The characteristic equation is \[\det\matrix{-17-\lambda & 12 \\ -16 & 11-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -17-\lambda & 12 \\ -16 & 11-\lambda \end{array} \right\vert &= (-17-\lambda)(11-\lambda)-12\cdot-16 \\
&= (-17-\lambda)(11-\lambda)+192 \\ &= \lambda^2+6\,\lambda+5
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+6\,\lambda+5 = (\lambda+5)(\lambda+1) \] So, the eigenvalues are \(\lambda_1 = -5\) and \(\lambda_2 = -1\).

Let \(\lambda = -5\) be an eigenvalue of the matrix \[A=\matrix{-17 & 12 \\ -16 & 11}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-5\vec{v}\), this is, \[(A+5 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+5 I\). We can do this through row reduction of the matrix \[A+5 I = \matrix{-17 & 12 \\ -16 & 11} - \matrix{-5 & 0 \\0& -5 }=\matrix{ -12 & 12 \\ -16 & 16}\] This can be done as follows:
\[\begin{aligned}
\matrix{-12&12\\-16&16\\}&\sim\matrix{1&-1\\-16&16\\}&{\blue{\begin{array}{c}-{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+16R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{1\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\1}\right\rangle\).

Let \(\lambda = -1\) be an eigenvalue of the matrix \[A=\matrix{-17 & 12 \\ -16 & 11}\] 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 \[A+I = \matrix{-17 & 12 \\ -16 & 11} - \matrix{-1 & 0 \\0& -1 }=\matrix{ -16 & 12 \\ -16 & 12}\] This can be done as follows:
\[\begin{aligned}
\matrix{-16&12\\-16&12\\}&\sim\matrix{1&-{{3}\over{4}}\\-16&12\\}&{\blue{\begin{array}{c}-{{1}\over{16}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{3}\over{4}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+16R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -1\) 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.