eigenvector q and cayley-hamilton

• Oct 12th 2011, 03:31 PM
shimara
eigenvector q and cayley-hamilton
hey,
I am going through artin 1st edition myself and I am having problems with 2 questions from chapter 4.

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• Oct 12th 2011, 03:38 PM
Drexel28
Re: eigenvector q and cayley-hamilton
Quote:

Originally Posted by shimara
hey,
I am going through artin 1st edition myself and I am having problems with 2 questions from chapter 4.

Attachment 22554

Attachment 22555

Let's go one at a time. For the second question, part a) you know that $\displaystyle p_A(A)=0$ but $\displaystyle p_A(x)=x^n+\cdots+\det(A)$ so this says $\displaystyle A^n+\cdots+\det(A)I=0$. How can you use this to get an expression for $\displaystyle A^{-1}$?
• Oct 12th 2011, 03:53 PM
shimara
Re: eigenvector q and cayley-hamilton
Quote:

Originally Posted by Drexel28
Let's go one at a time. For the second question, part a) you know that $\displaystyle p_A(A)=0$ but $\displaystyle p_A(x)=x^n+\cdots+\det(A)$ so this says $\displaystyle A^n+\cdots+\det(A)I=0$. How can you use this to get an expression for $\displaystyle A^{-1}$?

Thanks,
$\displaystyle A^n+\cdots+\det(A)I=0$
$\displaystyle A(A^{n-1}+\cdots)= -\det(A)I$

$\displaystyle A^{-1} = \frac{A^{n-1}+\cdots}{-\det(A)I}$

• Oct 12th 2011, 03:57 PM
HallsofIvy
Re: 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 PM
shimara
Re: eigenvector q and cayley-hamilton
Quote:

Originally Posted by HallsofIvy
Almost- you don't want that "I" in the denominator: $\displaystyle A\frac{A^{n-1}+\cdot\cdot\cdot}{-det(A)}= I$.

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 PM
Deveno
Re: 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 PM
shimara
Re: 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 PM
Deveno
Re: eigenvector q and cayley-hamilton
what do you have so far?
• Oct 12th 2011, 06:56 PM
shimara
Re: eigenvector q and cayley-hamilton
nothing, that is why I need a little help.
• Oct 12th 2011, 07:22 PM
Deveno
Re: 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 AM
shimara
Re: eigenvector q and cayley-hamilton
I have been trying to think about that but nothing comes to mind immediately.
• Oct 13th 2011, 03:08 PM
shimara
Re: 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