dimension of eigenvalue 1 subspace of a symmetric zero-diagonal matrix

Question

I am working over the finite field of two elements $\mathbb{F}_2$. Let $A\in \mathbb{F}_2^{n\times n}$ be a symmetric zero-diagonal matrix. I am trying to understand the eigenvalue 1 subspace of such matrices. In particular, what is the maximal dimension of the eigenvalue 1 subspace of such matrices? It can never be $n$, as that would imply that $A$ is the identity matrix. But how much smaller can it be? Thanks.

Answer 1

$A$ can have a unit eigenvalue of geometric multiplicity $n-1$. Let $\{e_1,e_2,\ldots,e_n\}$ be the standard basis of $\mathbb F_2^{n\times n}$. For $i=1,2,\ldots,n-1$, define $v_i=e_i+e_{i+1}$. These $v_i$s are linearly independent. When $A$ is the hollow matrix whose off-diagonal entries are all equal to $1$, we have $Av_i=v_i$ for each $i$.