For positive semidefinite matrix $A\in\mathbb R^{n\times n}$?

Question

For positive semidefinite matrix $A\in\mathbb R^{n\times n}:$

$ \frac{1}{n}\text{tr} A = (\text{det} A)^\frac{1}{n} $

Is this true?

Answer 1

Consider $$A=\begin{pmatrix}1&1\\1&1\end{pmatrix}.$$ We have that

$$(x\:\: y)\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=(x+y)^2\ge 0.$$ Thus it is positive semidefinite.

Now, $\det(A)=0$ and $\rm{tr}(A)=2.$ Thus the equality you claim is false in general.