# Rewrite $(\det A)^{1/n}=\min\left\{\frac{\mathrm{tr}(AC)}{n}:C \in \Bbb{C}^{n×n},C>0,\det C=1\right\}$ in terms of $\frac{\rm{tr}(CAC)}{n}$

Given $$(\det A)^{1/n} = \min \left\{\frac{\operatorname{tr}(AC)}{n} : C \in {\Bbb C}^{n \times n}, C > 0, \det C = 1\right\}. \label1\tag1$$ Question

1. Show that the formula can be rewritten as $$(\det A)^{1/n} = \min \left\{\frac{\operatorname{tr}(CAC)}{n} : C \in {\Bbb C}^{n \times n}, C > 0, \det C = 1\right\}. \label{2}\tag{2}$$
2. Then, show by an example that the formula \eqref{1} is false if $$A$$ is singular.

My approach, would love to get your opinions:

The determinant and the trace are two quite different beasts, little relation can be found among them.

If the matrix is not only symmetric (hermitic) but also positive semi-definite, then its eigenvalues are real and non-negative. Hence, given the properties $${\rm tr}(AC)=\sum \lambda_c$$ and $${\rm det}(A)=\prod \lambda_C$$, and recalling the AM GM inequality, we get the following (probably not very useful) inequality:

$$\frac{\operatorname{tr}(AC)}{n} \ge {\det}(A)^{1/n}.$$

(equality holds iff $$M = \lambda I$$ for some $$\lambda \ge 0$$)

My Solution:

1. If $$\det C = 1$$, could I say that $$C$$ is either an unity matrix or identity matrix, and hence it won't have any other max. or min. occurrences, thus the formula can be re-expressed as stated above?
2. Could I say that $$AC = A =$$ matrix with all zeros and just eigenvalues on the diagonal $$\lambda_1 ,..., \lambda_i,..., \lambda_n$$, then $$\frac{\lambda_1 +...+\lambda_n}{n} \ge (\lambda_1 \cdot ... \cdot \lambda_n)^{1/n}$$ and $$\frac{\lambda_1 +...+\lambda_n}{(\lambda_1 \cdot ... \cdot \lambda_n)^{1/n}} \ge n\:?$$

Then, I say that if $$A$$ is singular then $$\det A = 0$$ thus one of the $$\lambda_i = 0$$, therefore the product of all $$\lambda_i = 0$$, thus, could I say that $$\frac{0}{0} \ge n$$ (although it's undefined), therefore, it contradicts the assumption and the formula doesn't hold for a singular matrix?

What I'm struggling with: where I bolded "could I say that" - I'm not sure that it's correct and would love to know your opinion.

The statement is true if $$A$$ is positive definite or $$A=0$$.
Statement $$(2)$$ can be easily seen to be equivalent to statement $$(1)$$, because when $$C>0$$, $$\operatorname{tr}(AC)=\operatorname{tr}(AC^{1/2}C^{1/2})=\operatorname{tr}(C^{1/2}AC^{1/2})$$ and $$\det(C)=1$$ if and only if $$\det(C^{1/2})=1$$. In other words, the $$C$$ in $$(2)$$ is just the square root of the $$C$$ in $$(1)$$.
When $$A$$ is nonzero, singular and positive semidefinite, since $$C^{1/2}AC^{1/2}$$ is congruent to $$A$$, $$\operatorname{tr}(AC)=\operatorname{tr}(C^{1/2}AC^{1/2})$$ is always positive. Therefore both statements $$(1)$$ and $$(2)$$ are false, but they can be corrected by taking infima instead of minima (which do not exist).