# determinant inequality, $AB=BA$, then $\det(A^2+B^2)\ge \det(2AB)$

$A$ 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.

• It suffices to check for complex upper-triangular matrices .. since, $AB=BA$ it follows $A$ and $B$ are simultaneously upper triangulizable. – r9m Apr 22 '14 at 19:06
• @r9m The eigenvalues are not necessarily real. Your proof only works when the eigenvalues are real. – 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 ${\sf det}(2AB)=4$.
• so, for $n\times n$ matrices $A,B$, there also exist counter-example – ziang chen Apr 23 '14 at 3:14
• @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 $I_{n-2}$. – Ewan Delanoy Apr 23 '14 at 5:15