2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ If $AB$ is invertible, $B(AB)^{-1}=A^{-1}$ . $\endgroup$ – Berci Feb 18 '18 at 1:06
2
$\begingroup$

$A$ 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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $n$.

$\endgroup$
  • $\begingroup$ How can you say the first line without knowing B is investable? $\endgroup$ – Prince M Feb 18 '18 at 1:00
  • $\begingroup$ OP states "Let A be a matrix and B an invertible matrix" $\endgroup$ – Mauve Feb 18 '18 at 1:02
  • $\begingroup$ Oh, whoops. I only read the title. $\endgroup$ – Prince M Feb 18 '18 at 1:04
  • 1
    $\begingroup$ I was very concerned to see affirmative answers thinking the title was the proposition $\endgroup$ – Prince M Feb 18 '18 at 1:05
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.