Property of product of matrices with their transpose

Question

Let $A$ and $B$ be matrices of size $m\times n$.

Claim: $A^TA=B^TB$ iff there exists an orthogonal matrix $Q$ such that $QA=B$.

The backward implication is trivial, consider $B^TB=(QA)^T(QA)=A^TQ^TQA=A^TA$. But how do I show that the forward implication holds true?

Also are there other properties matrices $A$ and $B$ follow, given that $A^TA=B^TB$?

How do I compute such matrices? Is there an equivalent criterion which will a be basis for computing such matrices?

Answer 1

Let $r=\operatorname{rank}(A)$ and $\{Av_1,\ldots,Av_r\}$ be an orthonormal basis of $\operatorname{range}(A)$. Since $A^TA=B^TB$, the set $\{Bv_1,\ldots,Bv_r\}$ is an orthonormal basis of $\operatorname{range}(B)$. Thus there exists an orthogonal matrix $Q$ such that $QAv_i=Bv_i$. It follows that $QAv=Bv$ for every $v\in V=\operatorname{span}\{v_1,v_2,\ldots,v_r\}$.

However, as $A^TA=B^TB$, we also have $W:=\ker(A)=\ker(B)$. Hence $QAw=Bw=0$ on $W$ and we conclude that $QA=B$ because $\mathbb R^n=V\oplus W$.