1
$\begingroup$

What is the Basis of the eigenspace $\begin{pmatrix} 5 & 2 \\ 0 & 5 \end{pmatrix} $, for the eigenvalue $\lambda = 5.$

As far as I know I have to calculate $A-(\lambda-E)$, that would be $\begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix} $ . Now I guess you can "simplify" this by dividing the first linear equation by 2, which would then be $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $. Now is the eigenspace or rather eigenvector $\begin{pmatrix} 1 \\ 0 \end{pmatrix} $? Im a bit confused because the "solution" is not similar to the output of the Gauss-Algorithm.

$\endgroup$
1
$\begingroup$

The eigenspace associated to the eigenvalue $\lambda=5$ is $V_5\equiv 2x_2=0$ whose solutions are, $x_1=\alpha,\; x_2=0$ with $\alpha\in\mathbb{R}$: $$V_5=\{(\alpha,0):\alpha\in\mathbb{R}\}=\{\alpha(1,0):\alpha\in\mathbb{R}\}$$ hence a basis of $V_5$ is $B=\{(1,0)\}.$

$\endgroup$
  • $\begingroup$ Ah im starting to understand $\endgroup$ – WhatAMesh Jan 18 '17 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.