linear algebra - Finding eigenvalues for a $4times4$ 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?

Answer

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}