Ex. 8.5 – Mathematical Methods for Physics and Engineering (Riley)
By considering the matrices
$$ A = \left( \begin{matrix}
1 & 0 \\
0 & 0 \\
\end{matrix} \right)
\text{ , } B = \left( \begin{matrix}
0 & 0 \\
3 & 4 \\
\end{matrix} \right)
$$
show that $AB = 0$ does not imply that either $A$ or $B$ is the zero matrix, but that it does imply that at least one of them is singular.
So, my reasoning was the following:
It's not difficult to compute $AB = \left( \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right) $, in fact it's really even implied in the question.
So, assume that $A, B$ are each non-singular – i.e. they are invertible.
Thus, $A^{-1}AB=B$, and $ABB^{-1}=A$.
But $AB$ is a zero matrix, so $A=B=0$.
Thus proven that the initial assumption $A, B$ are non-singular is false.
Is my reasoning correct? I ask because the 'hints and answers' said simply "Use the property of the determinant of a matrix product."
While I don't expect there to be only one proof, it's a tad disconcerting for it to so flatly suggest a different method – I'm not nearly confident enough in my ability to disregard it.
Best Answer
Your proof is good. A product $AB$ can be the zero matrix with $A$ being invertible (or non-singular): just take $B=0$.
Your assignment is to prove that from $AB=0$ it follows that one among $A$ and $B$ is singular.
Now, if $A$ is invertible, then $AB=0$ implies $B=A^{-1}AB=A^{-1}0=0$, so $B$ is certainly singular. QED
Determinants are surely not needed for this.
You can prove more: if $AB=0$ and both $A$ and $B$ are non zero, then both are singular. Indeed, take a nonzero column of $B$, call it $v$; then $Av=0$ and so $A$ is singular. Then apply the same to $B^TA^T$, showing that $B^T$ is singular.