I am currently revising for my exam on Linear Algebra, and have reached a stump in a question regarding a $\lambda$-eigenspace taken from a $2\times2$ matrix with the bottom row consisting of zeroes.
The question says:
"Find the eigenvalues and eigenvectors of the linear transformation $\phi\,:\,\mathbb{R}^2\rightarrow\mathbb{R}^2$ with the matrix $\left(\begin{matrix}-2&&5\\-1&&4\end{matrix}\right)$."
I have found the eigenvalues to be $\lambda=-1,$ or $\lambda=3$.
For $\lambda=-1$, I obtained the system $\left(\begin{matrix} -1&&5\\-1&&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$.
and the coefficient matrix reduces to be $\left(\begin{matrix}-1&&5\\0&&0\end{matrix}\right)$.
In the solution to this question, my lecturer deduces that our $-1$-eigenspace is $\left\langle\left(\begin{matrix}5\\1\end{matrix}\right)\right\rangle$.
I am unsure as to how he came to this conclusion.
He defines in his notes the "$\lambda$-eigenspace of $f$" as the subspace $U_\lambda = \left\{u\in U\,:\,f(u)=\lambda u\right\}$. Now then, I know by the definition of span that $\left\langle\left(\begin{matrix}5\\1\end{matrix}\right)\right\rangle=\left\{\left(\begin{matrix}5a\\a\end{matrix}\right)\,:\,a\in\mathbb{R}\right\}$. I am however uncertain as to how this relates to the matrix I have acquired from the system.
My initial thoughts are: if $-x+5y=0$, then $x=5y$, so each vector takes a 'similar' form to his solution.
If anyone could offer insight I would be grateful. I would like to apologise in advance if a similar question has been asked before: I looked, but due to the 'niche' nature of the question, it was difficult to word effectively.