If the product of two non zero square matrices A and B of the same order is a zero matrix then

If A and B are non zero square matices, then AB=0 implies

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Let A and B be two non-zero square matrices. If the product AB is a null matrix, then (a) A and B are singular (b) B is non-singular (c) A is non-singular (d) None of these

I'm just browsing around looking at old questions but I thought I'd add to this. The question already has great answers, but like many questions in linear algebra there are many different (and often equivalent) ways to demonstrate some particular fact. For the sake of future readers, these might be interesting or helpful.

1. Canceling out

I'll start by boiling down the approach mentioned in the question to this:

If $AB=0$, then:

$A^{-1}$ exists $\Rightarrow$ $B=0$ (just "cancel out" the $A$ by left-multiplying $A^{-1}$)

$B^{-1}$ exists $\Rightarrow$ $A=0$ (just "cancel out" the $B$ likewise)

So the fact that $A\ne 0$ and $B\ne 0$ tells us that neither matrix is invertible (we are using the logical contrapositives of the two statements above). This all rests on the idea of "canceling out" a non-singular matrix.

2. Matrix Rank

Every matrix has a quantity called rank associated to it. The way rank is introduced in my experience is often as a consequence of row-reduction -- the rank is the number of pivot columns (same as the number of nonzero rows) after a matrix is put into row-echelon form.

In particular, the zero matrix has rank zero, so what we have tells us $\mathrm{rank}(AB)=0$. If $A$ is an invertible $n\times n$ matrix, then $\mathrm{rank}(A)=n$.

But there are two facts about rank that we can employ here:

Not only is $\mathrm{rank}(0)=0$, but the zero matrix is the only matrix with zero rank.
If $A$ is invertible, then $\mathrm{rank}(AB)=\mathrm{rank}(B)$.

We can make the same sort of deduction as before:

$A$ invertible $\Rightarrow$ $\mathrm{rank}(B)=\mathrm{rank}(AB)=0$ $\Rightarrow$ $B=0$

$B$ invertible $\Rightarrow$ $\mathrm{rank}(A)=\mathrm{rank}(AB)=0$ $\Rightarrow$ $A=0$

and thus, as above, we know neither matrix could be invertible.

3. Conclusions

As I mentioned, these answers (and those by other users) are all equivalent: A linear transformation has something called rank that is equivalent; The linear independence of columns/rows of a matrix can tell you about its rank... and so forth.

To try to take a stab at the heart of your question, you ask what is the essential or inherent reason why they must both be singular which I would actually rephrase:

What is the essential or inherent reason why one being non-singular forces the other to be zero?

If a square matrix is non-singular, it can be (in some way or another) removed from the equation, meaning the other one remains and is zero. This may be by cancellation, by using rank, by re-framing the question as linear transformations, or by breaking down a matrix product into a set of linear equations -- methods that are all really the same.

If you think of matrices as living in three possible categories, either non-singular, singular (but not zero), or zero, it's interesting that we know:

$AB$ being non-singular implies $A$ and $B$ are both non-singular.
$AB$ being singular implies one of $A$ or $B$ is singular.
$AB$ being zero implies both of $A$ and $B$ are singular unless one of them is zero.

The last statement is what we just discussed, the other two follow immediately from the fact that $\det(AB)=\det(A)\det(B)$.

These three rules often apply in other contexts with zero-divisors, like modular arithmetic.