eigenvector q and cayley-hamilton

**shimara** · October 12th 2011, 03:31 PM

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

eigenvector q and cayley-hamilton-artin.q17.png

eigenvector q and cayley-hamilton-artin.q16.png

Thanks for your help.

**Drexel28** · October 12th 2011, 03:38 PM

Originally Posted by shimara

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

Click image for larger version.

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Click image for larger version.

Name: artin.q16.png
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ID: 22555

Thanks for your help.

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

**shimara** · October 12th 2011, 03:53 PM

Originally Posted by Drexel28

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

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

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

I am sorry about this, I think it is wrong.

**HallsofIvy** · October 12th 2011, 03:57 PM

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

**shimara** · October 12th 2011, 04:01 PM

Originally Posted by HallsofIvy

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

Oh yeah, thanks, so

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

for part b, n=2

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

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

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

But this does not give me the correct $A^{-1}$ for a 2x2 matrix.

**Deveno** · October 12th 2011, 05:49 PM

you are not quite applying the formula correctly.

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

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

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

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

does that look any better?

**shimara** · October 12th 2011, 06:03 PM

Thank You Deveno, it works now.

I am still having trouble with the other question though.

**Deveno** · October 12th 2011, 06:28 PM

what do you have so far?

**shimara** · October 12th 2011, 06:56 PM

nothing, that is why I need a little help.

**Deveno** · October 12th 2011, 07:22 PM

well, you know x is an eigenvector for A, and y is an eigenvector for B^T, what can you do with that?

**shimara** · October 13th 2011, 05:48 AM

I have been trying to think about that but nothing comes to mind immediately.

**shimara** · October 13th 2011, 03:08 PM

Well I thought about doing this:

edit: sorry, stupid thought.

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

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