$A$ 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?
 A: If $A^2 = 0$, then the Jordan form of $A$ has blocks which are either size $1$ or size $2$. If $A$ is $n \times n$, and the dimension of the Null space is $m$, then the number of blocks of size $2$ is $(n - m)$. So the dimension of the Null space completely determines the Jordan form. Since $A$ and $B$ have a null space of the same dimension, they are similar. 
A: If $A^2 = 0$, then $A$ satisfies the polynomial $x^2$, and and similarly for $B$. So then the minimal polynomial for $A$ and $B$ both divide $x^2$. But since $A\neq 0$, $B \neq 0$, then $x^2$ must be the minimal polynomial for both $A$ and $B$. 
Then the Jordan Canonical Form of $A$ and $B$ will include at least one $2\times 2$ Jordan block of the form: $\left[ \begin{array}{cc}0 & 1 \\ 0 & 0 \end{array}\right]$, and if $A$ and $B$ have the same number of such blocks, they will be similar, which notably implies they would have the same rank. But you assured the dimension of the range, i.e. the rank, was the same, so they must be similar. 
A: 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.
A: that is surely possible. Take for example $A= \begin{pmatrix}0 & 0 \\ 1& 0 \end{pmatrix}$ and $ B=\begin{pmatrix}0 & 1 \\ 0& 0 \end{pmatrix}$ and then it holds that
$$
A^2=B^2=0
$$
and
$$
A=SBS^{-1} \text{ with } S=\begin{pmatrix}0 & 1 \\ 1& 0 \end{pmatrix}
$$
so $A$ and $B$ are indeed similar. 
