Unique solution for $AA^T$?

Question

I am doing this in $\mathbb{R}$. Given $A$ and $B$ are two square matrices so that $AA^T=BB^T$. I do not think it leads to $A=B$, but I cannot give a counterexample.

If not true generally, my further question is that: under what conditions, do we have $A=B$?

Thanks!

Answer 1

Indeed, $AA^T=BB^T$ does not imply that $A=B$. Take $A=I_2$ and $$ B=\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix}. $$ Because you said "but I cannot give a counterexample", let me say that taking matrices of size $2$ in many cases already gives a counterexample.

For the second question, this has been discussed already at MSE:

$AA^t=BB^t \implies A=B$

Answer 2

Here is a counter example.

$$A=\begin{bmatrix} 1 & 2 \cr 3 & 4 \end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 1 \cr 4 & 3 \end{bmatrix}$$

Then $$AA^T =BB^T=\begin{bmatrix} 5 & 11 \cr 11 & 25 \end{bmatrix}$$

Other counter examples could be made by changing $1,2,3,4$ to arbitrary $a,b,c,d$