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.
Best Answer
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.