Show that if $\operatorname{trace}(AB) = 0$ and $\operatorname{rank} (A)=1$ then $ABA=0$

Question

I know that

$$AB-BA=A \iff A \text{ is singular}$$

$A$ and $B$ can be complex. Any hints?

Answer 1

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 $\mathrm{Tr}(AB)=0$.

Answer 2

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,

$$\mathrm A \mathrm B \mathrm A = \mathrm{u} \mathrm{v}^* \mathrm B \mathrm{u} \mathrm{v}^* = \mathrm{u} (\underbrace{\mathrm{v}^* \mathrm B \mathrm{u}}_{=0}) \mathrm{v}^* = 0 \cdot \mathrm A = \mathrm O_n$$

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linear-algebra matrices rank-1-matrices trace

Answer 3

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).