Counterexamples when outer products of matrices are the same other

Question

Let X and Y be two different $m\times n$ matrices, what are the cases/examples when $XX^T= YY^T$ other than $X=YQ$, where Q is an orthogonal matrix? In other words, is there an example such that $X\neq YQ$, yet $XX^T=YY^T$?

Answer 1

For notational convenience, suppose instead that $X^TX=Y^TY$ (where $X$ and $Y$ are $m\times n$). In this case, we claim that $X=QY$ for some $m\times m$ orthogonal matrix $Q$.

Since $X^TX=Y^TY$, we have $\|Xv\|=\|Yv\|$ for every vector $v\in\mathbb R^n$. It follows that $X$ and $Y$ have the same null space $K$ and the same rank.

Let $\mathcal B$ be a basis of $K^\perp$. Since $XK^\perp=X(K+K^\perp)=X\mathbb R^n=\operatorname{range}(X)$, $X\mathcal B$ is a basis of $\operatorname{range}(X)$. Likewise, $Y\mathcal B$ is a basis of $\operatorname{range}(Y)$. So, if we view $Y$ and $X$ as linear maps from $\mathbb R^n$ to $\mathbb R^m$ and define a linear map $f:\operatorname{range}(Y)\to\operatorname{range}(X)$ by $$ f(Yb)=Xb \ \text{ for each }\ b\in\mathcal B, $$ we will have $fY=X$. In addition, since $\|Yb\|=\|Xb\|=\|fYb\|$ for all $b\in\mathcal B$, by the polarisation identity, we have $\langle Yb,Yb'\rangle=\langle fYb,fYb'\rangle$ for all $b,b'\in\mathcal B$. In turn, $\|Yv\|=\|fYv\|$ for all $v\in\operatorname{span}(\mathcal B)=K^\perp$. This means $\|w\|=\|fw\|$ for all $w\in\operatorname{range}(Y)$. That is, $f$ is an isometry.

As $X$ and $Y$ have the same rank, there also exists a linear isometry $g:\left(\operatorname{range}(Y)\right)^\perp\to\left(\operatorname{range}(X)\right)^\perp$. Since $f$ and $g$ are two isometries with mutually orthogonal domains and mutually orthogonal codomains, the linear map $L\in\operatorname{Hom}(\mathbb R^m)$ defined by $L|_{\operatorname{range}(Y)}=f$ and $L|_{\left(\left(\operatorname{range}(Y)\right)^\perp\right)}=g$ is an isometry. The matrix representation $Q$ of $L$ with respect to the standard basis of $\mathbb R^m$ is thus an orthogonal matrix. Also, as $L|_{\operatorname{range}(Y)}=f$, we have $QY=fY=X$. Now we are done.

Answer 2

If $XX^{T}=YY^{T}$, then $X=YQ$ where $Q$ is an orthogonal matrix.

$\textbf{Proof}~$: You can see $X$ and $Y$ as operators from $\mathbb{R^n}$ to $\mathbb{R^m}$.

This proof is similiar to the singular value decomposition.

Because $XX^{T}=YY^{T}$, then we can find an orthogonal matrix $P=(e_1,\cdots,e_m)$ such that $$XX^{T}=YY^{T}=P^{T}\mathrm{diag}\{\lambda_1,\lambda_2,\cdots,\lambda_r,0,\cdots,0\}P,$$ where $r=\mathrm{XX^{T}}=\mathrm{X^{T}}$

Let $\sigma _i=\sqrt{\lambda _i},~~(1\le i\le r)$, then you can get $$\Vert X^{T}(e_i)\Vert=\sigma_i,~~1\le i\le r ,~~(X^{T}(e_i),X^{T}(e_j))=0,~~1\le i\ne j\le r.$$

For $r+1\le j\le m$,$$X^{T}(e_j)=0.$$

Let $f_i=\dfrac{1}{\sigma_i}X^{T}(e_i),~~i=1,2,\cdots,r$, thus $\Vert f_i\Vert=1$ and $(f_i,f_j)=0,~~i\ne j$.

You can expand $f_i$ into standard orthogonal basis of $\mathbb{R^n}.$

For $X^{T}$ and $Y^{T}$, you can get $$X^{T}(e_1,\cdots,e_m)=(f_1,\cdots,f_r,f_{i+1},\cdots,f_n)D$$ $$Y^{T}(e_1,\cdots,e_m)=(f_1,\cdots,f_r,f'_{i+1},\cdots,f'_n)D$$

where $D=\mathrm{diag}\{\sigma_1,\cdots,\sigma_r,0,\cdots,0\}.$

If we denote $(f_1,\cdots,f_r,f_{i+1},\cdots,f_n)=Q$ and $(f_1,\cdots,f_r,f'_{i+1},\cdots,f'_n)=Q_1$, so you can get

$$X=PD^{T}Q^{-1},$$ $$Y=PD^{T}Q_1^{-1}.$$

If we denote $Q=Q_1Q^{-1}$, you can get $X=YQ$ where $Q$ is an orthogonal matrix.