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?
 A: 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.
A: Good method, bad execution.  You should find that
$$
AX = \pmatrix{x - 2z & y-2w\\-2x + 4z & -2y + 4w}
$$
Now, the system of equations for $AX = 0$ is equivalent to
$$
x = 2z, \quad y = 2w
$$
With that, we find that the nullity of $T$ is indeed $2$, and so the rank of $T$ is $4-2 = 2$, as you had stated.
I think you have found the quickest way to do this problem, but in this case it was easy to "eyeball" the solution.
A: A slightly different approach to determining $T$’s nullity:  
Each column of $AX$ is the result of multiplying the corresponding column of $X$ by $A$ and does not depend on any other columns of $X$, so if $AX=0$, then each column of $X$ considered as an element of $\mathbb R^2$ must lie in $\ker(A)$. This means that the nullity of $T$ is twice the nullity of $A$. The kernel of $A$ is easily computed by inspection or via the usual methods and is one-dimensional, so $\dim\ker(T)=2$ and $T$’s rank is thus $2$.
