# Show that AB is singular if A is singular

Actually I need to show that $\det(AB) = \det(A)\det(B)$ if $A$ is a singular matrix.

The determinant of $A$ is $0$ if $A$ is singular, so $\det(AB)$ has to be $0$ as well, but I have problems showing that $AB$ is singular if $A$ is singular. How can I show that?

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You've exactly written the reason there. The identity $det(AB) = det(A)det(B)$ is the key – BlueBuck May 3 '14 at 16:47
But that's what I need to prove. – eager2learn May 3 '14 at 16:47
some proofs are provided here:proofwiki.org/wiki/Determinant_of_Matrix_Product – Fermat May 3 '14 at 16:53

One approach is this: That a matrix $C$ is singular gives us in particular that its null space is non-trivial, that is, for some vector $x\ne0$ we have $Cx=0$. That $C$ is nonsingular, on the other hand, gives us in particular that the column space of $C$ has full rank, that is, for any vector $b$ there is a vector $a$ such that $Ca=b$.

Now, suppose $A$ is singular. If $B$ is also singular, then for some $x\ne 0$ we have $Bx=0$, but then $(AB)x=A(Bx)=A0=0$, and we conclude that $AB$ is also singular.

If, on the other hand, $B$ is nonsingular, use that $A$ is singular to find $b\ne 0$ such that $Ab=0$. Now, use that $B$ is nonsingular to find $a$ such that $Ba=b$. Clearly $a\ne0$ since $b\ne0$. But now we have that $(AB)a=A(Ba)=Ab=0$, and we conclude (again) that $AB$ is singular.

This completes the proof. Notice, by the way, that we also showed that 1) $AB$ is singular if $B$ is the one assumed singular. On the other hand, since $A,B$ being nonsingular gives us that $AB$ is nonsingular, then we also have that 2) if $AB$ is singular, then at least one of $A$ and $B$ must be singular as well.

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I didn't see your answer!+1 – user63181 May 3 '14 at 17:10

If $A$ is singular then it isn't injective: there's $y\ne0$ such that $$Ay=0$$ Now

• if $B$ is invertible then let $x$ such that $Bx=y$ and then $$ABx=Ay=0$$ and
• if $B$ is also singular then there's $z\ne0$ such that $Bz=0$ and then $$ABz=0$$ so we prove that $AB$ isn't injective which's equivalent to $AB$ is singular.
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This feature request is definitely needed... – Andrés E. Caicedo May 3 '14 at 17:07
Nice answer(s), Sami and Andres. +1 to you both. – user1551 May 3 '14 at 17:09
It's wrong to precipitate and say that its inverse is $B^{-1}A^{-1}$. Rather you should say that its inverse say $C$ and prove that $C= B^{-1}A^{-1}$@BCLC – user63181 May 3 '14 at 17:18
@BCLC To elaborate on Sami's comment, if $AB$ is invertible, it has an inverse $C$. Therefore $I=(AB)C=A(BC)$ and hence $A$ is invertible with its inverse equal to $BC$. – user1551 May 3 '14 at 17:27
@BCLC To elaborate on user1551's comment, if $C$ is the inverse of $AB$, then indeed $BC$ is a right inverse of $A$. One then needs an additional argument to conclude that also $(BC)A=I$, from which one can finally successfully conclude that $A$ is indeed invertible. Once we have that both $AB$ and $A$ are invertible, then we can conclude that $B$ is also invertible (but this also needs a proof, of course). Once we have that $A$ and $B$ are invertible, and only then, can we conclude that $C=B^{-1}A^{-1}$. – Andrés E. Caicedo May 3 '14 at 19:03

Hint: if $x^TA=0$ for some nonzero vector $x$, then ...

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Either A is a zero matrix or all the rows of A are the same. Sorry this doesn't help me. Can you give another hint? – eager2learn May 3 '14 at 16:57
@eager2learn With the aforementioned $x$, what is $x^TAB$? – user1551 May 3 '14 at 16:59
I think it's also 0. – eager2learn May 3 '14 at 17:01
@eager2learn Have you learnt that a matrix $M$ is singular if and only if $x^TM=0$ (or equivalently, $M^Tx=0$) for some nonzero vector $x$? – user1551 May 3 '14 at 17:06
Well we had defined a matrix A to be singular if rank(A)<n, where we defined the rank as dim(im(f)) where f is the linear map that corresponds to A. So if A is regular then f is injective and so Mx=0 <=> x=0. Then if A is singular it's not injective and there are non-zero vectors x such that Mx=0. So I guess we did indirectly cover that last semester, but I didn't think of this explicitly. So I guess if $x^TA=0$ for some non-zero vector x then it follows that A cannot be injective and thus has to be singular. Is that what you were trying to lead me towards? – eager2learn May 3 '14 at 17:17

Contrapositive: If AB is not singular, then A is not singular. If AB is not singular, then it has an inverse. Its inverse is $B^{-1}A^{-1}$ which implies that B and A are not singular.

Expansion of my answer: if AB is not singular, then A is not singular because if AB is not singular, then AB has an inverse. AB's inverse is $B^{-1}A^{-1}$ which implies that B and A are not singular which implies A is not singular.

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Why is A not singular then? – eager2learn May 3 '14 at 16:58
If B and A are not singular then A is not singular. I think? – BCLC May 3 '14 at 16:58
You wrote if AB is not singular then A is not singular. Why is that the case? You didn't write if A and B are not singular then A is singular. – eager2learn May 3 '14 at 16:59
This argument is incorrect. To write $B^{-1}A^{-1}$ makes no sense, until you prove that both $A$ and $B$ are indeed invertible, which you did not do. Your argument is very similar to saying that the zero matrix $0$ is invertible because $0^{-1}0=I$. – Andrés E. Caicedo May 3 '14 at 19:00
"if $A$ does not have an inverse, then we cannot obtain the inverse of $AB$" Well, yes. This is exactly what the question is asking you to prove. So, it is not that your argument is incorrect, but rather that there is no argument. You are just repeating the statement of the question, and claiming that it holds. It may be better to study the sketch we gave you on the other answer. The issue seems to be that you are assuming (as part of the background) too much already, almost as if the relevant statements were axioms, while the question is precisely to verify some of these assumptions. – Andrés E. Caicedo May 4 '14 at 0:24