$A = \begin{bmatrix} 3 & -1& -3 & 1& 0\\ 5 & -2 & -4 & -1 & 0 \\ 2 & - 1 & -2 & 1 & 0 \\ -1 & 0 & 1 & 1 & 0\\ -2 & 1 & 3 & -1 & 1 \end{bmatrix}$
I've found eigenvalues $\lambda_1 = 1$, $\lambda_2 = -1$
I've found Jordan form (it's correct, I've checked it on computer)
$J = \begin{bmatrix} -1 & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$
Now I was trying to find the Jordan basis, I've found eigenvectors for $\lambda_1$: $\begin{bmatrix} 1\\ 0\\1\\1\\0 \end{bmatrix}, \begin{bmatrix} 0\\0\\0\\0\\1 \end{bmatrix}$, and for $\lambda_2$: $\begin{bmatrix} 1\\1\\1\\0\\-1 \end{bmatrix}$
Also I completed these sets to a basis of $\ker (A - \lambda_1 E)^2$ and $\ker (A - \lambda_2 E)^2$ with $\begin{bmatrix} 0\\2\\-1\\-2\\0 \end{bmatrix}$ and $\begin{bmatrix} 0\\1\\0\\0\\0 \end{bmatrix}$.
At the end I have a matrix $C = \begin{bmatrix} 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 2 & 1 & 1\\ 1 & 0 & -1 & 1 & 0\\ 1 & 0 & -1 & 0 & 0\\ 0 & 1 & 0 & -1 & 0\\ \end{bmatrix}$, but $C^{-1}AC \ne J$ and I can't find the mistake. Can you help me, please?
To get the members of the Jordan basis corresponding to a Jordan block of size $\ n\ $ with eigenvalue $\ \lambda\ $, it's not enough merely to find a basis of $\ \ker\left(A-\lambda E\right)^n\ $. The eigenvector $\ e\ $ chosen must belong to both $\ \ker\left(A-\lambda E\right)\ $ and the range of $\ \ker\left(A-\lambda E\right)^{n-1}\ $. The remaining basis vectors, $\ v_1, v_2, \dots, v_{n-1}\ $ must satisfy the equations $\ \left(A-\lambda E\right)v_1=e\ $ and $\ \left(A-\lambda E\right)v_i=v_{i-1}\ $ for $\ i=2,3,\dots, n-1\ $.
For the vectors you've found, you instead have $\ \ker\left(A-\lambda_1E\right)v_1 = e_1^\prime - e_1\ $, and $\ \left(A-\lambda_2E\right)v_2 = - e_2\ $ where $\ e_1, e_1^\prime\ $ are your eigenvectors corresponding to the eigenvalue $\ \lambda_1\ $, and $\ e_2\ $ your eigenvector corresponding to $\ \lambda_2\ $. In fact, neither of the eigenvectors $\ e_1\ $ or $\ e_1^\prime\ $ is in the range of $\ A-\lambda_1E\ $, so there's no vector $\ v\ $ that will satisfy either of the equations $\ \left(A-\lambda_1E\right)v = e_1\ $ or $\ \left(A-\lambda_1E\right)v = e_1^\prime\ $. To get the Jordan block corresponding to your eigenvalue $\ \lambda_1\ $, the eigenvector you choose can only be a multiple of your $\ e_1^\prime - e_1\ $.
To get precisely the same Jordan form, $\ J\ $, that you initially give, you also have to put the eigenvalues and other basis vectors into $\ C\ $ in the proper order. So, if you take \begin{align} C&=\pmatrix{e_2,-v_2, e_1^\prime-e_1,v_1, e_1^\prime}\\ &=\pmatrix{1&0&-1&0&0\\ 1&-1&0&2&0\\ 1&0&-1&-1&0\\ 0&0&-1&-2&0\\ -1&0&1&0&1 }\ , \end{align} for instance, you'll find that then $\ C^{-1}AC=J\ $.