<s>[INST] Is $0$ an eigen value of $A$ 
Consider the following matrix:
$A=\begin{bmatrix} 9&1&1&1&1&1&2&2\\1& 9&1&1&1&1&2&2\\1&1&9&1&1&1&2&2\\1&1&1&9&1&1&2&2\\1&1&1&1&9&1&2&2\\1&1&1&1&1&9&2&2\\2&2&2&2&2&2&13&1\\2&2&2&2&2&2&1&13\end{bmatrix}$.

I got the eigen values of the matrix
\begin{bmatrix} 9&1&1&1&1&1\\1& 9&1&1&1&1\\1&1&9&1&1&1\\1&1&1&9&1&1\\1&1&1&1&9&1\\1&1&1&1&1&9\end{bmatrix}
and the eigen values of the matrix
\begin{bmatrix} 13&1\\1&13\end{bmatrix}.
I have the following two questions:

*

*Is $0$ an eigen value of $A$

*Find the eigen values of $A$.

Note that $0$ is not an eigen value of both the submatrices listed above.
Can we conclude that $0$ is not an eigen value of $A$ from above?If yes,How?
I am unable to solve Part(2) of the question.Please help.
 [/INST] Subtracting $8$ and $12$ from the main diagonal, it is easily seen that $8$ is an eigenvalue of multiplicity $5$ and $12$ is an eigenvalue of multiplicity $1$. Hence we are still missing two eigenvalues. The trace tells you that the sum of the remaining two eigenvalues has to be $80-52=28$. If one is $0$, then the other has to be $28$. But $28$ cannot be an eigenvalue as the diagonal of $A-28I$ is dominating. Therefore $0$ is not an eigenvalue of $A$. This answers 1. To get the remaining two eigenvalues, you'd have to compute the characteristic polynomial and divide out $(\lambda-8)^5(\lambda-12)$. This gets you (using computer algebra) $\lambda^2-28\lambda+148$, which can be solved for $\lambda$.
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