If $(AA^t)^r=I$ then $A^tA$ is invertible?

Question

Let $m,n,r$ be natural numbers. Let $A$ be an $m\times n$ matrix with real entries such that $(AA^t)^r=I$, where $I$ is the $m\times m$ identity matrix and $A^t$ is the transpose of the matrix $A$. We can conclude that

$m=n$
$AA^t$ is invertible
$A^tA$ is invertible
If $m=n$, then $A$ is invertible

If $r=1$, $AA^t=I$ hence $\vert{AA^t}\vert=\vert{I}\vert=1$ hence option 2 correct

If $m=n$, $\vert{AA^t}\vert=\vert{A}\vert\vert{A^t}\vert=1$ hence $A$ is invertible hence option 4 correct

Let $$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} $$ then $AA^t=I$ but $A$ is not square matrix hence option 1 wrong

how to conclude for option 3

Blanco · Answer 1 · 2017-07-25T07:20:37.043

Let $$ A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} $$ then $A^{t}A = I$ but $m \neq n$, option 1 is wrong. and $$ AA^{t}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \neq I $$ thus, option 3 is wrong.

$(AA^{t})^r=I$, then $\det (AA^{t})^r = 1$, it implies that $\det(AA^t) \neq 0$, thus option 2 is correct.

if $n=m$, then $0 \neq \det(AA^{t}) =\det(A)^2$, thus $\det(A) \neq 0$, option 4 is correct.

However, if $n=m$, then we can conclude that $A$, $A^{t}A$ and $AA^{t}$ are invertible.

José Carlos Santos · Answer 2 · 2017-07-25T08:27:27.863

0

Your proof that conclusion 1. holds is wrong; you can't suppose that $r=1$. What happens is that$$(AA^t)^r=\operatorname{Id}\Longrightarrow\det\bigl((AA^t)^r\bigr)=1\Longleftrightarrow\det(AA^t)^r=1$$and therefore $\det(AA^t)\neq0$. Therefore $AA^t$ is invertible.

If you don't want to use determinants, you can argue like this: if $AA^t$ wasn't invertible, there would be a vector $v$ such that $(AA^t).v=0$ But then $(AA^t)^r$ would be $0$ too.

edited Jul 25 '17 at 08:27

answered Jul 25 '17 at 07:00

José Carlos Santos

427,504

You only have only multiplicativity for square matrices, thus $\det(A^tA)=\det(AA^t)$ doesn't always hold. – Mathematician 42 Jul 25 '17 at 07:56
@Mathematician42 Right. I forgot that the matrices didn't have to be square matrices. My argument is correct only in that case. – José Carlos Santos Jul 25 '17 at 08:26
If $B^r=I$, with $r\ge1$, then $BB^{r-1}=I$, so $B$ is invertible. No need for determinants or null space. – egreg Jul 25 '17 at 08:33
@egreg Right you are! – José Carlos Santos Jul 25 '17 at 08:34

score 0 · Answer 3 · answered Jul 25 '17 at 08:42

I assume that $m,n,r\ge1$.

Set $B=AA^t$. Since $B^r=I$, we have $B^{r-1}B=BB^{r-1}=I$, so $B$ is invertible and $B^{-1}=B^{r-1}$.

Now note that $AA^t$ and $A^tA$ have the same rank as $A$ (it's a very interesting exercise, use the rank-nullity theorem). Therefore

if $m=n$, then $A$ is invertible, so also $A^tA$ is invertible;
if $m\ne n$, then $m<n$ and $A^tA$ is not invertible.

A simple example with $m\ne n$ is $$ A=\begin{bmatrix} 1 \\ 0 \end{bmatrix} $$

If $(AA^t)^r=I$ then $A^tA$ is invertible?

3 Answers3