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Math Help - Find the eigenvalues of the linear transformation:

  1. #1
    Junior Member krtica's Avatar
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    Find the eigenvalues of the linear transformation:

    Find the eigenvalues of the linear transformation L : R2x2 ! R2x2 defi ned by L(A) = A^T + A.

    I'm having trouble with LaTex, but are the eigenvalues 2a and 2d?
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  2. #2
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    Quote Originally Posted by krtica View Post
    Find the eigenvalues of the linear transformation L : R2x2 ! R2x2 defi ned by L(A) = A^T + A.

    I'm having trouble with LaTex, but are the eigenvalues 2a and 2d?


    It seems to be L:M_2(\mathbb{R})\rightarrow M_2(\mathbb{R})\,,\,\,L(A):= A^t+A , which is clearly linear. Let us choose now the following basis of M_2(\mathbb{R}) :

    B=\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix}\,,\,  \begin{pmatrix}0&1\\0&0\end{pmatrix}\,,\,\begin{pm  atrix}0&0\\1&0\end{pmatrix}\,,\,\begin{pmatrix}0&0  \\0&1\end{pmatrix}\right\} , L\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatr  ix}2&0\\0&0\end{pmatrix}\,,\,L\begin{pmatrix}0&1\\  0&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatr  ix}\,,\,L\begin{pmatrix}0&0\\1&0\end{pmatrix} =\begin{pmatrix}0&1\\1&0\end{pmatrix}\,,\,L\begin{  pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\  0&2\end{pmatrix} , so

    the matrix of L\,\,\,wrt\,\,\,B is:

    [L]_B=\begin{pmatrix}2&0&0&0\\0&1&1&0\\0&1&1&0\\0&0&0  &2\end{pmatrix} , and this matrix's char. pol. is:

    p(x):=\det(xI-A)=\left|\begin{pmatrix}x-2&0&0&0\\0&x-1&-1&0\\0&-1&x-1&0\\0&0&0&x-2\end{pmatrix}\right| =x(x-2)^3

    Tonio
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  3. #3
    Junior Member krtica's Avatar
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    Thanks Tonio!
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    Junior Member krtica's Avatar
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    I think your equation for the characteristic polynomial is in reverse though. I believe lamba multiplied with the identity should be negative.
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    Quote Originally Posted by krtica View Post
    I think your equation for the characteristic polynomial is in reverse though. I believe lamba multiplied with the identity should be negative.

    That never minds, of course, since all we're interested in is the polynomial's roots, so we can take it either way...

    Anyway, I find my way easier since then the leading coefficient is always +1, what makes things easier for me.

    Tonio
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  6. #6
    Junior Member krtica's Avatar
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    True. Thank you again for your help Tonio.
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