If $A \in M_2(\mathbb{C})$ and $A^2 = 0$ then what are the possible forms of A?

Question

Let $A \in M_2(\mathbb{C})$

1. If $A^2 = 0$ determine all of the JCF's possible.
2. If $A^2 = 0$ determine all of the possible A's.
3. Show that $A^2 = 0 \;\exists n \geq 2 \Leftrightarrow A^2 = 0 \;\forall n \geq 2$.

For 1, the possible JCF's correspond to the number of possible number partitions together with the restriction that the diagonal have at least one zero. So the possible JCF's correspond to $|J_{\lambda = 0}| = 2$, $|J_{\lambda = 0}| + |J_{\lambda = 0}| = 2$ and $|J_{\lambda = 0}| + |J_{\lambda = a}| = 2$ (up to permutation).

This is actually wrong. Because $Av = \lambda v \Rightarrow \lambda^2 v = 0$, any eigenvalue of such a nilpotent matrix must be equal to zero. This gives $|J_{\lambda = 0}| = 2$, $|J_{\lambda = 0}| + |J_{\lambda = 0}| = 2$ as the only correct JCF's.

For 3, one way, assume $A^n = 2 \;\exists n \geq 2$. Express $A = Q^{-1}J_AQ$. Then $\{J_A, (J_A)^2, ..., (J_A)^{n-1}\}$ are non-zero. Now J_A can't have a zero-free diagonal, because that would mean that the sequence of powers of $J_A$ with no zero matrix would be infinite, so there must be at least one zero. But we have shown that all of the possible sequences of Jordan forms with one zero are such that $J_A^2 = 0$. So $A^n = 2 \;\forall n \geq 2$

For 3, other way, assume $A^n = 2 \;\forall n \geq 2$. Let $n = 2$. Then $A^n = 2 \;\exists n \geq 2$.

However, I do not know how to answer two. May I have some assistance?

Answer 1

Your answer to (1) is wrong. The only possible eigenvalue is $0$. So it's either one Jordan block of size $2$ or two of size $1$, all for eigenvalue $0$.

To see that the only possible eigenvalue is $0$, note that if $A v = \lambda v$, $A^2 v = \lambda^2 v$, and $\lambda^2 \ne 0$ if $\lambda \ne 0$.

For (2), if you have two blocks of size $1$ the matrix is all $0$. If you have one block of size $2$ and an eigenvector for $0$ is $\pmatrix{a\cr b\cr}$, show that $A$ is a scalar times $\pmatrix{ab & -a^2\cr b^2 & -ab}$.

For (3), again these are the only possible Jordan canonical forms if $A^n = 0$: the matrix being $2 \times 2$, you can't have blocks of any size except $1$ and $2$, and again the only possible eigenvalue is $0$.