The characteristic equation is \[\det\matrix{-22-\lambda & -63 \\ 6 & 17-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -22-\lambda & -63 \\ 6 & 17-\lambda \end{array} \right\vert &= (-22-\lambda)(17-\lambda)+63\cdot6 \\
&= (-22-\lambda)(17-\lambda)+378 \\ &= \lambda^2+5\,\lambda+4
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+5\,\lambda+4 = (\lambda+1)(\lambda+4) \] So, the eigenvalues are \(\lambda_1 = -1\) and \(\lambda_2 = -4\).

Let \(\lambda = -1\) be an eigenvalue of the matrix \[A=\matrix{-22 & -63 \\ 6 & 17}\] 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 \[A+I = \matrix{-22 & -63 \\ 6 & 17} - \matrix{-1 & 0 \\0& -1 }=\matrix{ -21 & -63 \\ 6 & 18}\] This can be done as follows:
\[\begin{aligned}
\matrix{-21&-63\\6&18\\}&\sim\matrix{1&3\\6&18\\}&{\blue{\begin{array}{c}-{{1}\over{21}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-6R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -1\) equals \(\left\{ r \cv{3\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\-1}\right\rangle\).

Let \(\lambda = -4\) be an eigenvalue of the matrix \[A=\matrix{-22 & -63 \\ 6 & 17}\] 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 \[A+4 I = \matrix{-22 & -63 \\ 6 & 17} - \matrix{-4 & 0 \\0& -4 }=\matrix{ -18 & -63 \\ 6 & 21}\] This can be done as follows:
\[\begin{aligned}
\matrix{-18&-63\\6&21\\}&\sim\matrix{1&{{7}\over{2}}\\6&21\\}&{\blue{\begin{array}{c}-{{1}\over{18}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{7}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-6R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -4\) equals \(\left\{ r \cv{7\\-2} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{7\\-2}\right\rangle\).
If possible, we avoided fractions in the solution.