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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?

Michael Hardy
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eager2learn
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5 Answers5

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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.

Andrés E. Caicedo
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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.
  • This feature request is definitely needed... – Andrés E. Caicedo May 03 '14 at 17:07
  • Nice answer(s), Sami and Andres. +1 to you both. – user1551 May 03 '14 at 17:09
  • I'm guessing x $\neq$ 0 since x = 0 => y = 0? – BCLC May 03 '14 at 17:10
  • Btw, user1551, Sami and @AndresCaicedo, anything wrong with my attempt?

    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.

     – BCLC May 03 '14 at 17:12
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    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 –  May 03 '14 at 17:18
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    @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 03 '14 at 17:27
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    @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 03 '14 at 19:03
  • Thanks SamiBenRomdhane user1551 and AndresCaicedo! @AndresCaicedo, do we still need to worry about right and left inverses? We're getting determinants of A, B and AB so I think they are all square and of the same size... – BCLC May 03 '14 at 22:57
  • Wait @user1551, that's brilliant! Why not just submit that as the answer? It's seems elegant. – BCLC May 03 '14 at 23:02
  • @BCLC To elaborate on Andres's comment (when are our comments elaborate enough?), one can find many proofs of the statement "$XY=I\ \Rightarrow\ YX=I$ for square matrices $X,Y$" in q3852. One can see that finite dimensionality is crucial here. – user1551 May 03 '14 at 23:10
  • @user1551 I don't follow. Cmiiw, but getting the determinant of a matrix implies that it is square? If not, why? If so, why not just use the May 3 17:27 and/or May 3 19:03 arguments? – BCLC May 08 '14 at 14:33
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Hint: if $x^TA=0$ for some nonzero vector $x$, then ...

user1551
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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 03 '14 at 16:57
  • @eager2learn With the aforementioned $x$, what is $x^TAB$? – user1551 May 03 '14 at 16:59
  • I think it's also 0. – eager2learn May 03 '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 03 '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 03 '14 at 17:17
  • Ok so if $x^TAB=0$ for some non-zero x then it follows that AB must be singular. Thanks for your help. – eager2learn May 03 '14 at 17:18
  • @eager2learn It's more or less what you've said. The point is that a matrix is singular if and only if its transpose is singular. If you use matrix rank to define (non-)singularity, the previous statement holds because row rank = column rank. Similarly if you use null space to definie singularity. If determinant is used to define singularity, note that $M$ and $M^T$ have identical determinants. – user1551 May 03 '14 at 17:35
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Let $A$ be a singular matrix. Suppose by way of contradiction that $AB$, for some matrix $B$, has an inverse. Then, there exists some matrix C such that $(AB)C=C(AB)=I$, where I is the identity matrix. But then, $(AB)C=A(BC)$ by associativity. Then imposing $A(BC)=I$ leads us to $BC=A^{-1}$. Which is a contradiction since A was singular by the problem.

Cuhrazatee
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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.

BCLC
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  • Why is A not singular then? – eager2learn May 03 '14 at 16:58
  • If B and A are not singular then A is not singular.

    I think?

     – BCLC May 03 '14 at 16:58
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    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 03 '14 at 16:59
  • Oh sorry for the confusion. "if AB is not singular then A is not singular." is the contrapositive of what we're trying to prove "AB is singular if A is singular." – BCLC May 03 '14 at 17:00
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    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 03 '14 at 19:00
  • @AndresCaicedo Okay forget the $B^{-1} A^{-1}$. Cmiiw, but if AB is given to be invertible then does it not follow that A has an inverse because if A does not have an inverse, then we cannot obtain the inverse of AB which we know to exist, and, hence, A has an inverse?

    In what way is my argument "very similar to saying that the zero matrix $0$ is invertible because $0^{-1} 0 = I$"?

    If $00$ is invertible then $0$ is invertible because if $00$ is invertible then $00$ has an inverse. $00$'s inverse is $0^{-1}0^{-1}$ which implies that $0$ and $0$ are invertible which implies $0$ is invertible.

     – BCLC May 03 '14 at 22:48
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    "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 04 '14 at 00:24
  • @AndresCaicedo THANKS :)) – BCLC May 08 '14 at 14:36