# Two $m \times n$ matrices$\ A$ and$\ B$ satisfy that$\ A^TB=B^TA$. How to find a matrix $\ Q$ s.t.$\ A=QB$? [closed]

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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• no idea why there should be such a $Q.$ You have some examples? – Will Jagy Jan 11 at 1:51
• 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.. – 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.

• $QB=Q^TB \Rightarrow Q=Q^T$ is so not true, but it does give some insight. – Quang Hoang Jan 11 at 3:57
• Can you give a counterexample? thx. In the meanwhile, I have deleted the last part – pendermath Jan 11 at 4:03
• 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 $U$. – 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 ...?".

Counterexamples.

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

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