Determine that a vector is in the column space of a matrix I need hep with the following problem.
We will work only in $\mathbb{F}_2$. Let say $A$ is a symmetric matrix with its main diagonal consisting of only $1$s. I need to prove that $1$ (the vector with all ones) is in the column space (or row-space as is symmetric) of $A$.
Any help is appreciated, thanks.
 A: We may prove a more general statement: if $A\in M_n(\mathbb F_2)$ is symmetric, then its diagonal $\mathbf a^\top=(a_{11},a_{22},\ldots,a_{nn})$ lies inside the row space of $A$.
When $\mathbf a^\top$ is zero, clearly it lies inside the row space of $A$. Suppose $\mathbf a^\top$ is nonzero. Then $A$ represents a non-alternating symmetric bilinear form (i.e. $x^\top Ax$ is not always zero). Therefore it can be diagonalised by congruence (cf. Irving Kaplansky, Linear Algebra and Geometry: a Second Course, p.23, theorem 20), i.e. $A=P(I_m\oplus0)P^\top$ for some invertible matrix $P$, where $m$ denotes the rank of $A$.
Let $X$ be the $m\times n$ submatrix taken from the first $m$ rows of $P^\top$. Thus $A=X^\top X$ where $X$ has full row rank. Hence there exists some vector $u$ such that $Xu=\mathbf1_m$, the vector of ones. However, as the underlying field is $\mathbb F_2$, the $j$-th diagonal entry $a_{jj}=x_{\ast j}^\top x_{\ast j}$ of $A$ is equal to the parity of $x_{\ast j}$ and in turn also equal to $\mathbf 1_m^\top x_{\ast j}$. It follows that $\mathbf a^\top=\mathbf 1_m^\top X=(Xu)^\top X=u^\top X^\top X=u^\top A$.
