Linear maps satisfying $T^2 = I_n$ and Jordan Form of $T(X) = AX - XA$ 
Let $V$ be any $n$-dimensional vector space and $W = M_2(\mathbb{C})$.
(a) Construct all linear maps $T: V \to V$ such that $T^2 = I_n$.

For $T^2 = I_n$, taking $p(X) = X^2 - 1$ we must have $p(T) = 0$. Since the minimal polynomial of $T$ divides $p$ and contains all eigenvalues, $T$ must be diagonalizable with eigenvalues $1$ and $-1$. Since the minimal polynomial is of degree $\leq 2$, the largest Jordan block of the Jordan Form of $T$ must have order $2$. Then the maps satisfying $T^2 = I_n$ are the maps with eigenvalues $1$ and $-1$ such that the Jordan Normal Form is $J = (J^1,...,J^1)$ where each Jordan block $J^{i}$ have order $\leq 2$.
I'm not sure about my answer to this item.

(b) Find the Jordan Normal Form of $T: W \to W$ given by $T(X) = AX - XA$ with
$$A = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right).$$

We have to find the characteristical polynomial of $T$. If
$$X = \left(\begin{array}{cc} x & y \\ z & w \end{array}\right),$$
then
$$T(X) = \left(\begin{array}{cc} z & w-x \\ 0 & -z \end{array}\right)$$
I have problem to find the matrix associate to $T$ so, I tried to use $W \simeq \mathbb{R}^4$ and define $T: \mathbb{R}^4 \to \mathbb{R}^4$ by
$$T(x,y,z,w) = (z, w-z,0,-z).$$
Thus,
$$T(1,0,0,0) = (0,0,0,0) = 0e_1 + 0e_2 + 0e_3 + 0e_4,$$
$$T(0,1,0,0) = (0,0,0,0) = 0e_1 + 0e_2 + 0e_3 + 0e_4,$$
$$T(0,0,1,0) = (1,-1,0,-1) = 1e_1 -1e_2 + 0e_3 -1e_4,$$
$$T(0,0,0,1) = (0,1,0,0) = 0e_1 + 1e_2 + 0e_3 + 0e_4.$$
So, the matrix is
$$T = \left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \end{array}\right).$$
Is that matrix correct? If yes, I can continue. If no, I would like some help.
 A: For (a), TomTom314 already mentioned that $T$ is diagonalizable with eigenvalues $\pm1$.  In the case that all eigenvalues are the same, it is clear that $T=\pm I$.  But for the rest, I think the best you can say about $T$ is $T=XDX^{-1}$ where $X$ is invertible and $D$ is a diagonal matrix of the form $$\operatorname{diag}(-1,-1,\ldots,-1,1,1,\ldots,1)$$ where $-1$ appears $p$ times and $1$ appears $q$ times ($p,q$ are positive integers such that $p+q=n$).
However, if this is what the problem means, we can find a 1-1 correspondence between all such $T$ and all projections on $V$.  For a projection $P:V\to V$, set $T_P=2P-I$.  Then $T_P^2=I$.  For a map $T$ such that $T^2=I$, define $P_T=\frac{1}{2}(T+I)$ so $P_T^2=P_T$, making $P_T$ a projection.
For (b), the matrix of $T$ in your chosen basis is correct, but I'd like to point out that $T$ is nilpotent.  So all eigenvalues are $0$.  Note that
$$T^2X=A(AX-XA)-(AX-XA)A=A^2X-2AXA-XA^2$$
and since $A^2=O$, we have
$$T^2X=-2AXA.$$
That is,
$$T^3X=A(-2AXA)-(-2AXA)A=-2A^2XA+2AXA^2=O.$$
Since $T^2A^t=-2AA^tA\ne O$, $T$ is nilpotent of depth $3$.  That is $T$ has a Jordan block of size $3$, and since the space $M_2(\Bbb C)$ is $4$-dimensional, $T$ has another Jordan block of size $1$.  That is, a Jordan normal form of $T$ is
$$\left(\begin{array}{ccc|c}0&1&0&0\\0&0&1&0\\0&0&0&0\\\hline0&0&0&0\end{array}\right).$$
A: Hints
(1) As you've written, $\require{cancel}T^2 = \operatorname{id}$ implies that the minimal polynomial $m(x)$ of $T$ divides $x^2 - 1 = (x - 1) (x + 1)$. But as pointed out in the comments, (at least provided that the field $\Bbb F$ underlying $V$ does not have characteristic $2$) the linear factors of $m(x)$ do not repeat, so $T$ must be diagonalizable.

Thus, there is a basis of $V$ with respect to which the matrix representation $[T]$ of $T$ has the form $$\pmatrix{I_k \\ & -I_{n - k}} .$$ Put another way, if we fix any basis of $V$, the matrix representations $[T]$ of the linear transformations $T$ satisfying $T^2 = \operatorname{id}$ are precisely those similar to one of the above three matrices, that is, $\pm I$ and those of the form $$P^{-1} \pmatrix{I_k \\ & -I_{n - k}} P .$$

Remark The restriction on the characteristic is necessary: In characteristic $2$, $x^2 - 1 = (x - 1)^2$, so for example, that the nondiagonalizable matrix $A = \pmatrix{1&1\\&1}$ satisfies $A^2 = I$.
(2) In this case we can compute the Jordan Normal Form without working directly with the entries of $X$--though if you're not familiar with Lie algebras (which is surely the most important context wherein transformations of this form appear), computing $T^k(X)$ in terms of entries helps motivate this approach:
Notice that $A^2 = 0$---this makes easier the computation of powers of $T$: We have
\begin{multline}T^2(X) = T(AX - XA) = A(AX - XA) - (AX - XA)A \\= \cancel{A^2 X} - 2 A X A + \cancel{X A^2} = -2 A X A,\end{multline}
and computing similarly gives $$T^3 = 0 .$$

Since $T$ is thus nilpotent, its only eigenvalue is $0$, and all that remains to find is the size of the Jordan blocks. But since $T^3 = 0$ and $T^2 \neq 0$, the Jordan normal form contains a $3 \times 3$ block, $J_3(0)$, so the remaining block is $J_1(0) = \{0\}$. Thus, the Jordan normal form of $T$ is $$J_3(0) \oplus J_1(0) = \pmatrix{0&1&\cdot\\\cdot&0&1\\ \cdot&\cdot&0\\&&&0} .$$

Remark Our computations of $T^2, T^3$ used only the definition of multiplication and the fact that $A^2 = 0$ (and not the specific form of $A$). But $A$ is the Jordan form of any nonzero $2 \times 2$ nilpotent matrix, so our conclusion applies to any such matrix.
You might find it instructive to repeat the exercise replacing $A$ with a general $2 \times 2$ matrix. The observation implicit in the previous remark shows that it's enough to consider just the $A$ in Jordan form, so those of the forms $$\lambda I, \quad \pmatrix{\lambda&1\\&\lambda}, \quad \operatorname{diag}(\lambda, \mu) .$$
