Determinant of AB and BA, where A and B are $5\times 7$ and $7\times 5$ matrices over the real numbers.

Question

I was doing a multiple-choice exercise where the only option left is $$\det(BA)=0.$$

Could someone explain why this is true?

Answer 1

Since $BA$ is a $7 \times 7$ matrix,

$$\det(BA) \neq 0 \iff \mathrm{rank}(BA)=7. $$

However,

$$ \mathrm{rank}(BA) \leq \min \{ \mathrm{rank}(A), \mathrm{rank}(B) \} \leq 5$$

since $\mathrm{rank}(A) \leq 5$.

Edit: Arthur, thank you for the hint.

Answer 2

Since $A$ is $5\times 7$ and $5<7$ it is not injective means there is $x\neq 0$ such that $Ax=0$ so $BAx=o$ so $BA$ isnot invertible hence $\det(BA)=0.$

Answer 3

$BA$ is a $7 \times 7$ matrix. It therefore has zero determinant if and only if its image has dimension $7$. On the other hand, $A$ can have rank no higher than $5$, because its image is at most $5$-dimensional. It should be clear that the rank of a matrix cannot be increased by multiplying it by another matrix, since $$ \dim \operatorname{Im}(BA) = \dim (B\operatorname{Im}(A)) \leqslant \dim \operatorname{Im}(A): $$ $B$ can't create independent vectors from nowhere. Thus the rank of $BA$ is $\leqslant 5<7$, so the determinant cannot be nonzero.

Answer 4

Let the reduced-row echelon form of $B$ be $\left(\begin{matrix}R\\0 \dots0\end{matrix}\right)$. This is possible because $B$ has more rows than columns.

As such, $B=E\left(\begin{matrix}R\\0 \dots0\end{matrix}\right)$ for $E$ being some product of elementary matrices.

When we multiply it by $A$ on the right, we get:

$\left(\begin{matrix}RA\\0 \dots0\end{matrix}\right)$, which is clearly singular.

As such, $E\left(\begin{matrix}RA\\0 \dots0\end{matrix}\right)=BA$ is singular and $\det(BA)=0$.