Exercise 5.2.13

Answers

1.

For matrix

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

, corresponding to the same eigenvalue

0

we have

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

is the eigenspace for

A

while

E_{0} = span {(1, - 1)}

is the eigenspace for

A^{t}

2.

Observe that

\dim (E_{λ}) = null (A - λI) = null ({(A - λI)}^{t})

= null (A^{t} - λI) = \dim (E_{λ}^{'}) .

3.

A

is diagonalizable, then its characteristic polynomial splits and the multiplicity meets the dimension of the corresponding eigenspace. Since

A

and

A^{t}

has the same characteristic polynomial, the characteristic polynomial of

A^{t}

also splits. And by the precious exercise we know that the multiplicity meets the dimension of the corresponding eigenspace in the case of

A^{t}

jephianlin

2011-06-27 00:00