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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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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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