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I am given two matrices $A$ and $B$. I know that of $det(A)=1$, $det(B)=0$, $rank(B)=3$.

$A = \begin{bmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{bmatrix}$

$B = \begin{bmatrix} 2 & -1 & 0 & -1\\ 0 & 1 & 0 & -1\\ 0 & -1 & 2 & -1\\ 0 & -1 & 0 & 1 \end{bmatrix}$

I have to find the Eigenvalues and Eigenvectors of both Matrices. The thing is, I know how to do it generally by solving $det(A-\lambda I) = 0$, but this is a past exam question and I don't have unlimited time.

So is there any clues, any information about these two matrices that can help me find the Eigenvalues as fast as possible?

For example, I know that $A$ only "switches" axes, so would it be correct to say that all Eigenvalues have to be $1$ or $-1$ since 4D-Space is not being "scaled" by such a linear transformation?

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The first matrix is such that $$ A^2=I, $$ the corresponding linear operator $$ \varphi_A(x)=Ax,\qquad (x \in \mathbf R^n) $$ is called an involution, and it is known that eigenvalues of an involution of $\mathbf R^n$ are $\pm 1,$ and it is diagonalizable in some basis of $\mathbf R^n.$ You then try to guess what this basis may be for your matrix (it is easy, and you have almost done it).

For the second matrix you immediately see that $e_1$ and $e_3$ are eigenvectors of $B$ that belong to the eigenvalue $2,$ and you may with luck realize quickly enough that $(e_2-e_4)$ is also an eigenvector which belongs to $2.$ Finally, with some luck you can guess that $$ B(e_1+e_2+e_3+e_4)=0. $$ Thus you'll have a basis in which $B$ is diagonalizable, eigenvalues of $B,$ and eigenvectors/eigenspaces of $B.$

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    Once you have the first two eigenvectors of $2$, you know that this eigenvalue has algebraic multiplicity 3 since the trance of $B$ is 6, so there’s either a third independent eigenvector or you’ll have to look for generalized eigenvectors. – 2017-02-19