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Eigenspace (E's question at Yahoo! Answers)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello E,

The eigenspace associated to $\lambda=9$ is $\ker (A-9I)$, that is: $$\ker (A-9 I)\equiv{}\begin{bmatrix}{-9}&{\;\;0}&{-9}\\{\;\;0}&{-9}&{\;\;0}\\{-9}&{\;\;0}&{-9}\end{bmatrix} \begin{bmatrix}{x_1}\\{x_2}\\{x_3}\end{bmatrix}= \begin{bmatrix}{0}\\{0}\\{0}\end{bmatrix} \Leftrightarrow \left \{ \begin{matrix} x_1+x_3=0\\x_2=0\end{matrix}\right.\Leftrightarrow \left \{ \begin{matrix} x_1=-\alpha\\x_2=0\\x_3=\alpha\end{matrix}\right. \;(\alpha \in\mathbb{R})$$ so, the general form of the eigenvectors corresponding to $\lambda=9$ is $(x_1,x_2,x_3)^T=\alpha(-1,0,1)^T$ with $\alpha \in\mathbb{R}$.