A question about nilpotent matrix

Question

Assume $A,B$ are two $2\times2$ nilpotent matrices with $AB=BA$, and the entries of $A$ and $B$ are from a field $F$. How to show there is a nonzero-vector $v$ from $F^2$ s.t. $Av=Bv=0$

Answer 1

[This part is no longer relevant, in light of the correction to the question. Interpreting $vA$ as $v^T A$, the result is not true. For example, take $F= \mathbb{C}$ and

$$A=B=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}.$$

If $v=(x, y)$ then $Bv=0$ forces $y=0$ whilst $v^T A=0$ forces $x=0$.]

So I'll assume you mean $Av$, and hence that we want a non-zero $v \in \ker A \cap \ker B$. This can be deduced from the following two facts:

1. A nilpotent matrix has non-trivial kernel, otherwise it would be invertible and hence not nilpotent.

2. Commuting matrices act on each other's eigenspaces. So if $V \subset F^2$ is an eigenspace of $A$ then for all $v \in V$ we have $Bv \in V$.

Applying 1 to $A$ we see that $\ker A$ is non-trivial. Now applying 2 (with $V$ the $0$-eigenspace, i.e. kernel, of $A$) we see that $B_{\ker A}$ is a nilpotent matrix acting on $\ker A$. Applying $1$ again we see that $\ker B|_{\ker A}$ is non-trivial, so there exists non-zero $v \in \ker B_{\ker A} = \ker A \cap \ker B$.

Answer 2

$A$ is nilpotent, hence roots of its characteristic polynomial are $0,0$, and the Characteristic polynomial of $A$ will be $x^2$. By Cayley-Hamilton theorem, $A^2=0$. Similarly $B^2=0$.

If one of $A$ and $B$ is zero matrix (operator on $F^2$) then your conclusion is obvious.

Assume that $A\neq 0\neq B$.

Let $W=\langle w\rangle$ be an eigenspace of $A$ with eigenvalue $0$. (Note that $A$ vanishes on $W$ and it cannot vanish out side $W$, otherwise, if $Av=0$ for some $v\in F^2-W$, then $A$ vanishes on $\langle w,v\rangle=F^2$ i.e. $A=0$).

We show that $W$ is also an eigenspace of $B$ with eigenvalue $0$.

Now $Aw=0$. Hence $BAw=0$, i.e. $ABw=0$.

• If $Bw=0$ then $w$ is a required vector.

• If $Bw\neq 0$, then $ABw=0$ implies that $Bw$ is an eigen-vector of $A$ with eigen-value $0$; it should be in $\langle w\rangle$, say $Bw=\alpha w$. Apply $B$ again, $BBw=B^2w=0w=0$. $A$ also vanishesh on $Bw$, since $\langle Bw\rangle=\langle \alpha w\rangle=\langle w\rangle=$ eigen-space of $A$ (with eigen-value $0$).

Hence $k=Bw$ is required vector.

(The argument above is independent of basis, although your question is written with a specific basis (standard basis).)