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Thread: eigenvector q and cayley-hamilton

  1. #1
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    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.

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

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

    Thanks for your help.
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  2. #2
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    Re: eigenvector q and cayley-hamilton

    Quote Originally Posted by shimara View Post
    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 $\displaystyle p_A(A)=0$ 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 $\displaystyle A^{-1}$?
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  3. #3
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    Re: eigenvector q and cayley-hamilton

    Quote Originally Posted by Drexel28 View Post
    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,
    $\displaystyle A^n+\cdots+\det(A)I=0$
    $\displaystyle A(A^{n-1}+\cdots)= -\det(A)I$

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

    I am sorry about this, I think it is wrong.
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  4. #4
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    Re: eigenvector q and cayley-hamilton

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

    Quote Originally Posted by HallsofIvy View Post
    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.
    Last edited by shimara; Oct 12th 2011 at 04:19 PM.
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  6. #6
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    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?
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  7. #7
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    Re: eigenvector q and cayley-hamilton

    Thank You Deveno, it works now.

    I am still having trouble with the other question though.
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  8. #8
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    Re: eigenvector q and cayley-hamilton

    what do you have so far?
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    Re: eigenvector q and cayley-hamilton

    nothing, that is why I need a little help.
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  10. #10
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    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?
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  11. #11
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    Re: eigenvector q and cayley-hamilton

    I have been trying to think about that but nothing comes to mind immediately.
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  12. #12
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    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
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