How to prove that $AB$ is invertible if and only if $A$ is invertible?

Question

Let $A$ be a matrix and $B$ an invertible matrix. Show that $AB$ is invertible if and only if $A$ is invertible.

I know how to do this using determinants, but how else could you prove this?

Answer 1

$A$ is invertible $\implies$ $AB$ is invertible: This is because $(AB)^{-1}=B^{-1}A^{-1}$.

$A$ is invertible $\impliedby$ $AB$ is invertible: use the proven implication ($\implies$) above, applied to the matrices $AB$ and $ABB^{-1}=A$, with the fact that, since $B$ is invertible, $B^{-1}$ is also invertible.

Answer 2

If $A$ is invertible, $(AB)^{-1} = B^{-1}A^{-1}$. If $A$ is not invertible, its rank is less than $n$ (the size of the matrices). Therefore the rank of $AB$ is less than $n$.

Answer 3

For the converse, matrix multiplication is associative so that if C is the inverse of AB, then (AB)C= A(BC) = I.

Answer 4

Let $A$ and $B$ be $n\times n$ matrices. $B$ is invertible, so its rank is $n$.

If $A$ is invertible, then its rank is $n$. Then the rank of $AB$ is also $n$. Therefore $AB$ is invertible.

If $AB$ is invertible, then its rank is $n$. The rank of $A=ABB^{-1}$ is therefore also $n$ (the identity matrix has full rank).

We use the fact that multiplying two $n\times n$ matrices that have full rank results in a full rank $n\times n$ matrix. That is so, because each matrix will map $n$ linearly independent vectors onto another set of $n$ linearly independent vectors.

Answer 5

A direct proof for $AB$ invertible $\implies A$ invertible (knowing $B$ is):

If $AB$ is invertible, there exists a matrix $U$ such that $(AB)U=I$. However, $(AB)U=A(BU)=I$, hence $BU$ is a right-inverse for $A$.

But also, $U(AB)=(UA)B=I$, so $UA=B^{-1}$, and, multiplying on the left by $B$: $$B(UA)=(BU)A=BB^{-1}=I,$$ which shows $BU$ is also a left inverse for $A$.

Answer 6

Lets say that $A, B \in \mathbf F^{n \times n}$ for some field $\mathbf F$. If $A$ is not invertible then $\operatorname{span}A = \{Ax : x \in \mathbf F^n\}$

is a proper subset of $\mathbf F^n$. That is, there exists a vector $v \in \mathbf F^n$ such that $Ax = v$ has no solution for any $x \in \mathbf F^n$.

If $AB$ were invertible, then for some $y \in \mathbb F^n$, $ABy=v$; in particular, $y = (AB)^{-1}v$. But then $x=By$ would solve $Ax = v$. Hence $AB$ is not invertible.