Show that $0$ has multiplicity $3$ in $M-4I$? Question from an exam:
Consider the matrix $M=$
\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3\end{bmatrix}

Show that $4$ is an eigen value of the above matrix with multiplicity $3$.

I considered $M-4I$ from which I got 
\begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
By elementary row operations:
$$M-4I=$$
\begin{bmatrix} 1&1&1&1&1&1\\0&0&0&0&0&0\\0&0&0&0&0&0\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
So $0$ is an eigen value with multiplicity at least $2$ since $M-2I$ has two zero rows.

How to show that $0$ has multiplicity $3$ in $M-4I$?

If one can show how to proceed after this,I will be really grateful.
Please help 

If it only had multiplicity $2$, $M-4I$ would have rank $4$, i.e. four of its rows would be linearly independent.  Since the top three rows are the same, that would have to be the last four rows.  But $(row\ 3) - (row\ 4) - (row\ 5) + (row\ 6) = 0$, so they are linearly dependent.
