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Question: $A\in\mathbb{R}^{n\times n}$ is a positive definite matrix with constant trace, i.e., $A>0$ and $\mbox{tr} A=k$. Let $\lambda_1\ge\cdots\ge\lambda_n>0$ be the eigenvalues of $A$. Can we say maximizing $\det A=\prod_{i=1}^n \lambda_i$ is equivalent to minimizing $\mbox{tr} A^2=\sum_{i=1}^n \lambda_i^2$?

Remark: It seems a trivial problem. But maybe it is not so easy to solve. Since $\mbox{tr} A=\sum_{i=1}^n\lambda_i=k$, it is obvious that $\det A=\prod_{i=1}^n \lambda_i\le \left(\frac{\sum_{i=1}^n\lambda_i}{n}\right)^n=\left(\frac{k}{n}\right)^n$ and the maximum is achieved iff $\lambda_i=k/n, \forall i$. On the other hand, we have $\mbox{tr} A^2=\sum_{i=1}^n \lambda_i^2\ge\frac{(\sum_{i=1}^n\lambda_i)^2}{n}=\frac{k^2}{n}$ and the minimum is achieved iff $\lambda_i=k/n, \forall i$.

It seems $\det A$ and $\mbox{A}$ are equivalent because they reach optima simultaneously when $\lambda_i=k/n$. But in some cases $\lambda_i=k/n, \forall i$ is impossible, do they still reach optima simultaneously? For example, consider the constraints: for some $\lambda_i\le\alpha_i and the other $\lambda_i\ge\beta_i>k/n$.

My attempt: For a special case $n=2$, we have $\det A=((\mbox{tr}A)^2-\mbox{tr}(A^2))/2$. Since $\mbox{tr}A$ is constant, even if $\lambda_i=k/n, \forall i$ is not satisfied, we still can say $\mbox{tr}A^2$ and $\det A$ reach optimum at the same time. But for $n=3$ we have $\det A=((\mbox{tr}A)^3-3\mbox{tr}A\mbox{tr}A^2+\mbox{tr}A^3)/6$. Can we say $\det A$ and $\mbox{tr}A^2$ reach optima simultaneously? Thank you.

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    please add all relevant information to the text of the question itself.2011-05-31

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It seems to me that if any two of the eigenvalues are unequal, and if it is possible to move them closer without violating any constraint, then moving them closer will increase the determinant while decreasing the sum of squares. So if the determinant is maximized, then you can't move any two eigenvalues closer together, which implies you can't decrease the sum of squares, so that sum is minimized when the product is maximized.

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    @Shiyu, if $a\gt a-h\gt b+h\gt b\gt0$, then $(a-h)(b+h)\gt ab$, and $(a-h)^2+(b+h)^2\lt a^2+b^2$. Seems pretty rigorous to me.2011-06-01