# Let $A$ and $B$ be $n\times n$ matrices such that $AB$ is invertible. Prove A and B are invertible.

Let $$A$$ and $$B$$ be $$n\times n$$ matrices such that $$AB$$ is invertible.

1. Prove that $$A$$ and $$B$$ are invertible.

2. Give an example to show a product of nonsquare matrices can be invertible even though the factors are not.

1. If $$AB$$ is invertible, then there exists $$(AB)^{-1}$$ such that $$(AB)(AB)^{-1}=I_n$$. So we find inverses for both $$A$$ and $$B$$, hence $$A$$ and $$B$$ are invertible.

2. Let $$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix}$$ and $$B=\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$$ Then $$AB$$ is invertible while $$A$$ and $$B$$ are not.

Is it fine?

• In (a) you are not showing anything. How do you find the inverses for $A$ and $B$ ? Also, (b) is fine! Feb 13, 2020 at 4:14
• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. See also the section on titles in How to ask a good question. (The part entitled "Make your title your question" is especially relevant to this.) Feb 13, 2020 at 4:16
• In (b), what is a non-invertible nonsquare matrix? If it means "not full rank" then $A$ and $B$ are not valid. Feb 13, 2020 at 4:20
• Are you saying that $A^{-1}$ exists because $(AB)^{-1}$ exists? That's exactly the problem. Feb 13, 2020 at 4:21

Hints. Let $$E$$ be a square matrix. Then, the following are equivalent:

$$\rm (a)$$ $$E$$ is invertible.

$$\rm (b)$$ The only solution for $$E\vec{x} = \vec0$$ is $$\vec{x} = \vec0$$.

$$\rm (c)$$ For any vector $$\vec y$$, there exists a vector $$\vec x$$ such that $$\vec{y} = E\vec{x}$$.

Now, use $$\rm (b)$$ to show that $$B$$ is invertible, and use $$\rm (c)$$ to show that $$A$$ is invertible.

If you know some things about determinants: if $$AB = I$$ then

$$\det(A)\det(B)=\det(AB)=\det(I)=1$$

So, $$\det(A),\det(B)\neq 0$$ and so both are invertible.

For Part 1, think about how you might use the fact that $$\det(AB) = \det(A)\det(B)$$ to prove that statement in a more rigorous fashion.

Part 2 looks good.