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).

  • $\begingroup$ Thanks, is my approach correct? $\endgroup$ – Ilan Aizelman WS Apr 22 at 19:30

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