Got it, let's work through this step by step. Since your matrix has an eigenvalue $\lambda=2$ with algebraic multiplicity 3 but only one linearly independent ordinary eigenvector, we need to use generalized eigenvectors to build a Jordan chain—this will give us the full set of linearly independent solutions for the system.

First, let's recap what we already have:

• Eigenvalue $\lambda=2$, algebraic multiplicity 3, geometric multiplicity 1
• Ordinary eigenvector: $v_1 = \begin{bmatrix} 0 \ -1 \ 1 \end{bmatrix}$ (we can take $c=1$ for simplicity)

Step 1: Compute $(A - 2I)$ #

First, calculate the matrix $A - \lambda I$ where $\lambda=2$:
$$A - 2I = \begin{bmatrix} 1-2 & 1 & 1 \ 2 & 1-2 & -1 \ -3 & 2 & 4-2 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \ 2 & -1 & -1 \ -3 & 2 & 2 \end{bmatrix}$$

Step 2: Find the first generalized eigenvector $v_2$ #

We need a vector $v_2$ such that $(A - 2I)v_2 = v_1$. Let $v_2 = \begin{bmatrix} a \ b \ c \end{bmatrix}$, substitute into the equation to get the system:
$$\begin{cases}
-a + b + c = 0 \
2a - b - c = -1 \
-3a + 2b + 2c = 1
\end{cases}$$
From the first equation: $b + c = a$. Substitute this into the second equation: $2a - a = -1 \implies a = -1$. Then $b + c = -1$ (we can pick a free variable to simplify—let's choose $b=0$, so $c=-1$). This gives us:
$$v_2 = \begin{bmatrix} -1 \ 0 \ -1 \end{bmatrix}$$

Step 3: Find the second generalized eigenvector $v_3$ #

Next, we need $v_3$ such that $(A - 2I)v_3 = v_2$. Let $v_3 = \begin{bmatrix} d \ e \ f \end{bmatrix}$, substitute into the equation:
$$\begin{cases}
-d + e + f = -1 \
2d - e - f = 0 \
-3d + 2e + 2f = -1
\end{cases}$$
Add the first two equations together: $d = -1$. Substitute back into the first equation: $1 + e + f = -1 \implies e + f = -2$. Let's choose $e=0$, so $f=-2$. This gives:
$$v_3 = \begin{bmatrix} -1 \ 0 \ -2 \end{bmatrix}$$

Step 4: Construct the general solution #

For a 3x3 Jordan block corresponding to $\lambda=2$, the three linearly independent solutions are:

1. $x_1(t) = e^{2t} v_1$
2. $x_2(t) = e^{2t} \left( t v_1 + v_2 \right)$
3. $x_3(t) = e^{2t} \left( \frac{t^2}{2} v_1 + t v_2 + v_3 \right)$

The general solution is a linear combination of these three solutions:
$$
x(t) = C_1 e^{2t} \begin{bmatrix} 0 \ -1 \ 1 \end{bmatrix} + C_2 e^{2t} \left( t \begin{bmatrix} 0 \ -1 \ 1 \end{bmatrix} + \begin{bmatrix} -1 \ 0 \ -1 \end{bmatrix} \right) + C_3 e^{2t} \left( \frac{t^2}{2} \begin{bmatrix} 0 \ -1 \ 1 \end{bmatrix} + t \begin{bmatrix} -1 \ 0 \ -1 \end{bmatrix} + \begin{bmatrix} -1 \ 0 \ -2 \end{bmatrix} \right)
$$

Where $C_1, C_2, C_3$ are arbitrary constants. You can expand this to write each component explicitly if needed, but this form is the standard way to present solutions built from Jordan chains.

内容的提问来源于stack exchange，提问作者trich42