Let \(\lambda = 2\) be an eigenvalue of the matrix \[A=\matrix{-8 & -5 \\ 10 & 7}\] 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 \[A-2 I = \matrix{-8 & -5 \\ 10 & 7} - \matrix{2 & 0 \\0& 2 }=\matrix{ -10 & -5 \\ 10 & 5}\] This can be done as follows:
\[\begin{aligned}
\matrix{-10&-5\\10&5\\}&\sim\matrix{1&{{1}\over{2}}\\10&5\\}&{\blue{\begin{array}{c}-{{1}\over{10}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-10R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 2\) equals \(\left\{ r \cv{-1\\2} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

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