A permutation matrix of order $n$ is a matrix of size $n \times n$ composed of 0 and 1 such that the sum (in the field of real numbers) of elements for each column and each row is 1. Let λ1, λ2, ..., λ5 be the eigenvalues of the matrix of the permutation with order 5:
$$P = \begin{pmatrix} 0 & 1 & 0 & 0&0 \\ 0 & 0 & 0 &0&1 \\ 1&0&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \end{pmatrix} $$
Find λ = min | λi |
My solution:
So i calculated eigenvalues in this way:
$P \cdot x = \lambda \cdot x$
$x_{2} - \lambda \cdot x_{1} = 0$
$x_{5} - \lambda \cdot x_{2} = 0$
$x_{1} - \lambda \cdot x_{3} = 0$
$x_{3} - \lambda \cdot x_{4} = 0$
$x_{4} - \lambda \cdot x_{5} = 0$
We get matrix:
$$P = \begin{pmatrix} -\lambda & 1 & 0 & 0&0 \\ 0 & -\lambda & 0 &0&1 \\ 1&0&-\lambda&0&0 \\ 0&0&1&-\lambda&0 \\ 0&0&0&1&-\lambda \end{pmatrix} = 0 $$
Using Kramers method I calculated that $\lambda^{5} = 1$, so it means that $\lambda = 1$. Is my solution correct. I have doubts because it problem they write about λ2, ..., λ5, but i get only one answer
What you are missing, and what the comments have alluded to, is that you need to consider the 5th roots of unity. See https://en.wikipedia.org/wiki/Root_of_unity for example (look at the picture on the right!). Basically, the equation $z^n = 1$ has $n$ solutions, some of which may be complex. For example, the equation $z^4 = 1$ has solutions $z = 1,-1,i,-i$ (check this!). So here you have $\lambda^5 = 1$, which we solve as follows: $$\lambda = 1^{1/5} = (e^{2\pi ki})^{1/5} = e^{2\pi k i/5}=\cos(2 \pi k/5) + i\sin(2 \pi k/5),$$ where $k \in \mathbb Z$. So you can get the solution in the form $a+bi$ by plugging in $k=0,\ldots, 4$. If this is new to you, you should be able to check out the first chapter of any complex analysis book, where this method of solving equations is explained.