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$A$ 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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 ${\sf det}(2AB)=4$.

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  • $\begingroup$ 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

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