$\matrix {AB}=\matrix 0$ and $\matrix A\not= \matrix 0$, then $|\matrix B|=0$ ?!

Question

Let $A$ and $B$ be two square matrices such that $\matrix {AB}=\matrix 0$, where $\matrix 0$ denotes the Null Matrix.

My textbook mentions that [without proof]: if $\matrix A\not= \matrix 0$, then $|\matrix B|=0$.

I am not sure if this is even correct:

Consider the scenario: $|\matrix B|=k$, where $k$ is a non-zero constant and $|\matrix A|=0$ with $\matrix A\not =\matrix 0$, then this above property predicts that the product $\matrix {AB}$ cannot be $\matrix 0$. Though I could not find counter-examples, this seems improbable.

• Is this property true?
• If not, can you provide counter-example of the kind I mentioned?

Answer 1

Note that saying $|B|=0$ is very different from saying $B=0$ (assuming that $|B|$ denotes the determinant of $B$).

If $\det B\ne0$, then $B$ is invertible, so from $AB=0$ you get $$ A=ABB^{-1}=0B^{-1}=0 $$

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On the other hand it is easy to find nonzero matrices with zero product: $$ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$ but this doesn't contradict the above: indeed both matrices have zero determinant.

Answer 2

If $AB=0$ and $A\ne0$, then $aB=0$ for every nonzero row $a$ of $A$. Hence $B$ has linearly dependent rows (because $aB$ is a non-trivial linear combination of the rows of $B$) and its determinant must be zero.

Note that this proof is very basic: it works over any commutative ring (not just over fields) and the notion of determinant was devised to detect linear dependence in the first place.