I am trying to find eigen values of following matrix.Following matrix is positive semi definite matrix(i.e. All of its eigen values are non negative). I had applied several rows operations to find eigen values but I don't know how go for further calculation.
$A= \begin{bmatrix} 2 & -1 & -1 & 0 & 0 & 0 & 0 & 0 & ... & 0 \\ -1 & 3 & -1 & -1 & 0 & 0 & 0 & 0 & ... & 0 \\ -1 & -1 & 4 & -1 & -1 & 0 & 0 & 0 & ... & 0 \\ 0 & -1 & -1 & 4 & -1 & -1 & 0 & 0 & ... & 0 \\ 0 & 0 & -1 & -1 & 4 & -1 & -1 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & -1 & -1 & 4 & -1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 3 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 2 \\ \end{bmatrix}_{n \times n}$
Characteristic Polynomial is $|\lambda I - A| =0 $
$\therefore \begin{vmatrix} \lambda - 2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & ... & 0 \\ 1 & \lambda - 3 & 1 & 1 & 0 & 0 & 0 & 0 & ... & 0 \\ 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & 0 & ... & 0 \\ 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & ... & 0 \\ 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 2 \\ \end{vmatrix} = 0 $
Taking $R_1+R_i$ for each $i=2,3,4...n-1$
$\therefore \begin{vmatrix} \lambda & \lambda & \lambda & \lambda & \lambda & \lambda & \lambda & \lambda & ... & \lambda \\ 1 & \lambda - 3 & 1 & 1 & 0 & 0 & 0 & 0 & ... & 0 \\ 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & 0 & ... & 0 \\ 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & ... & 0 \\ 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 2 \\ \end{vmatrix} = 0 $
Taking $R_1(\frac{1}{\lambda})$
$\therefore \lambda \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & ... & 1 \\ 1 & \lambda - 3 & 1 & 1 & 0 & 0 & 0 & 0 & ... & 0 \\ 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & 0 & ... & 0 \\ 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & ... & 0 \\ 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 2 \\ \end{vmatrix} = 0 $
Taking $R_2-R_1$ and $R_3-R_1$
$\therefore \lambda \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & ... & 1 \\ 0 & \lambda - 4 & 0 & 0 & -1 & -1 & -1 & -1 & ... & -1 \\ 0 & 0 & \lambda - 5 & 0 & 0 & -1 & -1 & -1 & ... & -1 \\ 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & ... & 0 \\ 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 2 \\ \end{vmatrix} = 0 $
$\therefore \lambda \begin{vmatrix} \lambda - 4 & 0 & 0 & -1 & -1 & -1 & -1 & ... & -1 \\ 0 & \lambda - 5 & 0 & 0 & -1 & -1 & -1 & ... & -1 \\ 1 & 1 & \lambda - 4 & 1 & 1 & 0 & 0 & ... & 0 \\ 0 & 1 & 1 & \lambda - 4 & 1 & 1 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 4 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \lambda - 2 \\ \end{vmatrix}_{n-1 \times n-1} = 0 $
Now what to do for further calculation?
Thanks in advance
I define eigenvalues by being roots of $det(A-\lambda I)=0$. The first size which has the given structure, i.e., $5 \times 5$ (as advised by @Gerry Myerson)
$A=\begin{pmatrix}2 & -1 & -1 & 0 & 0 \\ - 1 & 3 & -1 & -1 & 0 \\ - 1 & -1 & 4 & -1 & -1 \\ 0 & -1 & -1 & 3 & -1 \\0 & 0 & -1 & -1 & 2\end{pmatrix}$
I have submitted the issue to Mathematica. The eigenvalues of $A$ are $0, 3 \pm \sqrt{2}, 3, 5$. Beyond $5 \times 5$, no simple set of eigenvalues can be found.
(the eigenvalue $0$ is present for each dimension, as your computation proves it)
Here are the characteristic polynomials for the size $5 \times 5$ to $10 \times 10$. One can notice certain "regularities" on certain coefficients.
$\chi_5(x)=-105 x + 146 x^2 - 70 x^3 + 14 x^4 - x^5$
$\chi_6(x)=-330 x + 611 x^2 - 404 x^3 + 124 x^4 - 18 x^5 + x^6$
$\chi_7(x)=-1008 x + 2396 x^2 - 2064 x^3 + 870 x^4 - 194 x^5 + 22 x^6 - x^7$
$\chi_8(x)=-3016 x + 8954 x^2 - 9670 x^3 + 5265 x^4 - 1608 x^5 + 280 x^6 - 26 x^7 + x^8$
$\chi_9(x)=-8883 x + 32216 x^2 - 42443 x^3 + 28696 x^4 - 11270 x^5 + 2682 x^6 - 382 x^7 + 30 x^8 - x^9$
$\chi_{10}(x)=-25840 x + 112428 x^2 - 176972 x^3 + 144524 x^4 - 70114 x^5 + 21391 x^6 - 4156 x^7 + 500 x^8 - 34 x^9 + x^{10}$