Let $A$, $B$ be two $3\times3$ commuting matrices, where $A$ is nilpotent and $\operatorname{tr}B = 0$. Prove that $ABA = O$ Let $A$ and $B$ be two $3\times3$ commuting matrices, where $A$ is nilpotent and $\operatorname{tr}B = 0$.
Prove that $ABA = 0$.
Progress
I know that $ABA=0 \implies A^2B=0$. Here
$$A^2=\begin{pmatrix} 0 & 0& a_{12}a_{23} \\ 0&0&0\\0&0&0\end{pmatrix}$$
or
$$A^2=\begin{pmatrix} 0 & 0& 0 \\ 0&0&0\\a_{21}a_{32}&0&0\end{pmatrix}$$
$b_{11}+b_{22}+b_{33}=0$. So I have to prove that $b_{11}a_{12}a_{23}=0$ or $b_{33}a_{21}a_{32}=0$.
 A: The progress from the OP is incorrect.  There is no reason to believe that $A$ has zeroes in certain places.  For example, take $$A=\left(\begin{smallmatrix}10&3&7\\-4&-1&-3\\-13&-4&-9\end{smallmatrix}\right)$$
We have $A^3=0$ and $$A^2=\left(\begin{smallmatrix}-3&-1&-2\\3&1&2\\3&1&2\end{smallmatrix}\right)$$
which has no zero entries at all.
We know that all of the eigenvalues of $A$ are zero, since $A$ is nilpotent.  We also know that the eigenvalues of $B$ sum to zero, since that is the trace of $B$, although this information is not needed.  
Now, consider the minimal polynomial of $A$.  It is either $x, x^2$, or $x^3$.  If it is either $x$ or $x^2$, then $A^2=0$ and hence $A^2B=0$.  If instead it is $x^3$ then its minimal polynomial coincides with its characteristic polynomial.  Hence, since $A,B$ commute, we may conclude that $B$ is a polynomial of $A$, i.e. there is some polynomial $p(x)$ such that $B=p(A)$. 
Now, because $A,B$ commute they can be simultaneously triangulated, assuming that your ground field is algebraically closed (such as $\mathbb{C}$).  Choose $S$ so that $SAS^{-1}$ and $SBS^{-1}$ are each upper triangular.  They have their eigenvalues on the diagonals, so in particular $SAS^{-1}$ has zeroes on the diagonal.  Since $B=p(A)$, $SBS^{-1}=p(SAS^{-1})$ and in particular $SBS^{-1}$ also has zeroes along the diagonal.
Hence each of $SAS^{-1}$ and $SBS^{-1}$ is of the form $$\left(\begin{smallmatrix}0&\star &\star\\0&0&\star\\0&0&0\end{smallmatrix}\right)$$
and the product of any three such matrices is $0$; in particular $A^2B=0$.
A: If $A^2=0$ then $ABA=BA^2=0$ and we are done, so assume $A^2\neq0$. Since the degree of the minimal polynomial of cannot exceed the size of the matrix (by Cayley-Hamilton if you like), one has $A^3=0$.
Chose a vector $v$ with $A^2\cdot v\neq 0$, then $[A^2\cdot v, A\cdot v,v]$ is a linearly independent family, hence a basis. Call the basis $\def\B{\mathcal B}\B$, then after base change to this basis we get
$$
  A'=\operatorname{Mat}_\B(A)=\pmatrix{0&1&0\\0&0&1\\0&0&0}.
$$
Similarly base-change $B$ to $B'=\operatorname{Mat}_\B(A)$, then $A',B'$ satisfy the same hypotheses as given for $A,B$. From commutation of $B'$ with $A'$ one easily sees that the three diagonal entries of $B'$ are equal, say to$~b$. We then have $\operatorname{tr} B=3b=0$. Then provided the field$~K$ this is over does not have characteristic$~3$, we conclude $b=0$ in particular the top left entry of$~B'$ is$~0$, and this implies $B'(A')^2=0$, and $ABA=0$.

In characteristic$~3$ the result is false, as $B'=A'+I$ shows.

A less coordinate-oriented approach would be to set $V=\ker(A)$ and $W=\ker(A^2)$, then by commutation of $A,B$ these subspaces are $B$-stable. Hence $B$ acts on the quotient spaces $V/\{0\}\simeq V$, $W/V$ and $K^3/W$, all of dimension$~1$, so the action of $B$ is by scalars. The action of$~A$ induces isomorphisms between these quotients (from $K^3/W$ to $W/V$ to $V/\{0\}$), which intertwine (commute with) the action of $B$, showing that $B$ acts by the same scalar$~b$ on all three spaces. The trace of $B$ then equals $3b=0$, and the rest proceeds as above.
