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For the matrix $A=\begin{pmatrix} 1 & 1\\ -1 & 3 \\ \end{pmatrix}$, find a basis for the generalized eigenspace of $L_A$ consisting of a union of disjoint cycles of generalized eigenvectors.

When I take the determinant of $(A-tI)$, I get that $\lambda_1=2$ with a multiplicity of 2. So, I get that the eigenvector is $\begin{pmatrix} 1\\ 1\\ \end{pmatrix}$, but looking at the answer online, I'm supposed to also get the vector $\begin{pmatrix} -1\\ 0\\ \end{pmatrix}$. Although, when I compute $(A-2I)v$ I get the zero vector. So I have no idea where I'm going wrong.

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  • $\begingroup$ Which $A$? Try to be a little more comprehensive about your ressources. $\endgroup$ Commented Aug 6, 2018 at 22:32

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You have to find a vector $v$ such that$$A.v=2v+\begin{pmatrix}1\\1\end{pmatrix}.$$In other words, you're after numbers $a,b\in\mathbb R$ such that$$A.\begin{pmatrix}a\\b\end{pmatrix}=2\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}1\\1\end{pmatrix}.$$This is equivalent to the system$$\left\{\begin{array}{l}a+b=2a+1\\-a+3b=2b+1\end{array}\right.$$of which $\binom{-1}0$ is indeed a solution.

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  • $\begingroup$ Why is necessary to use $A. v = \lambda v + v_0$? Is always working? $\endgroup$ Commented Nov 12, 2022 at 15:12
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    $\begingroup$ @BlueTomato If the dimension of the eigenspace corresponding to $\lambda$ is smaller that the multiplicity of $\lambda$ as a root of the characteristic polynomial, then, yes, that's a way of doing things that works. $\endgroup$ Commented Nov 12, 2022 at 15:51

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