How to find eigenvalues of this given matrix

Question

Consider the matrix $M$=\begin{bmatrix}9&1&1&1&1&1&1&1&1&1\\ 1&9&1&1&1&1&1&1&1&1\\ 1&1&9&1&1&1&1&1&1&1\\ 1&1&1&9&1&1&1&1&1&1\\ 1&1&1&1&9&1&1&1&1&1\\ 1&1&1&1&1&9&1&1&1&1\\ 1&1&1&1&1&1&6&0&0&0\\ 1&1&1&1&1&1&0&6&0&0\\ 1&1&1&1&1&1&0&0&6&0\\ 1&1&1&1&1&1&0&0&0&6\end{bmatrix}

Find the eigen values of $M$.

Can someone please check where is my calculation going wrong ?

I need to find $\det(xI-M)$.

I took $A=\begin{bmatrix}9&1&1&1&1&1\\1&9&1&1&1&1\\\dots \dots \\1&1&1&1&1&9\end{bmatrix}$

and $D=6I$

Thus $M=\begin{bmatrix} A&B\\B^T&D\end{bmatrix}$

Thus $\det (xI-M)=\det (xI-D)\det ((xI-A)-B(xI-D)^{-1} B^T)$

By following this I am getting the roots as $6$ with multiplicity $3$ and $8$ with multiplicity $5$ and other two as roots of $x^2-20x+72$ but after doing it in software I am getting the eigenvalues as $16.32455532 $ and $ 3.67544468$.

Where am I wrong?

Can someone please do it using the method I have used to find the correct ones.

NOTE:

$\det((xI-A)-B(xI-D)^{-1}B^T)=\det$ $\begin{bmatrix} (x-9)(x-6)-2& -(x-4)&\ldots& -(x-4)\\-(x-4)& (x-9)(x-6)-2& \ldots &-(x-4)\\\ldots&\ldots \\ \ldots&\ldots \\ -(x-4)&-(x-4)& \ldots & \ldots(x-9)(x-6)-2\\ \end{bmatrix}$

I get on adding all the rows to the first row :

$(x-9)(x-6)-2+5(4-x)=x^2-15x+54-2+20-5x=x^2-20x+72$

Thus all the elements of the first row becomes $x^2-20x+72$.

Thus $x^2-20x+72$ is a factor of the determinant and hence its roots are the eigenvalues of $M$.

But as shown in the below answer by [email protected]$ the factor must be $x^2-20x+60$ but where am I going wrong?

Kindly help me. I am in serious trouble.

Answer 1

It will be: $$\small{\begin{vmatrix}M-xI\end{vmatrix}= (6-x)^4 \cdot\det\left(\begin{pmatrix}9-x&1&\cdots&1\\ \vdots&\ddots&&1\\ 1&1&\cdots&9-x\end{pmatrix}- \frac1{6-x}B\cdot B^T\right)}=\\ =(6-x)^4\begin{vmatrix}9-x-\frac{4}{6-x}&1-\frac4{6-x}&\cdots&1-\frac4{6-x}\\ \vdots&\ddots&&1-\frac4{6-x}\\ 1-\frac4{6-x}&\cdots&&9-x-\frac4{6-x}\end{vmatrix}_{6\times 6}=\\ =\frac1{(6-x)^2}\begin{vmatrix}x^2-15x+50&2-x&\cdots&2-x\\ \vdots&\ddots&&2-x\\ 2-x&\cdots&&x^2-15x+50\end{vmatrix}_{6\times 6}\stackrel{\text{sum rows}}=\\ =\frac{x^2-20x+60}{(6-x)^2}\begin{vmatrix}1&1&\cdots&1\\ \vdots&\ddots&&2-x\\ 2-x&\cdots&&x^2-15x+50\end{vmatrix}_{6\times 6}\stackrel{R_1(x-2)+R_2,...,R_6}=\\ =\frac{x^2-20x+60}{(6-x)^2}\begin{vmatrix}x^2-14x+48&0&\cdots&0\\ \vdots&\ddots&&0\\ 0&\cdots&&x^2-14x+48\end{vmatrix}_{5\times 5}=\\ =\frac{x^2-20x+60}{(6-x)^2}\cdot (x-6)^5\cdot (x-8)^5=\\ =(x^2-20x+60)(x-6)^3(x-8)^5.$$

Answer 2

The characteristic polynomial of $M$ is indeed $$ \chi_M(t)=(t^2 - 20t + 60)(t - 6)^3(t - 8)^5. $$ So you are almost right. The quadratic equation has constant term $60$ and not $72$ (check this!), and has solutions $$ 2(5\pm\sqrt{10}). $$