Is maximizing $\det A$ equivalent to minimizing $\mbox{tr} A^2$?

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 $\lambda_i\le\alpha_i<k/n$ 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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I don't understand what you mean when you say that $\lambda_i = \frac{k}{n}$ is impossible. You haven't added any other constraints to the problem. –  Qiaochu Yuan May 31 '11 at 9:59
There are some constraints that make $\lambda_i=k/n$ impossible. For example, some $\lambda_i\le\alpha_i<k/n$. In addition, are the constraints important? I thought their equivalence is not related to the constraints. For example, when $n=2$, no matter what kind of constraints on $\lambda_i$, as aforementioned $\det A$ and $\mbox{tr} A$ reach optima simultaneously. –  Shiyu May 31 '11 at 10:08
Yes, but you haven't specified any such constraints. Do you have some in mind, or are you asking a question about what happens when you add an arbitrary additional constraint? I have a hard time believing the answer doesn't depend on the constraint in general. –  Qiaochu Yuan May 31 '11 at 10:09
I think we can consider the following constraints: for some $\lambda_i$, they are upper bounded, i.e., $\lambda_i\le\alpha_i<k/n$. For some $\lambda_i$, they are lower bounded, i.e., $\lambda_i\ge\beta_i>k/n$. But $\sum\lambda_i=k$ is always satisfied. –  Shiyu May 31 '11 at 10:12
please add all relevant information to the text of the question itself. –  Mariano Suárez-Alvarez May 31 '11 at 12:34

Thank you for the answer. Yes, intuitively $\det A$ and $\mbox A^2$ reaches optima simultaneously when all eigenvalues approach to the mean $k/n$ as close as possible. But can we prove it rigorously? –  Shiyu May 31 '11 at 14:32
@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. –  Gerry Myerson Jun 1 '11 at 0:39