There has two $m \times n$ matrices$\ A$ and$\ B$ satisfy that$\ A^TB=B^TA$. Find a matric$\ Q$ s.t.$\ A=QB$


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  • $\begingroup$ no idea why there should be such a $Q.$ You have some examples? $\endgroup$ – Will Jagy Jan 11 at 1:51
  • $\begingroup$ To the OP. The use on this site is to upvote or (and) to give a green chevron to the answer that satisfies you; if no answer is satisfactory, then write why.. $\endgroup$ – loup blanc Jan 16 at 18:36

$A=QB \Rightarrow A^T = B^T Q^T \Rightarrow A^T B = B^T Q^T B \Rightarrow$ (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.

  • $\begingroup$ $QB=Q^TB \Rightarrow Q=Q^T$ is so not true, but it does give some insight. $\endgroup$ – Quang Hoang Jan 11 at 3:57
  • $\begingroup$ Can you give a counterexample? thx. In the meanwhile, I have deleted the last part $\endgroup$ – pendermath Jan 11 at 4:03
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    $\begingroup$ What if $B=0$? More complex example is that pick any antisymmetric matrix $U=Q-Q^T$ and $B$ composes of the columns which are vectors in the kernel of $U$. $\endgroup$ – Quang Hoang Jan 11 at 4:12

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 ...?".


$m=3,n=5,r=2,s=1$. $A=\begin{pmatrix}1&2&0&1&2\\0&0&0&0&0\\0&0&0&0&0\end{pmatrix},B=diag(0_{1,3},I_2)$.

$m=5,n=3,r=2,s=1$. $A=\begin{pmatrix}1&2&3\\0&0&0\\0&0&0\\0&0&0\\0&0&0\end{pmatrix},B=diag(0_{3,1},I_2)$.


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