<s>[INST] Rank of a linear transformation? I'm given a linear trasnformation: $T:M_2\rightarrow M^{\:}_2$ which is defined such as $T\left(X\right)=AX$, where $A$ is:
$$A=\begin{pmatrix}1&-2\\ -2&4\end{pmatrix}$$
Find the rank of T?
My idea was to find the nullity of T and then use the rank-nullity theorem.
$$AX=A\begin{pmatrix}x&y\\ z&w\end{pmatrix}=\begin{pmatrix}x-2y&z-2w\\ 0&0\end{pmatrix}=0$$
$$x=2y $$ 
 $$ z=2w
$$
Thus concluding that the Nullity of T is 2, and by that theorem rankT + nullityT = n = 4, and finally rankT = 2.
Is my solution correct and could it have been done quicker?
 [/INST] Expanding on my comment, using $M_2\cong \mathbb{R}^4$ via $\begin{pmatrix}x & y\\z & w\end{pmatrix}\mapsto \begin{pmatrix}x\\y\\z\\w\end{pmatrix}$. Then
\begin{align*}&T:\mathbb{R}^4\rightarrow\mathbb{R}^4\\
&\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\mapsto \begin{pmatrix}x-2z\\y-2w\\-2x+4z\\-2y+4w\end{pmatrix}=\begin{pmatrix}1&0&-2&0\\0&1&0&-2\\-2&0&4&0\\0&-2&0&4\end{pmatrix}\begin{pmatrix}x\\y\\z\\w\end{pmatrix}
\end{align*}
Now compute the Rank of $T$ as usual (for example via Gauss Elimination plus Dimension Theorem)
NOTE: your entries of $AX$ are unfortunately all wrong.
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