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I am trying to find the general solution to $X' = AX$, where $A = \begin{pmatrix} 1&0&-1 \\ -1 & 0&-1 \\0&0&1 \end{pmatrix}$. I know that the characteristic polynomial of $A$ is $p(\lambda) = -\lambda (1-\lambda)^{2}$, giving us eigenvalues $\lambda_{1} = 0$ and $\lambda_{2} = 1$. I have found that the corresponding eigenvectors are $V_{1} = \begin{pmatrix} 0\\1\\0\end{pmatrix}$ and $V_{2} = \begin{pmatrix}-1\\1\\0\end{pmatrix}$. I also know that the block-diagonal matrix for $A$ should be $T^{-1}AT = \begin{pmatrix}0&0&0\\0&1&1\\0&0&1\end{pmatrix}$. However, I'm not sure how to go about finding the actual matrix $T$, because I'm not sure how to find a third eigenvector for $A$. Similarly, I know that for this block-diagonal matrix, some solutions are $y_{2} = e^{t} + c_{3}te^{t}$ and $y_{3} = c_{3}e^{t}$, but I'm not sure how to continue on from these to the general solution with the original matrix. How do I proceed from here to find the general solution of this system?

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The vector $V_2$ satisfies $AV_2=V_2.$ Now, we only need a vector $V_3$ such that $\{V_1,V_2,V_3\}$ are linearly independent and satisfying $AV_3=V_2+V_3$ (or equivalently $(A-I)V_3=V_2$). So, $$\left \{ \begin{matrix}AV_1=0V_1\\AV_2=V_2\\ AV_3=V_2+V_3 \end{matrix}\right.$$

and $T=[V_1,V_2,V_3]$ satisfies $$T^{-1}AT = \begin{pmatrix}0&0&0\\0&1&1\\0&0&1\end{pmatrix}.$$

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