Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{5 & 0 \\ -12 & 1}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=5\vec{v}\), that is, \[(A-5 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-5 I\).
We can do this through row reduction of the matrix \[A-5 I = \matrix{5 & 0 \\ -12 & 1} - \matrix{5 & 0 \\0& 5 }=\matrix{ 0 & 0 \\ -12 & -4}\] This can be done as follows:
\[\begin{aligned}
\matrix{0&0\\-12&-4\\}&\sim\matrix{-12&-4\\0&0\\}&{\blue{\begin{array}{c}R_2\\R_1\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}-{{1}\over{12}}R_1\\\phantom{x}\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{1\\-3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(5\) equals \(\left\langle\cv{1\\-3}\right\rangle\)