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Exercise 5.1.11

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1.

It’s just because

A = B^{- 1} λIB = λI

for some invertible matrix $B$ .

2.

Let

M

be the matrix mentioned in this question and

λ

be that only eigenvalue of

M

. We have that the basis to make

M

diagonalizable if consisting of vectors

v_{i}

such that

(M - λI) v_{i} = 0

. This means

(M - λI) v = 0

for every

v

since

{v_{i}}

forms a basis. So

M

must be

λI

3.

It’s easy to see that

1

is the only eigenvalue of the matrix. But since nullity of

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

is one, we can’t find a set of two vector consisting of eigenvectors such that the set is independent. By Theorem 5.1, the matrix is not diagonalizable.

jephianlin

2011-06-27 00:00

Exercise 5.1.11

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