$A$ and $B$ are two $n\times n $ real matrices, $AB=BA$. Can we conclude that

$$ \det \Big(A^2+B^2\Big)\ge \det(2AB) $$

is right?

Well, the inequality is interesting. if $A,B$ are upper triangular matrices, it is obvious right. If $AB\ne BA$, $ \det \Big(A^2+B^2\Big)\ge \det(AB+BA) $ is wrong.

  • $\begingroup$ It suffices to check for complex upper-triangular matrices .. since, $AB=BA$ it follows $A$ and $B$ are simultaneously upper triangulizable. $\endgroup$ – r9m Apr 22 '14 at 19:06
  • 3
    $\begingroup$ @r9m The eigenvalues are not necessarily real. Your proof only works when the eigenvalues are real. $\endgroup$ – Ewan Delanoy Apr 22 '14 at 19:08

The answer is NO. Take for example $A=I_2$ and

$$ B=\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) $$

We then have $B^2=-I_2$, $A^2+B^2=0$ and $2AB=2B$, so ${\sf det}(A^2+B^2)=0$ and ${\sf det}(2AB)=4$.

  • $\begingroup$ so, for $n\times n$ matrices $A,B$, there also exist counter-example $\endgroup$ – ziang chen Apr 23 '14 at 3:14
  • $\begingroup$ @ziangchen Indeed. Take $A=I_n$ and $B=$ the matrix made of two blocks, $\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$ and $I_{n-2}$. $\endgroup$ – Ewan Delanoy Apr 23 '14 at 5:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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