1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ You esssentially need to find the column space of the matrix $M - 4I$ i.e. what is the dimension of the vector space which is spanned by the columns of the matrix $M - 4I$. Look at the columns of $M-4I$, and try to see why the dimension is $3$. You have two rows of zeros, which is very good, but you can go one dimension better. The below answers will serve as hints. Now, by the rank nullity theorem, the kernel dimension is the space dimension minus the rank, which is $6 - 3 = 3$. $\endgroup$ – Teresa Lisbon Aug 29 '18 at 6:47
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Will u please explain how $3$ leading ones give the existence of $3$ dependent solutions $\endgroup$ – user572425 Aug 29 '18 at 6:28
  • $\begingroup$ The equations are $$x_1+x_2+x_3-x_6=0,\ \ \ x_4+x_6=0,\ \ \ x_5+x_6=0. $$Taking $x_2=r, x_3=s, x_6=t $ as parameters you get to express the solutions as $$x_1=-r-s+t,\ \ \ x_4=x_5=-t.$$ Having three parameters means that the solution space has dimension three. If you have trouble with linear systems, you won't be able to do much with eigenvalues. $\endgroup$ – Martin Argerami Aug 29 '18 at 6:39
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Sir will you please go through my edits I did in the question,it will help me a lot $\endgroup$ – user572425 Aug 29 '18 at 6:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy