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

Question

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.

Answer 1

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 $B=\{(1,0)\}.$