I have spent an hour and a half on this problem, just cannot solve for eigenvalues of this $4\times 4$ matrix $A$. Any help will be appreciated.
\begin{bmatrix} 7 & 1 & 2& 2\\ 1&4&-1&-1\\ -2&1&5&-1\\ 1 &1&2 &8 \end{bmatrix}
I get $\det(A-\lambda I)$ is equivalent to the determinant of
$$\begin{bmatrix} 7-\lambda & 1 & 2& 2\\ 0&9-2\lambda&3-\lambda&-3\\ -2&1&5-\lambda&-1\\ 0 &3&9-\lambda &15-2\lambda \end{bmatrix}.$$
And then, everything is super messy. I even did not get an integer at the end.
In general, you would expect to need some luck to find the eigenvalues of a $4\times 4$-matrix by hand. Here, there are some similar elements in the upper right hand corner of $A$. By subtracting the third column from the fourth column, we get that $$\det(A-\lambda I)=\begin{vmatrix} 7-\lambda & 1 & 2 & 2 \\ 1 & 4-\lambda & -1 & -1 \\ -2 & 1 & 5-\lambda & -1 \\ 1 & 1 & 2 & 8-\lambda\end{vmatrix}=\begin{vmatrix} 7-\lambda & 1 & 2 & 0 \\ 1 & 4-\lambda & -1 & 0 \\ -2 & 1 & 5-\lambda & \lambda-6 \\ 1 & 1 & 2 & 6-\lambda\end{vmatrix}$$ Add the fourth row to the third row, and expand by the last column: $$\det(A-\lambda I)=(6-\lambda)\begin{vmatrix} 7-\lambda & 1 & 2 \\ 1 & 4-\lambda & -1 \\ -1 & 2 & 7-\lambda \end{vmatrix}$$ Add the second row to the third row: $$\det(A-\lambda I)=(6-\lambda)\begin{vmatrix} 7-\lambda & 1 & 2 \\ 1 & 4-\lambda & -1 \\ 0 & 6-\lambda & 6-\lambda \end{vmatrix}=(6-\lambda)^2\begin{vmatrix} 7-\lambda & 1 & 2 \\ 1 & 4-\lambda & -1 \\ 0 & 1 & 1\end{vmatrix}$$ Subtract the third row twice from the first row, and add the third row to the second row, before you expand by the last column: $$\det(A-\lambda I)=(6-\lambda)^2\begin{vmatrix}7-\lambda & -1 \\ 1 & 5-\lambda\end{vmatrix}=(6-\lambda)^4$$