0
$\begingroup$

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

$\endgroup$

closed as unclear what you're asking by Arnaud D., Abcd, metamorphy, José Carlos Santos, user91500 Jan 12 at 10:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $QB=Q^TB \Rightarrow Q=Q^T$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

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

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.