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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.

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You are right that $v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $2$; i.e $ v_1 \in \ker(A-2I)$. But like you mentioned, the dimension of this kernel is $2$, so you need to find another linearly independent eigenvector. It is easy to verify that $\xi = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} $ is such a vector. In other words, $\{v_1, \xi\}$ form a basis for $\ker(A-2I)$. Hence, your general solution will be \begin{equation} x(t) = C_1 e^{2t} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + C_2 e^{2t} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + C_3 e^{9t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \end{equation}

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Your eigenvectors are not right for $\lambda=2.$ You have the right system, but you should get two solutions: $v_1=(1,1,0)$ and $v_2=(1,0,1).$

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The system $$x'= \begin{bmatrix} 2 &0 &0 \\ -7&9 &7 \\ 0&0 &2 \end{bmatrix}$$ is simply $$ x_1'=2x_1$$

$$ x_2'=-7x_1+9x_2+7x_3$$

$$ x_3'=2x_3$$

Therefore $$x_1=c_1e^{2t}$$ and $$x_3=c_2 e^{2t}$$

Solving for $$x_2'=9x_2+7(c_2-c_1)e^{2t}$$

we get $$x_2=c_3e^{9t}+(c_1-c_2)e^{2t}$$

Thus we have $$x(t) = c_1 e^{2t} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 0\\-1 \\ 1 \end{pmatrix} + c_3 e^{9t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} $$

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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.

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