<s>[INST] Finding general solution of $3{\times}3$ system I am given the following:
$$
x'=
\begin{bmatrix}
2 &0  &0 \\ 
 -7&9  &7 \\ 
 0&0  &2 
\end{bmatrix}
x
$$
Solving $\det(A-\lambda I)$, I get $\lambda = 2,2,9$. Solving $\det(A-2\lambda)$, I get 
\begin{bmatrix}
0&0  &0 \\ 
 -7&7  &7 \\ 
 0&0  &0 
\end{bmatrix}
So we have geom multi. = alg. multi, so our matrix is complete.
Take $v_1 = 
\begin{bmatrix}
1\\ 
 1\\ 
 0
\end{bmatrix}
$
Similarly, solve $\det(A-9\lambda)$ to get
$
\begin{bmatrix}
-7&0  &0 \\ 
 -7&0  &7 \\ 
 0&0  &-7 
\end{bmatrix}
$
So take $v_2 = 
\begin{bmatrix}
0\\ 
 1\\ 
 0
\end{bmatrix}
$
So the general solution should be $x(t) = C_1e^{2t}\begin{bmatrix}
1\\ 
 1\\ 
 0
\end{bmatrix}
+C_2e^{2t} \begin{bmatrix}
1\\ 
 1\\ 
 0
\end{bmatrix}
+ C_3e^{9t}\begin{bmatrix}
0\\ 
 1\\ 
 0
\end{bmatrix}$
However, according to the back of the book solution, this is incorrect. What am I missing here? Thank you.
 [/INST] For $\lambda=2$, $(A-2I)v=0$ gives $-x+y+z=0$.
Letting $y=k_1$ and $z=k_2$ you get $x=k_1+k_2$.
Then $v=\begin{pmatrix}k_1+k_2\\k_1\\k_2\end{pmatrix}=k_1\begin{pmatrix}1\\1\\0\end{pmatrix}=k_2\begin{pmatrix}1\\0\\1\end{pmatrix}$.
$\implies v_1=\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $v_2=\begin{pmatrix}1\\0\\1\end{pmatrix}$ are two corresponding eigenvectors.
</s>