rank of a linear combination of matrices Let $A_1,..., A_s \in M_n(\mathbb{R})$ be symmetric matrices and suppose they are linearly independent over $\mathbb{R}$. This means that
$$
m = \min_{(c_1, ..., c_s) \in \mathbb{R}^s \backslash \{0\}} rank( \sum_{i=1}^s c_i A_i ) > 0
$$
I am interested in the question how large can $m$ be?
I am not sure where to start looking... if someone could point me to a good reference it's very appreciated! Also any comments are appreciated!
 A: The best bound relating $m$, $n$, and $s$ (i.e., the best possible bound that does not take into account any structure of the $A_j$ matrices) is
$$
s \leq \binom{n - m + 2}{2}.
$$
To see that this bound is tight (i.e., you can achieve $s = \binom{n - m + 2}{2}$), consider the matrices $A_j$ that mostly consist of zeroes, except they either have non-zero entries on their main diagonal, or non-zero entries on a single matching super/sub-diagonal pair. By choosing those diagonals generically, you can control how many zero entries linear combinations of them have, and thus the rank of any linear combination of these matrices (since you can always find a triangular submatrix of appropriate size). Sorry for being light on details -- this construction is given in Section 4.2.1 of our paper https://arxiv.org/abs/2010.02876 (there is likely an earlier reference for this particular construction, but I unfortunately do not know of it)
For example, if $n = 3$ and $m = 2$, you can find $s = \binom{n - m + 2}{2} = 3$ matrices as follows:
$$
A_1 = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1\end{bmatrix}, A_2 = \begin{bmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4\end{bmatrix}, A_3 = \begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0\end{bmatrix}.
$$
Any linear combination of those $3$ symmetric matrices has a $2 \times 2$ triangular submatrix with non-zero entries on the diagonal, and thus has rank at least $2$.
That you can't do any better than this follows from facts about determinantal varieties, but unfortunately I do not know a great reference.
Edit: I apologize - this answer is incorrect. The original poster asked about subspaces of real symmetric matrices, but I answered the question for complex symmetric matrices. The bound $s \leq \binom{n-m+2}{2}$ that I said is not true in general for real matrices (though $s = \binom{n-m+2}{2}$ is still achievable for real matrices, via the same construction that I described).
To see that you can sometimes construct larger real subspaces than complex ones, consider the case $m = n = 2$. In the complex case, the largest such subspace has dimension $s = \binom{n-m+2}{2} = 1$, but in the real case there is such a subspace of dimension $2$, such as the span of the two matrices
$$
\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix} \ \ \text{and} \ \ \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}.
$$
The best reference for the real version of this problem that I have been able to find is "Spaces of symmetric matrices containing a nonzero matrix of bounded rank" by Friedland and Loewy.
