# Showing AB=0 does not imply either A,B=0, but that singular

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.

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To type $A^{-1}$, you need to put $-1$ inside a pair of curly brackets, i.e. A^{-1}. –  user1551 May 6 at 13:29
@user1551 Woops, didn't spot that error. Thanks. –  Ollie Ford May 6 at 13:30
@egreg True, but that's not the aim. See the blockquote. –  Ollie Ford May 6 at 13:31
@egreg To "show that ... at least one of them is singular" –  Ollie Ford May 6 at 13:33
Your proof seems correct. I think the hint just want to point out that $det(AB)=det(A)det(B) \neq 0$, so $AB$ is also non-singular and can't be null. You don't even need the $det$, since the product of two invertible elements in a ring is always invertible and zero is never...(except in the trivial field $\{0=1\}$ ) –  Yann Hamdaoui May 6 at 13:34

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.

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Surely proving both are singular follows from the same argument - just switch As for Bs and vice versa? –  Ollie Ford May 7 at 7:52
@OllieFord A condition for singularity of $B$ is that there is a nonzero row vector $w$ such that $wB=0$ (which is the same as saying that $B^T$ is singular). –  egreg May 7 at 8:41

No need for a contradiction. Matrix $C$ is singular iff $Cx=0$ for some nonzero $x$. $AB=0$ implies that for any $x\neq 0$, $ABx=0$. If $Bx=0$, you are done and $B$ is singular. If $Bx=y\neq 0$, then $Ay=ABx=0$ and $A$ is singular.

With determinants: $AB=0$ implies $\det(AB)=\det(A)\det(B)=0$, so either $\det(A)=0$ or $\det(B)=0$ or both.

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Why only for $x\ne0$? What is $0\cdot0$ if not $0$? –  Ollie Ford May 6 at 13:41
Because if $x=0$, then obviously $Cx=0$ for any $C$. –  Algebraic Pavel May 6 at 13:42
Ah of course. Thanks. –  Ollie Ford May 6 at 13:42

There's nothing wrong with your proof. It can be argued to be even "more basic" than the determinant one.

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You can do a bit better than this: if $AB=0$ then either both matrices are singular, or one of them is zero; of course a zero matrix is singular*. Because if $AB=0$, then if $A$ is non-singular, then one has $B=A^{-1}AB=A^{-1}0=0$; similarly for $B$ non-singular gets $A=0$.

*If like me, you happen to care not to ignore the existence of matrices without entries (which happens if one or both of their dimensions is$~0$), then this is wrong: the $0\times0$ matrix is both nonsingular (it is the identity) and zero. But note that "$AB=0$ implies $A$ or $B$ is singular" also fails for $0\times0$ matrices. So the reformulation I gave, which remains valid for empty matrices, is in fact better than the original formulation.

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