hey,

I am going through artin 1st edition myself and I am having problems with 2 questions from chapter 4.

Attachment 22554

Attachment 22555

Thanks for your help.

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- Oct 12th 2011, 03:31 PMshimaraeigenvector q and cayley-hamilton
hey,

I am going through artin 1st edition myself and I am having problems with 2 questions from chapter 4.

Attachment 22554

Attachment 22555

Thanks for your help. - Oct 12th 2011, 03:38 PMDrexel28Re: eigenvector q and cayley-hamilton
- Oct 12th 2011, 03:53 PMshimaraRe: eigenvector q and cayley-hamilton
- Oct 12th 2011, 03:57 PMHallsofIvyRe: eigenvector q and cayley-hamilton
Almost- you don't want that "I" in the denominator: $\displaystyle A\frac{A^{n-1}+\cdot\cdot\cdot}{-det(A)}= I$.

- Oct 12th 2011, 04:01 PMshimaraRe: eigenvector q and cayley-hamilton
Oh yeah, thanks, so

$\displaystyle A^{-1}= -(\det(A))^{-1} (A^{n-1} + \cdot \cdot \cdot)$

for part b, n=2

Let $\displaystyle A= \begin{bmatrix} a & b\\c & d \end{bmatrix}$

$\displaystyle p(A)=A^2-(a+d)A+(ad-bc)=0$

So $\displaystyle A^{-1}= -(\det(A))^{-1} (A^{2-1} +I)$

But this does not give me the correct $\displaystyle A^{-1}$ for a 2x2 matrix. - Oct 12th 2011, 05:49 PMDevenoRe: eigenvector q and cayley-hamilton
you are not quite applying the formula correctly.

$\displaystyle p(A) = A^2 - (a+d)A + \det(A)I = 0$

$\displaystyle A^2 - (a+d)A = -\det(A)I$

$\displaystyle \left(\frac{-1}{\det(A)}\right)(A - (a+d)I)A = I$, so

$\displaystyle A^{-1} = -(\det(A))^{-1}(A - (a+d)I)$

does that look any better? - Oct 12th 2011, 06:03 PMshimaraRe: eigenvector q and cayley-hamilton
Thank You Deveno, it works now.

I am still having trouble with the other question though. - Oct 12th 2011, 06:28 PMDevenoRe: eigenvector q and cayley-hamilton
what do you have so far?

- Oct 12th 2011, 06:56 PMshimaraRe: eigenvector q and cayley-hamilton
nothing, that is why I need a little help.

- Oct 12th 2011, 07:22 PMDevenoRe: eigenvector q and cayley-hamilton
well, you know x is an eigenvector for A, and y is an eigenvector for B^T, what can you do with that?

- Oct 13th 2011, 05:48 AMshimaraRe: eigenvector q and cayley-hamilton
I have been trying to think about that but nothing comes to mind immediately.

- Oct 13th 2011, 03:08 PMshimaraRe: eigenvector q and cayley-hamilton
Well I thought about doing this:

edit: sorry, stupid thought.

Can someone give me a good hint on how to do this question?

Thanks