There has two $m \times n$ matrices$\ A$ and$\ B$ satisfy that$\ A^TB=B^TA$. Find a matric$\ Q$ s.t.$\ A=QB$
closed as unclear what you're asking by Arnaud D., Abcd, metamorphy, José Carlos Santos, user91500 Jan 12 at 10:00
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$A=QB \Rightarrow A^T = B^T Q^T \Rightarrow A^T B = B^T Q^T B \Rightarrow$ (hypothesis) $\Rightarrow B^T A = B^T Q^T B \Rightarrow A= Q^TB$. Therefore, $QB=Q^TB$.
Also, $A^TB=B^TA$ means $A^TB=(A^TB)^T$, so $A^TB$ is symmetric.
Hope it helps to start working on your solution.
We assume that the matrices are real. Necessarily $rank(A)=s\leq rank(B)=r$.
$Q$ exists iff ($Bx=0\implies Ax=0$) iff $A=AB^+B$ where $B^+$ is the Moore Penrose inverse of $B$.
If it's true, then we may choose $Q=AB^+$.
Unfortunately, it's false.
For the OP: when one is not sure about a question, one writes: "is it true that ...?".