How to show the matrix has Rank $\le 5$ 
I want to show that the following matrix has Rank $\le 5$.

The matrix is 
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatrix in the matrix which has rank $ =4$ given by
$[2,1,1,1],[1,2,1,1],[1,1,2,1],[1,1,1,2]$.
I need to show the given matrix has at least 3 zero rows in order to show that Rank $\le 5$..
But I dont know how to show it.
Can someone help.

We will show that the rank of the given matrix is exactly $5$.
The matrix can be written as a block matrix such that (see LU-decomposition),
$$M:=\begin{bmatrix} A & B \\B &A \end{bmatrix}=\begin{bmatrix} I & 0 \\BA^{-1}&I \end{bmatrix}\cdot \begin{bmatrix} A&0\\0&A-BA^{-1}B \end{bmatrix}\cdot\begin{bmatrix} I&A^{-1}B \\ 0 & I\end{bmatrix}.$$
with 
$$A = \begin{bmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 &1 \\ 1 &1&2&1\\1&1&1&2\end{bmatrix}, \quad B = \begin{bmatrix} 0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0 \end{bmatrix}.$$
Now 
$$A^{-1}=\frac{1}{5}\begin{bmatrix} 4 & -1 & -1 & -1 \\ -1 & 4 & -1 &-1 \\ -1 &-1&4&-1\\-1&-1&-1&4\end{bmatrix}\quad
\text{and}\quad
A-BA^{-1}B=\frac{4}{5}\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 &1 \\ 1 &1&1&1\\1&1&1&1\end{bmatrix}.$$
Therefore we may conclude that
$$\text{rank}(M)=\text{rank}\left(\begin{bmatrix} A&0\\0&A-BA^{-1}B \end{bmatrix}\right)=\text{rank}(A)+\text{rank}(A-BA^{-1}B)
=4+1=5.$$
