$A$ and $B$ similar if $A^2=B^2=0$ and dimension of range $A$ and $B$ are equal Suppose $A$ and $B$ are linear transformations on finite dimensional vector space $V$,s.t. $A,B\neq 0$ and $A^2=B^2=0$. Suppose the dimension of range $A$ and $B$ are equal, can $A$ and $B$ be similar?

If $A^2 = B^2 = 0$, then all the eigenvalues of $A$ and $B$ are identically equal to $0$.  Now, if we put $A$ into Jordan normal form, the blocks can only be $1 \times 1$ or $2 \times 2$, else $A^2 \neq 0$; thus, $A$ is similar to $$ Q \left(\begin{array}{ccccccc} 
0 & 1 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 1 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 0 & 0  \\
0 & 0 & 0 & 0 & \cdots & 0 & 0 
\end{array} \right) Q^{-1}$$
for some $Q$, where the matrix in the middle consists of $r$ $2 \times 2$ blocks, with $r = \text{rank}(A)$; this is because the above is thethe Jordan Normal form for $A$.  If the rank of $B$ is also equal to $r$, then we have that $B = PDP^{-1}$ where $D$ is the same matrix above.  We then have $$A = QP^{-1} BPQ^{-1}.$$
Thus, if $r > 0$, the two matrices must be similar.
