# counterexamples to $\det \Big(A^2+B^2\Big)\ge \det(AB-BA)$

$n\geq3$. A and B are two $n\times n$ reals matrices. For $n\times n$, Could one give counterexamples to show that

$$\det \Big(A^2+B^2\Big)\ge \det(AB-BA) \tag{*}$$

is not necessarily true?

Well, I won't do more than $3\times3$: the following $A,B$ show (*) is not right

$$A=\left(\begin{array}{ccc}0&2&0\\0&0&1\\0&0&0\end{array}\right), B=\left(\begin{array}{ccc}0&0&0\\1&0&0\\0&1&0\end{array}\right)$$

What is the counterexample for $n\times n$? Thanks a lot!

• We, one is symmetric, and one is anti-symmetric when $n$ is odd, so if you found $A$ and $B$ such that $\det(A^2+B^2)<0$ you'd be done, at least if $n$ is odd. – Thomas Andrews Apr 19 '14 at 16:56
• With this counterexample you are proving that it does not work for all n. – YTS Apr 19 '14 at 16:58
• @YotasTrejos: Yes, I think the question is to prove that in fact it works for no $n \ge 3$. (Note, for instance, that for $n=1$ it is true.) – Nate Eldredge Apr 19 '14 at 17:19

Let $$A=\begin{bmatrix}I_{n-2} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & 0 & -1 \\ \mathbf{0} & 1 & 0\end{bmatrix},$$ then $$A^2=\begin{bmatrix} I_{n-2} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&-1 & 0 \\ \mathbf{0}& 0 & -1\end{bmatrix}.$$ Now make $B$ a matrix filled with zeros, except for the very last diagonal entry, make this entry $\sqrt{2}$. Then $B^2$ has all zeros except for the last entry on the diagonal which is $2$, and so we have $$A^2+B^2=\begin{bmatrix} I_{n-2} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&-1 & 0 \\ \mathbf{0}& 0 & 1\end{bmatrix}.$$ We then also have $$AB-BA=\begin{bmatrix} \mathbf{0} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&0 & -\sqrt{2} \\ \mathbf{0}& -\sqrt{2} & 0\end{bmatrix},$$ and so then det$(A^2+B^2)=-1$ and det$(AB-BA)=0$ for all $n \geq 3$.