Find the eigenvalues of the linear transformation:

**krtica** · April 26th 2010, 07:35 AM

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

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

**tonio** · April 26th 2010, 09:34 AM

Originally Posted by krtica

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

**krtica** · April 26th 2010, 10:09 AM

Thanks Tonio!

**krtica** · April 26th 2010, 10:12 AM

I think your equation for the characteristic polynomial is in reverse though. I believe lamba multiplied with the identity should be negative.

**tonio** · April 26th 2010, 06:12 PM

Originally Posted by krtica

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

**krtica** · April 26th 2010, 06:19 PM

True. Thank you again for your help Tonio.

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