Recall that a matrix $A$ is diagonalizable if there is a diagonal matrix $D$ and an invertible matrix $P$ such that \begin{equation*} A = PDP^{-1}. \end{equation*}
Next we study the possibility of diagonalizing the matrix \begin{equation*} A = \begin{pmatrix} 2&1&1\\ 1&2&1\\ -2&-2&-1 \end{pmatrix}. \end{equation*}
We look for the proper values of the matrix $A$, for this we take the roots of the characteristic polynomial given by the expression $\text{det}(A-\lambda I)$, that is \begin{equation*} \text{det}\begin{pmatrix} 2-\lambda&1&1\\ 1&2-\lambda&1\\ -2&-2&-1-\lambda \end{pmatrix} = -\lambda^{3} + 3\lambda^{2} - 3\lambda + 1 = -(\lambda - 1)^{3}, \end{equation*} therefore the only eigenvalue of $A$ is $\lambda = 1$ with algebraic multiplicity $3$.
To find the eigenvectors, it is enough to obtain a base for $\text{Ker}(A-\lambda I)$, this is obtained by solving the homogeneous system $ (A-\lambda I) x = 0 $, in this case we have to \begin{equation*} \left(\begin{array}{ccc|c} 2-1&1&1&0\\ 1&2-1&1&0\\ -2&-2&-1-1&0 \end{array}\right) = \left(\begin{array}{ccc|c} 1&1&1&0\\ 1&1&1&0\\ -2&-2&-2&0 \end{array}\right), \end{equation*} staggering \begin{equation*} \left(\begin{array}{ccc|c} 1&1&1&0\\ 1&1&1&0\\ -2&-2&-2&0 \end{array}\right) \to \left(\begin{array}{ccc|c} 1&1&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right), \end{equation*} thus \begin{equation*} \text{Ker}(A-I) = \left\langle\left\{ \begin{pmatrix} -1\\1\\0 \end{pmatrix}; \begin{pmatrix} -1\\0\\1 \end{pmatrix} \right\}\right\rangle, \end{equation*} therefore the geometric multiplicity of $\lambda = 1$ is $2$. Then two eigenvectors are \begin{equation*} v_{1} = \begin{pmatrix} -1\\1\\0 \end{pmatrix};\quad v_{2} \begin{pmatrix} -1\\0\\1 \end{pmatrix}. \end{equation*}
It is clearly non-diagonalizable, but can we find a third generalized eigenvector? Let's try to factor using Jordan matrix
First idea:
To recap, we have a single eigenvalue of algebraic multiplicity $3$ and geometric multiplicity $2$, the latter means that the eigenvalue associated with $\lambda$ is of dimension $2$, which generates a conflict, since we require $3$ eigenvectors li to get an invertible $P$ array, this means you can't write \begin{equation*} A = PDP^{-1}, \end{equation*} where $ D $ is a diagonal matrix made up of the eigenvalues of $ A $ and $ P $ is an invertible matrix made up of eigenvectors.
Due to the above, a Jordan decomposition is sought, the Jordan matrix in this case is of the form \begin{equation*} J = \begin{pmatrix} 1&0&0\\ 0&1&1\\ 0&0&1 \end{pmatrix}, \end{equation*} then to complete the matrix $P$ (since we need $3$ vectors) we study $AP = PJ $, this is to find a vector $(a, b, c)^{t}$ such that \begin{equation*} \begin{pmatrix} 2&1&1\\ 1&2&1\\ -2&-2&-1 \end{pmatrix} \begin{pmatrix} -1&-1&a\\ 1&0&b\\ 0&1&c \end{pmatrix} = \begin{pmatrix} -1&-1&a\\ 1&0&b\\ 0&1&c \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&1&1\\ 0&0&1 \end{pmatrix}, \end{equation*}
However, it is not necessary to carry out all the products of the matrix product, since it will be enough to study the equality of matrices in the third column (since it will not verify), this equality of column vectors is \begin{equation*} \begin{pmatrix} 2a+b+c\\ a+2b-c\\ -2a-2b-c \end{pmatrix} = \begin{pmatrix} -1+a\\ b\\ 1+c \end{pmatrix}, \end{equation*} so it is enough to solve the system of equations \begin{equation*} \left(\begin{array}{ccc|c} 1&1&1&-1\\ 1&1&-1&0\\ -2&-2&-2&1 \end{array}\right), \end{equation*} but as it turns out that there is no solution for this system, then there is no vector $ v_{3} = (a, b, c)^{t}$ such that $P = (v_{1} | v_{2} | v_{3})$ and check $A = PJP^{-1}$.
Second idea: To complete the matrix $P$ we look for a generalized eigenvector, this is a vector $v$ such that for some $k\in\mathbb{N}$ it is found in the space $\text{Ker}((A-I)^{k})$).
For example, for $k = 2$, we study the system of equations \begin{equation*} \left(\begin{array}{ccc|c} 1&1&1&0\\ 1&1&1&0\\ -2&-2&-2&0 \end{array}\right)^{2} = \left(\begin{array}{ccc|c} 0&0&1&0\\ 0&0&1&0\\ -2&-2&-3&0 \end{array}\right) \to \left(\begin{array}{ccc|c} 1&1&0&0\\ 0&0&1&0\\ 0&0&0&0 \end{array}\right), \end{equation*} later \begin{equation*} \text{Ker}((A-I)^{2}) = \left\langle\left\{ \begin{pmatrix} 1\\-1\\0 \end{pmatrix} \right\}\right\rangle, \end{equation*} but let's note that $\text{Ker}((A-I)^{2})\subseteq\text{Ker}(A-I)$, so we see what happens with $k = 4$, this is the system \begin{equation*} \left(\begin{array}{ccc|c} 1&1&1&0\\ 1&1&1&0\\ -2&-2&-2&0 \end{array}\right)^{4} = \left(\begin{array}{ccc|c} -2&-2&-3&0\\ -2&-2&-3&0\\ 6&6&5&0 \end{array}\right) \to \left(\begin{array}{ccc|c} 6&6&5&0\\ 0&0&-4/3&0\\ 0&0&0&0 \end{array}\right) \to \left(\begin{array}{ccc|c} 1&1&0&0\\ 0&0&1&0\\ 0&0&0&0 \end{array}\right), \end{equation*} later \begin{equation*} \text{Ker}((A-I)^{4}) = \left\langle\left\{ \begin{pmatrix} 1\\-1\\0 \end{pmatrix} \right\}\right\rangle, \end{equation*}
Due to the above, I don't know how to get the third (generalized) eigenvector. How can I get it?