Determining rank of a matrix over $\mathbb{F}_2$. We are given a $(2n,2,2n) \times (2n,2,2n)$ matrix $A$ over $\mathbb{F}_2$ as follows: 
$$A= \begin{bmatrix} I & 0 & 1 \\ 0 & I & 1 \\ 1 & 0 & I \end{bmatrix}$$
Where $I$ is the identity matrix, $0$ is the zero matrix and $1$ is the matrix containing all $1$s. 
The authors state (without proof) that $A$ has full rank over $\mathbb{F}_2$. Furthermore, the matrix obtained by deleting the first and last row, and first and last column of $A$ has rank $4n-1$. 
I have been looking for a proof all day. I am thinking in terms of row operations and reducing the matrix to a simple form but I can't get a proof. How do I prove this? Am I missing some basic point?
 A: I write $E$ for $1$. This is less confusing. Consider a matrix of the form
$$
B=\pmatrix{I_m&0&E_{m\times m}\\ 0&I_2&E_{2\times m}\\ E_{m\times m}&0&I_{m}}.
$$
Subtract the third block row by $E_{m\times m}$ times the first block row, we get
$$
\pmatrix{I_m&0&E_{m\times m}\\ 0&I_2&E_{2\times m}\\ 0&0&I_m-E_{m\times m}^2}
=\pmatrix{I_m&0&E_{m\times m}\\ 0&I_2&E_{2\times m}\\ 0&0&I_m-mE_{m\times m}}=:C.
$$
Therefore
$$
\operatorname{rank}(B)=\operatorname{rank}(C)=m+2+\operatorname{rank}(I_m-mE_{m\times m}).
$$
Note that


*

*$I_m-mE_{m\times m}=I_m$ is nonsingular when $m$ is even;

*$I_m-mE_{m\times m}$ has rank $m-1$ when $m$ is odd, because by the identity $\det(I+uv^T)=1+v^Tu$, we see that $I_m-mE_{m\times m}$ has determinant $1-m^2=0$ and its leading principal $(m-1)\times(m-1)$ submatrix $I_{m-1}-mE_{(m-1)\times(m-1)}$ has determinant $1-m(m-1)=1$.


Now, return to your question. When $m=2n$, we have $B=A$. Since $I_m-mE_{m\times m}$ is nonsingular, $C$ and in turn $A$ are nonsingular.
In the second part, by deleting the first row and column and the last row and column of $A$, we get a matrix in the form of $B$ with $m=2n-1$. Since $I_m-mE_{m\times m}$ has rank $m-1$, its rank deficiency is $1$. Therefore $\operatorname{rank}(B)=4n-1$.
