Exercise 5.2.1

Answers

1.

No. The matrix

(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

has only two distinct eigenvalues but it’s diagonalizable.

2.

No. The vectors

(\begin{matrix} 1 \\ 0 \end{matrix})

and

(\begin{matrix} 2 \\ 0 \end{matrix})

are both the eigenvectors of the matrix

(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

corresponding the same eigenvalue

1

3.

No. The zero vector is not.

4.

Yes. If

x \in E_{λ_{1}} \cap E_{λ_{2}}

, we have

λ_{1} x = Ax = λ_{2} x

. It’s possible only when

x = 0

5.

Yes. By the hypothesis, we know

A

is diagonalizable. Say

A = P^{- 1} DP

for some invertible matrix

P

and some diagonal matrix

D

. Thus we know that

Q^{- 1} AQ = {(PQ)}^{- 1} D (PQ) .

6.

No. It need one more condition that the characteristic polynomial spilts. For example, the matrix

(\begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix})

has no real eigenvalue.

7.

Yes. Since it’s a diagonalizable operator on nonzero vector space, it’s characteristic polynomial spilts with degree greater than or equal to

1

. So it has at least one zero.

8.

Yes. Because we have

W_{i} \cup \sum_{i \neq k} W_{k} = {0}

and

\sum_{i \neq k} W_{k} = {0} \supset W_{j}

for all $j \neq i$ , we get the desired answer.

9.

No. For example, take

W_{1} = span {(1, 0)}

W_{2} = span {(0, 1)}

, and

W_{3} = span {(1, 1)}

jephianlin

2011-06-27 00:00