Find the range of $(1/2)T^2$ Let $V=M_{2\times 2}(\mathbb{R})$ be the vector space consisting of all $2 \times 2$ real matrices. Define a transformation $T : V → V$ by
$$             
                 T(A)=AP-PA, \mbox{ where } 
P = \begin{pmatrix}
0 & 1 \\ 
1 & 0 \\ 
\end{pmatrix}.
$$
The transformation $T^2: V → V$ is defined to be $T^2(A) = T(T(A))$, for each $A ∈ V$. Given $c ∈ \mathbb{R}$, the transformation $cT$ is defined to be $(cT)(A) = cT(A)$, for each $A ∈ V$.
Find the Range of $(1/2)T^2$.  
This is a question my professor gave me for practice. There were some other questions about proving that $T$ is linear and stuff like that but I have no idea where to even begin with this. 
(ps. I apologize if my formatting is not proper but this is the first time I have asked a question on this website).
 A: Say $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then
$$
  T(A) = AP - PA = \begin{pmatrix} b & a \\ d & c \end{pmatrix} - \begin{pmatrix} c & d \\ a & b \end{pmatrix} = \begin{pmatrix} b-c & a-d \\ d-a & c-b \end{pmatrix} .
$$
Continuing,
$$
  T^2(A) = T(T(A)) = \begin{pmatrix} (a-d)-(d-a) & (b-c)-(c-b) \\ (c-b)-(b-c) & (d-a)-(a-d) \end{pmatrix} = 2 \begin{pmatrix} a-d & b-c \\ c-b & d-a \end{pmatrix} .
$$
So
$$
  \frac{1}{2} T^2(A) = \begin{pmatrix} a-d & b-c \\ c-b & d-a \end{pmatrix} .
$$
Before going on, I'd like to point out a slightly different approach suggested in the comments. Note that $P^2 = I$. Then
$$
  T^2(A) = T(A)P - PT(A) = (AP-PA)P - P(AP-PA) = AP^2 - PAP - PAP + P^2A = 2(A-PAP),
$$
so $\frac{1}{2} T^2(A) = A-PAP$ (not $\frac{1}{2}T^2(A)=PAP$). One can compute
$$
  A - PAP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} - \begin{pmatrix} d & c \\ b & a \end{pmatrix}
$$
and get the same expression as before.
Now, having an expression for $\frac{1}{2} T^2$ is nice, but the question was to describe the range of this operator. What kind of description would you like? If you want equations for the range, some obvious ones pop out: set
$$
  \begin{pmatrix} a-d & b-c \\ c-b & d-a \end{pmatrix} = \begin{pmatrix} x & y \\ z & w \end{pmatrix},
$$
then it's clear that $x+w = 0$ and $y+z = 0$. It's clear that these equations are independent. You already know that $\frac{1}{2} T^2$ is linear (right?) so its range is a linear subspace in $M_{2\times 2}(\mathbb{R})$, which is $4$-dimensional. Since we've written $2$ independent equations, we know the range has dimension at most $2$. A priori, however, there could be additional equations.
To see that there are no other equations we could, for instance, write down $2$ independent elements of the range, such as $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ (can you see which matrices $A$ give these as $\frac{1}{2}T^2(A)$?).
So the range is at most $2$-dimensional (because we have $2$ equations and a $4$-dimensional ambient space) and it is also at least $2$-dimensional (because we have these $2$ independent elements). We can conclude that the range is precisely $2$-dimensional, spanned by the two elements given above. And that gives you a second description of the range, if you need a basis for it rather than equations.
For example, having this basis makes it very easy to check that $\frac{1}{2} T^2$ is a projection onto its range.
