If $A$ is a $4 \times 4$ matrix with $rank(A) = 1$, then either $A$ is diagonalizable or $A^2 = 0$, but not both 
If $A$ is a $4 \times 4$ matrix with rank$(A) = 1$, then either $A$ is diagonalizable (over $C$) or $A^2 = 0$, but not both (Note that $A$ has complex entries)

So far, the only thing I've tried is noting that if rank$(A) = 1$, then, letting $J$ be the JCF of $A$, rank$(J) = 1$. I then started writing out possible JCF's of a $4 \times 4$ matrix. But, I'm having some issues... if $A$ is diagonalizable, then $J$ will be diagonal, but then $J$ has $4$ linearly independent columns, so rank$(J) = 4$. So, $A$ cannot be diagonalizable. I realize that this can't be right, I've obviously made a mistake somewhere, but I don't know where...
 A: Another way to solve this: consider linear transformation $F: \mathbb C^4 \to \mathbb C^4$ corresponding to the matrix $A$. By assumption the image of $F$ is the span of some nonzero $w \in \mathbb C^4$, and $\ker F $ is a three-dimensional subspace, call it $V$. We have two cases:


*

*$w \in V$

*$w \notin V$


In the first case, $A^2 =0$. Indeed, for any $x \in \mathbb C^4$, $F(x) = \lambda w \in V = \ker F$ (for some $\lambda \in \mathbb C$), so that $F^2(x)=(F(\lambda w))=0$.
In case (2), let $a_1, a_2, a_3$ be a basis for $V$. Then $\{w, a_1, a_2, a_3\}$ is a basis for $\mathbb C^4$. With respect to this basis, $F$ takes the form
\begin{pmatrix} \alpha & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}
For some $\alpha \neq 0$. This is a diagonal matrix similar to $A$.
This highlights the mistake in your attempted proof: remember that the diagonal entries of a diagonal matrix can be zero!
It's clear that the two options are mutually exclusive, since the square of a nonzero diagonal matrix is never zero.
A: If $A^2=0$,  the minimal polynomial of $A$ is either $x^2$, and it is not diagonalisable since its minimal polynomial has a double root, or $x$, in which case $A=0$, and it has rank $0$, not $1$.
