Given $A_{5x7}$, $B_{7x5}$, prove that $BA$ is singular.
This is what I did:
Matrix $A$ is $5x7$. That means, there are 7 parameters with 5 equations. There are at least 2 free variables, so the solution for $A\underline{x}=\underline{0}$ is non-trivial. BA means each $i$ column of B multiplies each $i$ column of A, where $1≤i≤7$. The column with the free variables also multiplied by the corresponding B column. This leads to free variables multiplied by a scalar, so the outcome solution is still non-trivial.
I'm afraid that is not how the proof goes...
Hint:
$A$ is $5 \times 7$ implies $Ax=0$ has a non zero solution.