I'm reading Smirnov's Linear Algebra and Group Theory, here:
He says that is the characteristic equation but when I compute it on Mathematica, I obtained: $-(\lambda -2)^2 \left(\lambda ^3-27 \lambda -26\right)$. For reference, I computed it as follows:
I think the different result may come from Mathematica computing it in some different number domain perhaps? Or this is some typo in the book?
I assume there is a typo in the matrix $A$ and the issue is that $a_{41}$ should be negative, so lets use this and find the Jordan canonical form. We have
$$A = \begin{pmatrix} -2 & -1 & -1 & 3 & 2 \\ -4 & 1 & -1 & 3 & 2 \\ 1 & 1 & 0 & -3 & -2 \\ -4 & -2 & -1 & 5 & 1 \\ 4 & 1 & 1 & -3 & 0 \\ \end{pmatrix}$$
The eigenvalues are $\lambda = 2$ (a triple root) and $\lambda_2 = -1$ (a double root).
This is a deficient matrix with some generalized eigenvectors.
We find
$$P = \begin{pmatrix} -1 & 0 & 1 & -1 & 1 \\ -1 & 0 & 0 & -1 & 0 \\ 1 & 1 & -1 & 1 & -1 \\ -1 & 0 & 1 & -2 & 0 \\ 1 & 0 & 0 & 1 & 1 \\ \end{pmatrix}, ~~~J = \begin{pmatrix} -1 & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2 \\ \end{pmatrix}$$
This does not appear to match the authors results, but we can validate both.
Using the above, we verify
$$ A = P J P^{-1} = \begin{pmatrix} -2 & -1 & -1 & 3 & 2 \\ -4 & 1 & -1 & 3 & 2 \\ 1 & 1 & 0 & -3 & -2 \\ -4 & -2 & -1 & 5 & 1 \\ 4 & 1 & 1 & -3 & 0 \\ \end{pmatrix}~~~\Large\color{\green}{\unicode{10004}}$$
Using the authors' results, we can verify
$$A = V \Lambda V^{-1} = \begin{pmatrix} 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 \\ \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 \\ \end{pmatrix} \begin{pmatrix} 0 & -1 & 1 & 1 & -1 \\ 1 & -1 & 0 & 1 & -1 \\ 0 & 1 & -1 & 0 & 1 \\ 0 & -2 & 1 & 1 & -1 \\ 0 & 1 & 0 & -1 & 1 \\ \end{pmatrix}~~~\Large\color{\green}{\unicode{10004}}$$
A couple of items to note:
$1.$ Eigenvectors are not unique.
$2.$ The Jordan canonical form is unique up to order of permutation.
So, these answers are equivalent and we have proven using the authors' calculations that indeed there is a typo in the book.