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