I was solving problems on linear maps and came across this problem; Consider the map L(C):= AC+CB, where C represents set of 2x2 matrices and A=$\begin{bmatrix}2&1\\0&3\end{bmatrix}$ and B=\begin{bmatrix}1&1\\2&-1\end{bmatrix} What are the eigenvalues and eigenvectors of the linear map? There is no matrix C given here. I started by taking C as any arbitrary matrix and used the operator but didn't get any clue on how to get the eigen values.

Thanks in advance.

  • 1
    $\begingroup$ Doesn't $C'$ being an eigenvector of this map with eigenvalue $\lambda$ mean: $L(C')=AC'+C'B=\lambda C'$ $\endgroup$ – dimpol Nov 16 '16 at 11:14

Observe that for any matrix

$$C=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;,\;\;\text{we actually get the map}\;\;$$


Take now the usual basis of $\;M_{2\times2}(F)\;$ (I suppose $\;F=\Bbb R\;$ in this case), and find the matrix representation of $\;L\;$ wrt it:

$$\begin{align*}&L\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}3&1\\0&0\end{pmatrix}=3\begin{pmatrix}1&0\\0&0\end{pmatrix}+1\begin{pmatrix}0&1\\0&0\end{pmatrix}+0\begin{pmatrix}0&0\\0&0\end{pmatrix}+0\begin{pmatrix}0&0\\0&0\end{pmatrix}\\{}\\ &L\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}2&1\\0&0\end{pmatrix}=2\begin{pmatrix}1&0\\0&0\end{pmatrix}+1\begin{pmatrix}0&1\\0&0\end{pmatrix}+0\begin{pmatrix}0&0\\0&0\end{pmatrix}+0\begin{pmatrix}0&0\\0&0\end{pmatrix}\\{}\\ &L\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\4&1\end{pmatrix}=1\begin{pmatrix}1&0\\0&0\end{pmatrix}+0\begin{pmatrix}0&1\\0&0\end{pmatrix}+4\begin{pmatrix}0&0\\0&0\end{pmatrix}+1\begin{pmatrix}0&0\\0&0\end{pmatrix}\\{}\\ &L\begin{pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&1\\2&2\end{pmatrix}=0\begin{pmatrix}1&0\\0&0\end{pmatrix}+1\begin{pmatrix}0&1\\0&0\end{pmatrix}+2\begin{pmatrix}0&0\\0&0\end{pmatrix}+2\begin{pmatrix}0&0\\0&0\end{pmatrix}\end{align*}$$

Thus, $\;L\;$ has a matrix rep.:


If I'm not wrong, the eigenvalues are $\;3\pm\sqrt3\;,\;\;2\pm\sqrt3\;$

  • $\begingroup$ Thanks for the explanation, but L is a map from 2x2 to 2x2. Does the matrix representation above does this mapping ? $\endgroup$ – FAIZ Nov 16 '16 at 11:41
  • $\begingroup$ @FAIZ Yes, of course. The matrix representation of a linear map $\;T:V\to W\;$ ,with $\;\dim V=n\,,\,\,\dim W=m\;$ is an $\;m\times n\;$ matrix, and thus in this case we get as $\;4\times4\;$ matrix since $\;\dim M_{2\times2}(\Bbb R)=4\;$ ... $\endgroup$ – DonAntonio Nov 16 '16 at 11:44

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