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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 $\lambda_i\le\alpha_i<k/n$ 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

1 Answer 1

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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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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

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