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Let $X$ be an $n\times n$ symmetric matrix. Suppose $\lambda_1(X)\geq \lambda_2(X) \geq \cdots \geq \lambda_n(X)$ are eigenvalues of $X$. Let $r$ be any integer with $1\leq r\leq n$. It is well-known that $\sum_{i=1}^r \lambda_i(X)$ is convex. Now, my question is: Is the following function convex? $$\sum_{i=1}^r \max(0,\lambda_i(X))$$

Thanks!

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The answer is yes, because your function (let me call it $f$) is the maximum of convex functions. As such, it is convex. The formula : $$f(X)=\max\left(0,\max_{1\le r\le n}\sum_{j=1}^r\lambda_j(X)\right).$$

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  • $\begingroup$ Yes, your idea is correct. But the reason should be revised to be $$f(X) = \max\{0,\sum_{i=1}^1 \lambda_i(X), \cdots, \sum_{i=1}^r \lambda_i(X)\}.$$ $\endgroup$
    – user11870
    Commented May 30, 2012 at 12:42
  • $\begingroup$ @wmmiao. I agree. $\endgroup$
    – Denis Serre
    Commented May 30, 2012 at 12:54
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The matrix is assumed to be real, correct?

The answer seems to be trivially yes. Indeed, assume that $\lambda_{k} \geq 0$ and $\lambda_{k+1}<0$ and $r \geq k+1$. Then $\sum_{i=1}^{r}{\max(0,\lambda_{i})}=\sum_{i=1}^{k}{\lambda_{i}}$.

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    $\begingroup$ @Felix. Your argument does not work because your $k$ depends upon $X$. $\endgroup$
    – Denis Serre
    Commented May 30, 2012 at 12:21
  • $\begingroup$ Yes, assume to be real. But what we want to prove is for any 0≤α≤1, $$\alpha \sum_{i=1}^r \max(0, \lambda_i(X))+(1-\alpha)\sum_{i=1}^r \max(0,\lambda_i(Y)) \leq \sum_{i=1}^r \max(0,\lambda_i(\alpha X+(1-\alpha)Y)). $$It is not trivial. $\endgroup$
    – user11870
    Commented May 30, 2012 at 12:24
  • $\begingroup$ Oh dear. Well, I had a feeling something was amiss. :) $\endgroup$
    – Felix Goldberg
    Commented May 30, 2012 at 13:41

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