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In the textbook Contemporary Linear Algebra by Anton and Busby, there was a small question in section 3.2 page 101 concerning this. It asks if $A$ and $B$ are two non-zero square matrices such that $AB=0$, then $A$ and $B$ must both be singular. Why is this so?

I can prove that if $A$ is non-singular then $B=I_nB=A^{-1}AB=0$, implying $B$ must be the zero matrix which is a contradiction. Similarly if $B$ is non-singular, then $A$ must be the zero matrix. Hence, both must be singular. But this doesn't really answer why, it just shows a contradiction for any case and hence must be the negation of our supposition that at least one is non-singular.

I would like to know the essence and inherent property as to why they must be both singular (and why can't it be the case that only one is singular?) and what is the motivation for such a conclusion?

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What's wrong with the proof you've given? This shows that neither factor can be non-singular. –  Thomas Bloom Feb 21 '12 at 8:45
I guess in some sense the idea is that an invertible matrix can be thought of as a change of basis as it's just an isomorphism between two vector spaces of the same dimension. And changing the basis of the domain or codomain won't affect whether the other map is zero or not. That's a bit vague which is why I haven't given it as a full answer, but I think that might be the idea you're after. (The proof you have given is essentially showing this). –  Matt Pressland Feb 21 '12 at 8:50
Thanks for the answers. I guess part of the reason why I wasn't satisfied was probably because of the motivation behind the proof. In other words without being told that both must be singular, how would one conjecture or intuitively know that both must be singular. (Or possibly first conclude that at least one must be singular and then later realize that both must be singular)? –  tcmtan Feb 21 '12 at 22:26
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4 Answers

up vote 5 down vote accepted

As Thomas points out, your proof is fine, but if you want another way to look at it, consider the following:

Suppose $AB = 0$. What is the $j$-th column on either side of this equation? On the left, it is a linear combination of the columns $\{\mathbf a_j\}$ of $A$, with coefficients from the $j$-th column of $B$, and on the right is the 0 vector:

$$b_{1j}\mathbf a_1 + b_{2j} \mathbf a_2 + \cdots + b_{nj}\mathbf a_n = \mathbf 0$$

This is true for each $j$, and there must be at least one non-zero $b_{ij}$ coefficient, since $B\neq 0$, so the columns of $A$ are linearly dependent.

Similarly, we can ask what are the rows on each side of the equation? The $i$-th row is a linear combination of the rows of $B$, with coefficients from the $i$-th row of $A$. So you see that the rows of $B$ must be linearly dependent.

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That's a nice way of looking at it from a more "beginner" perspective of linear algebra like myself. –  tcmtan Feb 21 '12 at 22:23
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The argument that you gave does indeed work and proves the claim.

If you are not "philosophically" satisfied with it, you may look at it in this way. Recall that matrices can be identified to linear transformations (once bases are fixed) and that the product of matrices correspond to composition of transformations. Also, non singular matrices correspond to automorphisms.

Now you can translate the claim into the following. Suppose that you have linear transformations $f$ and $g$ such that the composition $$ V\stackrel{f}{\longrightarrow}V\stackrel{g}{\longrightarrow}V $$ is the $0$-map. Suppose that, say, $g$ is an automorphism. Then, if $f$ is not the $0$-map, the subspace $W={\rm im}(f)$ is not trivial. But now, since $g$ is an automorphism, we must have $g(W)=g\circ f(V)\neq\{0\}$ contradicting the assumption.

A symmetric argument deals with the case when $f$ is an automorphism. Thus we conclude that neither $f$, nor $g$ can be automorphisms.

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I have not yet done linear transformations formally, but will learn it this semester. Thanks for the answer, hopefully I will understand it soon after. –  tcmtan Feb 21 '12 at 22:14
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Perhaps it would be helpful to think in terms of the associated linear transformations $T_A$, $T_B$, and $T_{AB}$. You perhaps know that $T_{AB}=T_A\circ T_B$. If $AB$ is zero, then $T_{AB}$ is the zero transformation that sends every vector to $\vec 0$. That means that $T_A\circ T_B$ must do the same thing. This means that $T_A$ has to ‘kill off’ every non-zero vector in range of $T_B$, if there are any.

Since $T_B$ isn’t the zero transformation, there must be at least one vector that it doesn’t kill off, so $T_A$ has to send at least one non-zero vector to $\vec 0$. But this means that it can’t be invertible: it ‘collapses’ two vectors together, and there’s no way to ‘un-collapse’ them.

On the other hand, $T_A$ isn’t the zero transformation either, so there’s at least one vector $v$ such that $T_A(v)\ne\vec 0$. If $T_B$ were invertible, that $v$ would have to be in the range of $T_B$, i.e., there would have to be some $u$ such that $v=T_B(u)$. But then $T_A\circ T_B$ wouldn’t kill off $u$, so it wouldn’t be the zero transformation after all. Thus, $T_B$ can’t be invertible either.

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Again, thank you for another answer from the perspective of linear transformations. I will be looking forward to understanding this once I have covered linear transformations this semester. =) –  tcmtan Feb 21 '12 at 22:15
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as AB is a O matrix its determinant has to be 0 so |AB| = |A| |B| = 0 so either |A| = 0 or |B| = 0

see for example 1 2 3 4 5 6 7 8 9

is a non 0 matrix but its determinant is 0

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