I am trying to find a generalized eigenvector in this problem. (I understand the general theory goes much deeper, but we are only responsible for a limited number of cases.)
I have found eigenvectors $\vec {u_1}$ and $\vec {u_2}.$
When I try $u_1$ and $u_2$ as $u_3$ into this equation: $$ (A - I)u_4 = u_3$$ I get systems which are inconsistent.
How can I find the $u_3$? I've been told it has something to do with $(A - I)^3 = 0$, but that's about it.
We are given the matrix:
$$\begin{bmatrix}2 & 1 & 1\\1 & 2 & 1\\-2 & -2 & -1\\\end{bmatrix}$$
We want to find the characteristic polynomial and eigenvalues by solving
$$|A -\lambda I| = 0 \rightarrow -\lambda^3+3 \lambda^2-3 \lambda+1 = -(\lambda-1)^3 = 0$$
This yields a single eigenvalue, $\lambda = 1$, with an algebraic multiplicity of $3$.
If we try and find eigenvectors, we setup and solve:
$$[A - \lambda I]v_i = 0$$
In this case, after row-reduced-echelon-form, we have:
$$\begin{bmatrix}1 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\\\end{bmatrix}v_i = 0$$
This leads to the two eigenvectors as he shows, but the problem is that we cannot use that to find the third as we get degenerate results, like you showed.
Instead, let's use the top-down chaining method to find three linearly independent generalized eigenvectors.
Since the RREF of
$$[A - 1 I] = \begin{bmatrix}1 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\\\end{bmatrix}$$
We have $E_3 = kernel(A - 1I)$ with dimension $= 2$, so there will be two chains.
Next, since
$$[A - 1 I]^2 = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\\\end{bmatrix}$$
the space Kernel $(A-1I)^2$ has dimension $=3$, which matches the algebraic multiplicity of $\lambda=1$.
Thus, one of the chains will have length $2$, so the other must have length $1$.
We now form a chain of $2$ generalized eigenvectors by choosing $v_2$ in kernel $(A-1I)^2$ such that $v_2$ is not in the kernel $(A-1I)$.
Since every vector is in kernel $(A-1I)^2$, and the third column of $(A-1I)$ is non-zero, we may choose:
$$v_2 = (1, 0, 0) \implies v_1 = (A-1I)v_2 = (1,1,-2)$$
To form a basis for $\mathbb R^3$, we need one additional chain of one generalized eigenvector. This vector must be an eigenvector that is independent from $v_1$. Since
$$E_3 = ~\text{span}~ \left(\begin{bmatrix}0\\1\\-1\\\end{bmatrix}, \begin{bmatrix}-1\\0\\1\\\end{bmatrix}\right).$$
and neither of these spanning vectors is itself a scalar multiple of $v1$, we may choose either one of them. So let
$$w_1 = (0, 1, -1).$$
Now we have two chains:
$$v_2 \rightarrow v_1 \rightarrow 0$$
$$w_1 \rightarrow 0$$
So, to write the solution, we have:
$\displaystyle 1^{st}$ Chain
$$x_1(t) = e^t \begin{bmatrix}1\\1\\-2\\\end{bmatrix}$$
$$x_2(t) = e^t\left(t \begin{bmatrix}1\\1\\-2\\\end{bmatrix} + \begin{bmatrix}1\\0\\0\\\end{bmatrix}\right)$$
$\displaystyle 2^{nd}$ Chain
$$x_3(t) = e^t \begin{bmatrix}0\\1\\-1\\\end{bmatrix}$$
Thus:
$$x(t) = x_1(t) + x_2(t) + x_3(t)$$
Note, you can use this linear combination of $x(t)$ and verify that indeed it is a solution to $x' = Ax$.
