Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{-8 & 4 \\ -6 & 2}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-2\vec{v}\), that is, \[(A+2 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+2 I\).
We can do this through row reduction of the matrix \[A+2 I = \matrix{-8 & 4 \\ -6 & 2} - \matrix{-2 & 0 \\0& -2 }=\matrix{ -6 & 4 \\ -6 & 4}\] This can be done as follows:
\[\begin{aligned}
\matrix{-6&4\\-6&4\\}&\sim\matrix{1&-{{2}\over{3}}\\-6&4\\}&{\blue{\begin{array}{c}-{{1}\over{6}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{2}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+6R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{2\\3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(-2\) equals \(\left\langle\cv{2\\3}\right\rangle\)