Exercise 5.1.1

Answers

1.
No. For example, the identity mapping I2 has two eigenvalues 1, 1.
2.
Yes. If we have (A − λI)v = 0, we also have
(A − λI)(cv) = c(A − λI)v = 0

for all c ∈ ℝ. Note that this skill will make the statement false when the field is finite.

3.
Yes. For example, the matrix (0−1 1 0 ) means the rotation through the angle π 2 . And the matrix has no real eigenvalue and hence no real eigenvector.
4.
No. See definition.
5.
No. For the matrix I2, the vectors (1,0) and (2,0) are all eigenvectors but they are parallel.
6.
No. The matrix ( 1 0 −1 0 ) has only two eigenvalues 1 and − 1. But the sum 0 is not an eigenvalue of the same matrix.
7.
No. Let P be the space of all polynomial and T be the identity mapping from P to P. Thus we know 1 is an eigenvalue of T.
8.
Yes. That a matrix A is similar to a diagonal matrix D means there is some invertible matrix P such that P−1AP = D. Since P is invertible, P = [I]αβ for some basis α ,where β is the standard basis for 𝔽n. So the first statement is equivalent to that A is diagonalizable. And the desired result comes from Theorem 5.1.
9.
Yes. If A and B are similar, there is some invertible matrix P such that P−1AP = B. If Av = λv, we have
B(P−1v) = λP−1v.

And if Bv = λv, we have A(Pv) = Pv.

10.
No. It usually false. For example, the matrices (11 0 2 ) and (20 1 1 ) are similar since
(01 1 0 ) (11 0 2 ) (01 1 0 ) = (20 1 1 ).

But the eigenvector (1,0) of the first matrix is not a eigenvector of the second matrix.

11.
No. The vectors (1,0) and (0,1) are eigenvectors of the matrix ( 1 0 −1 0 ). But the sum of them (1,1) is not an eigenvector of the same matrix.
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2011-06-27 00:00
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