# Math Help - eigenvector q and cayley-hamilton

1. ## 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.

2. ## Re: eigenvector q and cayley-hamilton

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

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 $A^{-1}$?

3. ## Re: eigenvector q and cayley-hamilton

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

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

4. ## Re: eigenvector q and cayley-hamilton

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

5. ## Re: eigenvector q and cayley-hamilton

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.

6. ## Re: eigenvector q and cayley-hamilton

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?

7. ## Re: eigenvector q and cayley-hamilton

Thank You Deveno, it works now.

I am still having trouble with the other question though.

8. ## Re: eigenvector q and cayley-hamilton

what do you have so far?

9. ## Re: eigenvector q and cayley-hamilton

nothing, that is why I need a little help.

10. ## 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?

11. ## Re: eigenvector q and cayley-hamilton

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

12. ## 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