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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.