DP −^1 , where D is a diagonal matrix with λ 1 ,... , λn on the main diagonal, and P is an invertible matrix whose columns are x 1 ,... , xn. The eigenspace of A corresponding to each λk is one-dimensional. Indeed, A − λkIn = P DP −^1 − λkIn = P (D − λkIn)P −^1
and the diagonal matrix D − λkIn has n − 1 nonzero entries (pivots on the main diagonal), thus only one free variable. The solution set of the equation (D − λkIn)y = 0 is therefore one-dimensional. Then so in the solution set of the equation (A − λIn)x = 0 because for each its solution x, there is a solution y of the preceding equation such that x = P y. Next, by part (a) we have Bxk = 0 or Bxk is an eigenvector of A corre- sponding to the eigenvalue λk, for k = 1,... , n. In the first case, xk is an eigenvector of B, corresponding to the eigenvalue μk = 0. In the second case, Bxk must be proportional to xk because the eigenspace of A corresponding to the eigenvalue λk, is one-dimensional; therefore, there is a number μk such that Bxk = μkxk. In both cases, xk is an eigenvector of B. Since B and A have the same set of eigenvectors, B is also diagonalizable by means of the same invertible matrix P , i.e., B = P F P −^1 , where F is a diagonal matrix which has numbers μ 1 ,... , μn on the main diagonal.
