The characteristic equation is \[\det\matrix{-5-\lambda & 6 \\ -8 & 9-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -5-\lambda & 6 \\ -8 & 9-\lambda \end{array} \right\vert &= (-5-\lambda)(9-\lambda)-6\cdot-8 \\
&= (-5-\lambda)(9-\lambda)+48 \\ &= \lambda^2-4\,\lambda+3
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-4\,\lambda+3 = (\lambda-1)(\lambda-3) \] So, the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 3\).

Let \(\lambda = 1\) be an eigenvalue of the matrix \[A=\matrix{-5 & 6 \\ -8 & 9}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=1\vec{v}\), this 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 \[A-I = \matrix{-5 & 6 \\ -8 & 9} - \matrix{1 & 0 \\0& 1 }=\matrix{ -6 & 6 \\ -8 & 8}\] This can be done as follows:
\[\begin{aligned}
\matrix{-6&6\\-8&8\\}&\sim\matrix{1&-1\\-8&8\\}&{\blue{\begin{array}{c}-{{1}\over{6}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+8R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 1\) equals \(\left\{ r \cv{1\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\1}\right\rangle\).

Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{-5 & 6 \\ -8 & 9}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), that is, \[(A-3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-3 I\). We can do this through row reduction of the matrix \[A-3 I = \matrix{-5 & 6 \\ -8 & 9} - \matrix{3 & 0 \\0& 3 }=\matrix{ -8 & 6 \\ -8 & 6}\] This can be done as follows:
\[\begin{aligned}
\matrix{-8&6\\-8&6\\}&\sim\matrix{1&-{{3}\over{4}}\\-8&6\\}&{\blue{\begin{array}{c}-{{1}\over{8}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{3}\over{4}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+8R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{3\\4} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\4}\right\rangle\).
If possible, we avoided fractions in the solution.