Let $\mathcal A := \{ A_1, A_2, \dots, A_n \} \subset \Bbb R^{d \times d}$ be a set of symmetric and positive semidefinite matrices.

For a matrix $A_k \in \mathcal A$, denote its (real) eigenvalues by $\lambda_i (A_k)$, ordered such that

$$\lambda_i (A_k) \geq \lambda_{i+1}(A_k)$$

for $i \in \{1, 2,\dots, d-1 \}$. Hence, $\lambda_{1}(A_k)$ and $\lambda_{d}(A_k)$ denote the largest and smallest eigenvalues of $A_k$, respectively. For each matrix $A_k \in \mathcal A$, let $r_k := \mbox{rank}(A_k)$. Thus, the smallest non-zero eigenvalue of $A_k$ is $\lambda_{r_k}(A_k)$. We also have that

$$\mbox{Range} \left( \sum_{k=1}^n A_k \right) = \mathbb{R}^d$$

or, equivalently,

$$\mbox{rank} \left( \sum_{k=1}^n A_k \right) = d$$

I am trying to find an **upper bound** on

$$\sum_{k=1}^n \lambda_{r_k} (A_k)$$

in terms of the smallest eigenvalue of the sum of matrices, $\lambda_d \left( \displaystyle\sum_{k=1}^n A_k \right)$.

### Observations

I would like to obtain something like the following inequality for the smallest eigenvalue

$$\sum_{k = 1}^n \lambda_{d} (A_k) \leq \lambda_{d} \left( \sum_{k = 1}^n A_k \right),$$

generalized to the smallest non-zero eigenvalues instead of the minimum one for each of the matrices. I think that the answer should be something like

$$\sum_{k=1}^n \lambda_{r_k} (A_k) \leq \lambda_d {}\left(\sum_{k=1}^n A_k \right)\cdot \max_{k\in[n]}\lambda_{r_k}(A_k),$$

but I am not sure about it and I don't know how to prove it. Any help would be really appreciated.

### Example

As a simple example think of the case in which

$$ A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \qquad A_2 = \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}.$$

In this case my bound holds since $\lambda_1(A_1) = 1$, $\lambda_1(A_2) = 1$, and $\lambda_2(A_1+A_2) = 1$.

6more comments