Finding the Matrix of a linear map $T$. Let $T\colon \mathcal M_{22}(\Bbb R) \to \mathcal M_{22}(\Bbb R)$  be  defined by:
$
T\left(\begin{bmatrix} 
a & b\\
c & d
\end{bmatrix}\right) = \begin{bmatrix} 2c & a+ c\\
b-2c & d\\
\end{bmatrix}
$
a) Find the eigenvalues of $T$.
Sort of stuck on this. Is there a standard basis $\mathcal B$ that I can use to find $[T]_{\mathcal B}$? I've really never dealt with this sort of problem. Any sort of hints or help? :O
 A: The canonical basis is $$\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix}\;\begin{pmatrix} 0&1\\0&0\end{pmatrix}\;\begin{pmatrix} 0&0\\1&0\end{pmatrix}\;\begin{pmatrix} 0&0\\0&1\end{pmatrix}\right\}$$
In general, the canonical basis for the space $\Bbb R^{n\times m}={\bf M}_{n\times m}(\Bbb R)$ is the set of $n\cdot m$ matrices $\{E^{\ell k}:1\leq \ell\leq n,1\leq k\leq m\}$, defined by $$(E^{\ell k})_{ij}=\delta_{\ell i}\delta_{kj}$$ where $$\delta_{ab}=\begin{cases} 1&a=b\cr 0&\text{ otherwise }\end{cases}$$
that is, the matrix with a $1$ at the entry $(\ell,k)$ and $0$ elsewhere. As you're being suggested, maybe you feel more comfortable working in $\Bbb R^4$ where we can think of your transformation as $$(a,b,c,d)\mapsto (2c,a+c,b-2c,d)$$ under the canonical basis $\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}$. Then you can roll it back to $\Bbb R^{2\times 2}$.
A: There's an isomorphism between $\mathcal{M}_{22}$ and the vector space of $4$-tuples that simply unfolds a matrix. Then you can use the same standard basis you're used to, which simply corresponds to the matrices with a single nonzero entry set to $1$.
