Q:

Finding eigenvalues for a $4\times4$ matrix

I am trying to find the eigenvalues for this $4\times4$ matrix $A$, where $A$ is
\begin{bmatrix}
        2 & 1 & 1 & 1 \\
        1 & 2 & 1 & 1 \\
        0 & 0 & 2 & 1 \\
          0 & 0 & 1 & 2 \\
        \end{bmatrix}
I was wondering if there is an easy way to do it or do I have to find the determinant the long way?

A:

Result 1: Eigenvalues of the matrix in a triangular form is the entries of the principal diagonal. For example, if we have the following matrix (Upper triangular)
\begin{bmatrix}         2 & 1 & -1 & -1 \\         0 & 2 & 1 & 1 \\         0 & 0 & 2 & 1 \\           0 & 0 & 0 & 2 \\         \end{bmatrix}
Then eigenvalues of this matrix are given by the entries of principal diagonal which are $2, 2, ,2 ,2$. That is $2$ with algebraic multiplicity $4$.
Generalized form of above result can be summarized from the below example which may help you to solve your problem. 
Consider the following matrix
\begin{bmatrix}         A_1 & B  \\  \mathcal{O} & A_2 \\         \end{bmatrix}
where $A_1$ and $A_2$ are square matrices and $\mathcal{O}$ is the matrix with zero entries.
Eigenvalue of above matrix can be obtained by computing the eigenvalues of matrices $A_1$ and $A_2$. Since Characteristic polynomial of this matrix is the product of Characteristic polynomial of $A_1$ and $A_2$. 
In your problem you can take $A_1 $ to be
\begin{bmatrix}        2 & 1  \\  1 & 2 \\         \end{bmatrix}
and 
$A_2 $ to be
\begin{bmatrix}        2 & 1  \\  1 & 2 \\         \end{bmatrix}

