Let \(\lambda = -1\) be an eigenvalue of the matrix \[A=\matrix{9 & -4 \\ 30 & -13}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-1\vec{v}\), that is, \[(A+I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+I\).
We can do this through row reduction of the matrix \[A+I = \matrix{9 & -4 \\ 30 & -13} - \matrix{-1 & 0 \\0& -1 }=\matrix{ 10 & -4 \\ 30 & -12}\] This can be done as follows:
\[\begin{aligned}
\matrix{10&-4\\30&-12\\}&\sim\matrix{1&-{{2}\over{5}}\\30&-12\\}&{\blue{\begin{array}{c}{{1}\over{10}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{2}\over{5}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-30R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -1\) equals \(\left\{ r \cv{-2\\-5} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(-1\) equals \(\left\langle\cv{-2\\-5}\right\rangle\)