# Convexity of Sum of $k$-smallest Eigenvalue

If I have a real positive definite matrix $$A\in\mathbb{R}^{n\times n}$$, and denote its eigenvalues as $$\lambda_1\leq \lambda_2 \leq ... \leq \lambda_n$$.

Define the function as $$f(A)=\sum_{i=1}^{k} \lambda_i$$ for a constant $$k. What do we know about the convexity of $$f(A)$$? Is it convex or concave?

Given $$A \in {\bf S}^n$$ (a positive definite matrix) with eigenvalues $$\lambda_1 \leq \lambda_2 \leq \ldots \leq \lambda_n$$, then:

1. $$\displaystyle f_k(A)=\sum_{i=1}^{k} \lambda_i$$ is concave. Why?

$$f_k(A) = \inf \left\{ {\bf tr}(V^T A V) | V \in {\bf R}^{n \times k}, V^T V = I \right\}$$

This follows from the Poincare separation theorem (see e.g. Horn and Johnson's Matrix analysis, 2nd ed., corollaries 4.3.37 and 4.3.39). $$f_k$$ is the pointwise infimum of a family of linear functions $${\bf tr}(V^T A V)$$, hence it is concave (Boyd and Vandenberghe, section 3.2.3).

2. $$\displaystyle g_k(A)=\sum_{i=n-k+1}^{n} \lambda_i$$ is convex. Again, we can show that

$$g_k(A)=\sum_{i=n-k+1}^{n} \lambda_i(A) = \sup \left\{ {\bf tr}(V^T A V) | V \in {\bf R}^{n \times k}, V^T V = I \right\}$$

$$g_k$$ is the pointwise supremum of a family of linear functions $${\bf tr}(V^T A V)$$, hence it is convex (Boyd and Vandenberghe, section 3.2.3).

cvxpy treats that function as being concave (link).