Partial Sum to be invertible Let $A_1,\cdots,A_m$ be $n\times n$ matrices, satisfying 
$$m>n, A_1+\cdots+A_m=E_n,$$
where $E_n$ is the $n\times n$ identity matrix.
Show that there exists a subset $P\subset \{1,\cdots,m\}$ with cardinal $\leq n$ such that 
$$\sum_{k\in P}A_k$$
is invertible.
 A: Here we have a solution for $n=2$. I don't know how to do for $n>2$. I hope some of these ideas can help.
Let us prove this result by induction on $m$. 
First for $m=3$.
Suppose $A_1+A_2+A_3=Id_{2\times 2}$.
We may assume that each $A_i$ has a zero eingenvalue, otherwise one of these $A_i$ is invertible. 
Now, if each $A_i$ has eigenvalues $0$ and $1$ then the trace of each $A_i$ is 1 and 
$3=trace(A_1)+trace(A_2)+trace(A_3)=tr(Id)=2$, which is a contradiction.
Thus, there is $A_i$ with eingenvalues different from 1, suppose it is $A_1$, then $A_2+A_3=I-A_1$ and $I-A_1$ is invertible and we are done. 
Now, assume the result is true for $m=k>3$. 
Let $A_1+A_2+\ldots+A_{k+1}=Id_{2\times 2}$. Notice that $A_1+A_2+\ldots+A_{k-1}+B_k=Id$, where $B_k=A_k+A_{k+1}$. 
By induction hypothesis, or $A_i+A_j$ is invertible for some $i,j<k$ or $A_i+B_k$ is invertible for some $i<k$. 
If the first case occurs we are done. If the second occurs then $C=A_i+A_{k-1}+A_{k}$ is invertible. Thus, $Id=C^{-1}A_i+C^{-1}A_{k-1}+C^{-1}A_{k}$. 
By the case $m=3$ we can add two of these three matrices to obtain an invertible matrix and multiplying by $C$, we obtain the result. 
