Show that if $\operatorname{trace}(AB) = 0$ and $\operatorname{rank} (A)=1$ then $ABA=0$ I know that 
$$AB-BA=A \iff A \text{ is singular}$$
$A$ and $B$ can be complex. Any hints?
 A: Hint: Since $A$ has rank $1$, we can write
$$A = \mathbf{u}\mathbf{v}^\mathrm{T}$$
for some vectors $\mathbf{u}$ and $\mathbf{v}$. Now use the cyclicity of the trace with respect to its arguments and see what you get with $\mathrm{Tr}(AB)=0$.
A: Since $\operatorname{rank} (\mathrm A) = 1$, there are vectors $\mathrm{u}, \mathrm{v} \in \mathbb{C}^n$ such that $\mathrm A = \mathrm{u} \mathrm{v}^*$. Hence,
$$\operatorname{tr} (\mathrm A \mathrm B) = \operatorname{tr} (\mathrm{u} \mathrm{v}^* \mathrm B) = \operatorname{tr} (\mathrm{v}^* \mathrm B \mathrm{u}) = \mathrm{v}^* \mathrm B \mathrm{u}$$
If $\operatorname{tr} (\mathrm A \mathrm B) = 0$, then $\mathrm{v}^* \mathrm B \mathrm{u} = 0$. Thus,
$$\mathrm A \mathrm B \mathrm A = \mathrm{u} \mathrm{v}^* \mathrm B \mathrm{u} \mathrm{v}^* = \mathrm{u} (\underbrace{\mathrm{v}^* \mathrm B \mathrm{u}}_{=0}) \mathrm{v}^* = 0 \cdot \mathrm A = \mathrm O_n$$

linear-algebra matrices rank-1-matrices trace
A: The cyclicity of the trace is certainly the way to go here, however, I'll offer a solution that is more basic although a bit brute force. I'm  going to assume real entries, but the calculation is nearly identical if complex entries are allowed.
Since $n\times m$ matrix $A$ is rank one, it's rows can be written as multiples of a single row vector.
$$
A=
\left[\begin{matrix}
a_1r_1 & a_1r_2 & \cdots & a_1r_m \\
a_2r_1 & a_2r_2 & \cdots & a_2r_m \\
\vdots & \vdots & \ddots & \vdots \\
a_nr_1 & a_nr_2 & \cdots & a_nr_m \\
\end{matrix}\right]=
\left[\begin{matrix}
a_1 \\
a_2  \\
\vdots\\
a_n  \\
\end{matrix}\right]
\left[\begin{matrix}
r_1 \ \ r_2  \ \ \cdots \ \ r_m 
\end{matrix}\right]=\mathbf{a} \mathbf{r}^T .
$$
Let $m\times n$ matrix $B=[b_{ij}],$ and let $\mathbf{b}_i$ be the $i^{th}$ column of matrix $B$. Let's take a look at the product $AB.$
$$
\begin{aligned}
AB
=\mathbf{a} \mathbf{r}^T B
&=\left[\begin{matrix}
a_1 \\
a_2  \\
\vdots\\
a_n  \\
\end{matrix}\right]
\left[\begin{matrix}
r_1 \ \ r_2  \ \ \cdots \ \ r_m 
\end{matrix}\right]
\left[\begin{matrix}
\mathbf{b}_1 \ \ \mathbf{b}_2  \ \ \cdots \ \ \mathbf{b}_n 
\end{matrix}\right] \\
&=\left[\begin{matrix}
a_1\mathbf{r}\cdot\mathbf{b}_1 & a_1\mathbf{r}\cdot\mathbf{b}_2 & \cdots & a_1\mathbf{r}\cdot\mathbf{b}_n \\
a_2\mathbf{r}\cdot\mathbf{b}_1 & a_2\mathbf{r}\cdot\mathbf{b}_2 & \cdots & a_2\mathbf{r}\cdot\mathbf{b}_n \\
\vdots & \vdots & \ddots & \vdots \\
a_n\mathbf{r}\cdot\mathbf{b}_1 & a_n\mathbf{r}\cdot\mathbf{b}_2 & \cdots & a_n\mathbf{r}\cdot\mathbf{b}_n \\
\end{matrix}\right].
\end{aligned}
$$
Now we know that Tr$(AB)=0$ so that
$$
\sum_{k=1}^n a_k\mathbf{r}\cdot\mathbf{b}_k = 0.
$$
Now let's look at $AB\mathbf{a}$ and conclude that it is a zero column vector. The $i^{th}$ component is:
$$
\begin{aligned}
(AB\mathbf{a})_i&=
\left[\begin{matrix}
a_i\mathbf{r}\cdot\mathbf{b}_1 & a_i\mathbf{r}\cdot\mathbf{b}_2 & \cdots & a_i\mathbf{r}\cdot\mathbf{b}_n
\end{matrix}\right]
\left[\begin{matrix}
a_1 \\
a_2  \\
\vdots\\
a_n  \\
\end{matrix}\right] \\
&=a_i\sum_{k=1}^n a_k\mathbf{r}\cdot\mathbf{b}_k = 0.
\end{aligned}
$$
So not only do we have $ABA=\mathbf{a}\mathbf{r}^TB\mathbf{a}\mathbf{r}^T=O_n$ (the $n\times n$ zero matrix), we also have that $\mathbf{a}\mathbf{r}^TB\mathbf{a}=\mathbf{0}_n$ (the $n\times 1$ zero column vector). Likewise, we can show that $\mathbf{r}^TB\mathbf{a}\mathbf{r}^T=\mathbf{0}_m^T$ (the $1\times m$ zero row vector). And of course as noted in the other answers, $\mathbf{r}^TB\mathbf{a}=0$ (the number zero).
