Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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

4 Answers 4

up vote 5 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer
    
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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.