Matrix Determinant inequality.

Question

I'm having some trouble in proving this inequality:

$$\text{Det}\left(\frac{1}{n}\mathsf{A}^n\right) \leq \frac{1}{n-k}\text{Det}\left(k \mathsf{A}^{n-k}\right)$$

For every $k < n$, $k\neq 0$, assuming $n, k \in\mathbb{N}$, and then try to extend to whole $\mathbb{R}$ set.

$\mathsf{A}$ are obviously some square matrices whose determinant is non zero.

Any Hint? I clearly already tried with ordinary methods..

Answer 1

Suppose that $A$ is $m\times m$.

Then $$\det\left(\frac{1}{n}A^n\right)=\frac{1}{n^m}\det(A)^n$$

On the other hand, $$ \frac{1}{n-k}\det(kA^{n-k})=\frac{k^m}{n-k}\det(A)^{n-k} $$

Assuming that $\det(A)$ is positive, the inequality means that $$ \det(A)^k\leq\frac{(kn)^m}{n-k} $$

Since $\det(A)$ can be arbitrarily large and the RHS does not depend on $A$, it looks like this inequality does not hold.

Answer 2

Counterexample

Suppose that

$$ k=1, \;\;\;\; n=2, \;\;\;\; \text{and} \;\;\;\; \mathbf{A}=\left[ \begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}\right]$$

then

$$ \det\left(\frac{1}{n}\mathbf{A}^n\right)=\frac{9^2}{2^2} \gt 9 = \frac{1}{n-k}\det\left(k\mathbf{A}^{n-k}\right)$$

which violates the proposed inequality.