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Math Help - Eigenvalues

  1. #1
    Super Member Showcase_22's Avatar
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    Eigenvalues

    Suppose that A is a square matrix and \lambda is an eigenvalue of A.

    (i) Show that \lambda ^n is an eigenvalue of A^n for all positive integers n.

    (ii) Suppose A is invertible. Show \lambda is non-zero and that \lambda ^{-1} is an eigenvalue of A^{-1}
    My method is:

    Let |A|=f(x)

    Since \lambda, f(\lambda)=0

    I need to show that f(\lambda ^{n})= 0

    f ( \lambda)=0
    (f(\lambda))^n=0^n
    (f(\lambda)^n)=f(\lambda ^n)=0 which is what I needed to show.

    Is this right?

    For part 2:

    If A is invertible then A^{-1} exists.

    Like before f(\lambda)=0.

    A=\begin{pmatrix}<br />
{a_{11}}&{.....}&{a_{1n}}\\ <br />
{...}&{...}&{....}\\ <br />
{a_{n1}}&{....}&{a_{nn}}<br />
\end{pmatrix}

    Det(A- \lambda I)=0

    Hence:

    \begin{vmatrix}<br />
{a_{11}}&{....}&{a_{1n}}\\ <br />
{....}&{....}&{....}\\ <br />
{a_{n1}}&{....}&{a_{nn}}<br />
\end{vmatrix}-\lambda I=\begin{vmatrix}<br />
{a_{11}-\lambda}&{....}&{a_{1n}}\\ <br />
{....}&{....}&{....}\\ <br />
{a_{n1}}&{....}&{a_{nn}-\lambda}<br />
\end{vmatrix}=f(\lambda)=0

    This is where my method goes awry. f(\lambda) cannot equal zero or the matrix A is not invertible.

    What am I doing wrong??
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Showcase_22 View Post
    My method is:

    Let |A|=f(x)

    Since \lambda, f(\lambda)=0

    I need to show that f(\lambda ^{n})= 0

    f ( \lambda)=0
    (f(\lambda))^n=0^n
    (f(\lambda)^n)=f(\lambda ^n)=0 which is what I needed to show.

    Is this right?
    Let x be an eigen vector corresponding to \lambda, then:

     <br />
A^nx=A^{n-1}Ax=\lambda A^{n-1}x<br />


    Which can be used as the basic idea of proof by induction of the following theorem:

    Let \lambda be an eigen value of square matrix A with eigen vector x, then \lambda^n is an eigen value of A^n with eigen vector x.

    CB
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  3. #3
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Let x be an eigen vector corresponding to \lambda, then:

     <br />
A^nx=A^{n-1}Ax=\lambda A^{n-1}x<br />


    Which can be used as the basic idea of proof by induction of the following theorem:

    Let \lambda be an eigen value of square matrix A with eigen vector x, then \lambda^n is an eigen value of A^n with eigen vector x.

    CB
     <br />
A^nx=A^{n-1}Ax=\lambda A^{n-1}x<br />

    This implies that Ax=\lambda x. I thought that if the matrix A was multiplied by an eigenvector then only a multiple of this eigenvector was produced. How do you know that the scalar in front of the eigenvector will be it's eigenvalue? Since you know that it's going to be a multiple of the eigenvector, did you just decide to make this the eigenvalue since the eigenvector can just be scaled down?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Showcase_22 View Post
     <br />
A^nx=A^{n-1}Ax=\lambda A^{n-1}x<br />

    This implies that Ax=\lambda x. I thought that if the matrix A was multiplied by an eigenvector then only a multiple of this eigenvector was produced. How do you know that the scalar in front of the eigenvector will be it's eigenvalue? Since you know that it's going to be a multiple of the eigenvector, did you just decide to make this the eigenvalue since the eigenvector can just be scaled down?
    Every eigen value has at least one eigen vector. If you read what I wrote you will see that I said that x was an eigen vector corresponding to the eigen value \lambda of A.

    Then by definition:

    Ax=\lambda x

    You may have a different convention about this, the other one has:

     <br />
xA=\lambda x<br />

    but the argument will still work with a little adaptation

    CB
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