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.
Best Answer
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 $$\left(\begin{array}{ccc|c}0&1&0&0\\0&0&1&0\\0&0&0&0\\\hline0&0&0&0\end{array}\right).$$