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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 $\lambda_2(A_1+A_2) = 1$.

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    $\begingroup$ Do you have tried the very special case that additionally the matrices are diagonal matrices (of course these commute). Which estimation is of interest to you for this special case? $\endgroup$ Mar 10, 2021 at 10:07
  • $\begingroup$ I added an example at the end of the question @DieterKadelka. I am in general interested in the case in which $\text{Range}(\sum_{n=1}^N A_n) = \mathbb{R}^d$. I think in that case the bound (or a similar one) should hold. $\endgroup$
    – Apprentice
    Mar 10, 2021 at 10:48
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    $\begingroup$ If I may ask, why did you delete the question on Math SE? $\endgroup$ Mar 10, 2021 at 14:48
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    $\begingroup$ Your conjectured inequality is not homogeneous, thus it can not hold. In general it is very much possible that the minimal non-zero eigenvalues of all matrices are large but the minimal eigenvalue of their sum is small. Consider the projections onto close subspaces (for example, two projections to close lines on the plane). $\endgroup$ Mar 11, 2021 at 6:28
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    $\begingroup$ I see what you would like to have. My point is that it is not possible without further assumptions because of the following example: take two lines $\ell_1,\ell_2$ on the plane with a small angle $\delta$ between them. Let $n=2$ and $A_1,A_2$ be projections to $\ell_1,\ell_2$. So $r_1=r_2=1$ and $\lambda_1(A_1)=\lambda_1(A_2)=1$. Then $\lambda_2(A_1+A_2)$ is positive but small since $A_1+A_2$ is close to $2A_1$. $\endgroup$ Mar 11, 2021 at 7:47

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