Show that $0$ has multiplicity $3$ in $M-4I$?

Question

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

Answer 1

Since $M$ is symmetric, the geometric and algebraic multiplicity of its eigenvalues coincide.

If you finish your row reduction, you get $$ \begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix} \xrightarrow{\ \ \ \ \ \ \ \ \ \ } \begin{bmatrix} 1&1&1&0&0&-1\\0&0&0&1&0&1\\0&0&0&0&1&1\\0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\end{bmatrix} $$ As there are three leading ones, this shows that the set of solutions of $(M-4I)x=0$ has three dependent solutions, and three independent. Thus $\dim \ker(M-4I)=3$. This means that the eigenvalue 4 has geometric and algebraic multiplicity 3.

Answer 2

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.