# Zero Product Rule in Matrix Multiplication [duplicate]

So we all know that when we have two matrices A and B (AB=0) whose product is zero,it could imply any of the following,

1. A=0 and B≠0
OR
2. A≠0 and B=0 OR
3. A=0 and B=0 OR 4.A≠0 and B≠0

But there is one property mentioned in my textbook without any proof and I tried my best to prove it but wasn't able to. The property is as folows:

In genreal AB=0 does not imply A=0 or B=0. But if A is non singular and AB=0,then B=0. Similarly if B is non singular and AB=0,then A=0.

Therefore AB=0 Implies : Either both are singular or one of them is a null matrix. Could someone explain this?

A matrix is singular iff it's non invertible.

Suppose $$AB=0$$.
If $$A$$ is invertible then multiplying by $$A^{-1}$$ from the left we get $$B=0$$.
Similarly, if $$B$$ is invertible, we can multiply by $$B^{-1}$$ from the right to obtain $$A=0$$.
In all other cases both $$A$$ and $$B$$ are singular.

In the case of square matrices, if $$B$$ is non-singular you can think of it as a linear map that takes $$\mathbb{K}^N$$ to itself, where $$\mathbb{K}$$ is the field to which the entries of $$B$$ belong to. In particular, the only vector that is mapped to $$0$$ by $$B$$ is $$0$$ itself.

Now, if $$A$$ is not $$0$$ then there is a $$v \neq 0$$ such that $$A v = w \neq 0$$. Then, since a $$B$$ is not singluar the corresponding linear application is surjective: this means that there is a $$u$$ s.t. $$B u = v$$. Then $$AB u = w \neq 0$$, in contraddiction with $$AB = 0$$

First, note that saying that a matrix $$A$$ is nonsingular is equivalent to a number of statements referred to collectively as the Invertible Matrix Theorem. As a result, there are many routes you could take to prove both statements. I'm not sure the extent to which you are familiar with the more abstract concepts of linear algebra, so please let me know if anything is unfamiliar.

If $$A$$ is an $$m \times n$$ matrix whose columns are given by $$A = [\mathbf{a}_1 \dotsb \mathbf{a}_n]$$ and $$B$$ is an $$n \times p$$ matrix whose columns are given by $$B = [\mathbf{b}_1 \dotsb \mathbf{b}_p]$$, the columns of $$AB$$ are given by:

$$[A\mathbf{b}_1 \dotsb A\mathbf{b}_n]$$

That is, the columns of $$AB$$ are formed via linear combinations of the columns of $$A$$ using the columns of $$B$$ as weights.

If $$A$$ is nonsingular, then the columns of $$A$$ are linearly independent, and the equation $$A\mathbf{x}=\mathbf{0}$$ has only the trivial solution. Thus, if for each column of $$AB$$, $$A\mathbf{b}_k = \mathbf{0}$$ then each $$\mathbf{b}_k = \mathbf{0}$$ so $$B$$ is the null matrix.

This next one requires a bit more machinery. If $$AB=0$$, but $$B$$ is nonsingular, then the columns of $$B$$ must be linearly independent. Thus, the dimension of the null space of $$A$$ is $$n$$. By the Rank Theorem, that means that the dimension of the column space is $$\mathrm{rank} A = 0$$. The only subspace with rank zero is the zero matrix so $$A = 0$$.