How to determine the eigenvalue?

Question

I have this question:

Let $\{u_1, u_2, u_3, u_4\}$ be a basis for $\mathbb{R}^4$ and $B$ a $4 \times 4$ matrix such that: $$ Bu_1=u_2,~~ Bu_2=u_1,~~ Bu_3=u_4,~~ Bu_4=u_3.~~ $$ Find all eigenvalues of $B$ and determine whether $B$ is diagonalizable.
Justify your answers.

^{[image of problem statement]}

I am having troubles identifying all the eigenvalues of the matrix $B$ described in this question.

I believe that one of the eigenvalues is $-1$, which can be obtained from $$\begin{eqnarray} Bu_1 - Bu_2 &=& u_2 - u_1 \\ \Rightarrow~~ B(u_1 -u_2) &=& u_2 - u_1 \end{eqnarray}$$

The above can be true only if $\lambda = -1$.  This is the same for the $Bu_3 - Bu_4 = u_4 - u_3$

I can't seem to determine the rest or prove that this is the only eigenvalue.

Answer 1

Relative to the given basis, the matrix is $M=\begin {pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix} $.

Now find the roots of the characteristic polynomial: $\rm {det}(M-xI) $.

Get $x^2(x^2-1)+(1-x^2)=x^4-2x^2+1=(x+1)^2 (x-1)^2$.

Thus the eigenvalues are $\pm1$.