Quick method for finding eigenvalues and eigenvectors in a symmetric $5 \times 5$ matrix?

The matrix $B$:

$B = \pmatrix{ 0 & 0 & 0 & 0 & 0 \cr 0 & 8 & 0 & -8 & 0 \cr 0 & 0 & 8 & 0 & -8 \cr 0 & -8 & 0 & 8 & 0 \cr 0 & 0 & -8 & 0 & 8 \cr }$

Which has nonzero eigenvalues $\lambda_1=16$ and $\lambda_2=16$ and corresponding eigenvectors:

v$_1 = \pmatrix{ 0\cr \frac{1}{2} \sqrt2 \cr 0 \cr -\frac{1}{2} \sqrt2 \cr 0\cr }$ and v$_2 = \pmatrix{ 0\cr 0\cr \frac{1}{2} \sqrt2 \cr 0 \cr -\frac{1}{2} \sqrt2 \cr }$

What is the method for obtaining these eigenvalues and corresponding eigenvectors?

It's a large matrix and I'm hoping there's some kind of easy trick to it. From what I can remember of eigen decomposition, normally I'd do:

$Ax = \lambda x \implies|A-\lambda I|x = 0$

$\implies \det \pmatrix{ 0-\lambda & 0 & 0 & 0 & 0 \cr 0 & 8-\lambda & 0 & -8 & 0 \cr 0 & 0 & 8-\lambda & 0 & -8 \cr 0 & -8 & 0 & 8-\lambda & 0 \cr 0 & 0 & -8 & 0 & 8-\lambda \cr }$ $\pmatrix{ x_1 \cr x_2 \cr x_3 \cr x_4 \cr x_5 \cr }$ = $\pmatrix{ 0 \cr 0 \cr 0 \cr 0 \cr 0 \cr }$

So the determinant is

$\implies -\lambda \det \pmatrix{ 8-\lambda & 0 & -8 & 0 \cr 0 & 8-\lambda & 0 & -8 \cr -8 & 0 & 8-\lambda & 0 \cr 0 & -8 & 0 & 8-\lambda \cr }$

$\implies -\lambda * [ (8- \lambda)\det \pmatrix{ 8-\lambda & 0 & -8 \cr 0 & 8-\lambda & 0 \cr -8 & 0 & 8-\lambda \cr }-8 \det \pmatrix{ 0 & 8-\lambda & -8 \cr -8 & 0 & 0 \cr 0 & -8 & 8-\lambda \cr }]$

etc.

There's got to be an easier way?

Hint: if $A$ and $B$ are square of the same order, $$\det\begin{pmatrix} A & B \\ B & A\end{pmatrix}=\det(A-B)\det(A+B)$$ So setting $$A=\begin{pmatrix}8-\lambda & 0 \\ 0 & 8-\lambda\end{pmatrix}$$

and

$$B=\begin{pmatrix}-8 & 0\\0&-8\end{pmatrix}$$

shows that the determinant of your $4\times 4$ matrix is $(16-\lambda)^2\lambda^2$, and therefore the determinant of your original matrix is $$\boxed{-\lambda^3(16-\lambda)^2}$$

Here's a quick way to do the problem.

We can write this as a block matrix in the following form: $$ B = \pmatrix{ 0&0&0\\ 0&8I&-8I\\ 0&-8I&8I } = 8\pmatrix{ 0&0&0\\ 0&I&-I\\ 0&-I&I } $$ $B$ is a block diagonal matrix; one block is $0$ (corresponding to an eigenvalue of $0$ with corresponding eigenvector $[1,0,0,0,0]^T$), and the other block is $8 \pmatrix{I&-I\\-I&I}$. It suffices to find the eigenvalues of $A = \pmatrix{I&-I\\-I&I}$ and multiply the result by $8$.

In fact, if you are aware of tensor products, you may note that $A = \pmatrix{1&-1\\-1&1} \otimes I$ (in fact, I will be indirectly exploiting this).

Note that the matrix $\pmatrix{1&-1\\-1&1}$ has eigenvalues $0$,$2$ and corresponding eigenvectors $[1,1]^T$ and $[1,-1]^T$.

Now, suppose that $v$ is an eigenvector of $I$ (in other words, let $v$ be any non-zero $2 \times 2$ vector). We find that $$ A \pmatrix{v\\v} = 0 $$ and $$ A \pmatrix{v\\-v} = \pmatrix{2v\\-2v} $$ We may thus establish that $A$ has eigenvectors $0,0,2,2$ As corresponding eigenvectors we may take, for example, $$ v_{0,1} = \pmatrix{1\\0\\1\\0}; v_{0,2} = \pmatrix{0\\1\\0\\1}; v_{2,1} = \pmatrix{1\\0\\-1\\0}; v_{2,2} = \pmatrix{0\\1\\0\\-1} $$

Multiplying the eigenvectors of $A$ by $8$, we note that the matrix $B$ has eigenvectors $0,0,0,16,16$. We may take the eigenvectors $$ \pmatrix{1\\0\\0\\0\\0}; \pmatrix{0\\1\\0\\1\\0}; \pmatrix{0\\0\\1\\0\\1}; \pmatrix{0\\1\\0\\-1\\0}; \pmatrix{0\\0\\1\\0\\-1} $$

For simple matrices, you can often find the eigenvalues and eigenvectors by observation. Once you guess an eigenvalue, its easy to find the eigenvector by solving the linear system $(A-\lambda I)x=0$. Here, you already know that the matrix is rank deficient, since one column is zero. (The corresponding eigenvector is $[1~0~0~0~0]^T$.) So $\lambda=0$ is one eigenvalue.

Also, if you can guess the eigenvector (up to a constant) by looking at the structure of $(A-\lambda I)$, you can find the corresponding eigenvalue. Here, one possibility is $k[0~1~1~1~1~1]^T$, since adding up the columns of $A$ gives you a vector of zeros. The corresponding eigenvalue is zero. So 0 has multiplicity 2.

Another way to get zeros is by subtracting the 4th column from the 2nd column (with $\lambda=16$). The eigenvector is $k[0~1~0~-1~0~0]^T$. $k$ can be found by normalization as $1/\sqrt{2}$. We similarly get your second eigenvector by considering the 3rd and 5th columns.

Now that we know 4 of the 5 eigenvalues and eigenvectors, the last is easy to find using the eigenvalue decomposition. i.e. $\lambda_5 v_5 v_5^T = A-\sum_{i=1}^4 \lambda_i v_i v_i^T$