Eigenvalues and eigenspace of a linear transformation from $M_{2\times2}$ to $M_{2\times2}$ Let $T:{M_{2\times2}} \to {M_{2\times2}}$ be defined as
$$T(A) = \left( {\matrix{
   2 & 1  \cr 
   1 & 2  \cr 
 } } \right)A - A\left( {\matrix{
   2 & 1  \cr 
   1 & 2  \cr 
 } } \right)$$
Find all eigenvalues of $T$ and a basis for each eigenspace.
How would you do this?
 A: The map $T$ associates to a matrix $A$ its commutator with the fixed matrix $B=(\begin{smallmatrix}2&1\\1&2\end{smallmatrix})$. A first thing you can do is add to$~B$ a multiple of the identity that commutes with every$~A$, so that commutators will be unchanged; we can therefore define the same linear map as the commutator with $B'=(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})$.
Such commutator maps always have a nontrivial kernel (which forms the eigenspace for$~\lambda=0$), namely those matrices that commute with$~B$ (or$~B'$) which at least contains the polynomials in$~B$. Those polynomials form the subspace spanned by $B'$ and $I_2$, and this turns out to be the whole kernel. We'd like to concentrate next on a $T$-stable complementary subspace to the kernel, and if such a subspace exists, it is the image of$~T$. Indeed that image, which is spanned by $J=(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix})$ and $K=(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix})$ is complementary to the span of $I_2$ and $B'$. So we compute $T(J)=2K$ and $T(K)=2J$, so on the basis $[J,K]$ the matrix of the restriction of $T$ to its image is $(\begin{smallmatrix}0&2\\2&0\end{smallmatrix})$. The eigenvalues of that matrix, which are also the nonzero eigenvalues of$~T$, are $\lambda=-2$, with corresponding eigenvectors the scalar multiples of $J-K=(\begin{smallmatrix}1&1\\-1&-1\end{smallmatrix})$ and $\lambda=2$ with corresponding eigenvectors the scalar multiples of $J+K=(\begin{smallmatrix}1&-1\\1&-1\end{smallmatrix})$.
