The characteristic equation is \[\det\matrix{-1-\lambda & -12 \\ 0 & -5-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -1-\lambda & -12 \\ 0 & -5-\lambda \end{array} \right\vert &= (-1-\lambda)(-5-\lambda)+12\cdot0 \\
&= (-1-\lambda)(-5-\lambda)
\end{aligned}\] So, the eigenvalues are \(\lambda_1 = -5\) and \(\lambda_2 = -1\).

Let \(\lambda = -5\) be an eigenvalue of the matrix \[A=\matrix{-1 & -12 \\ 0 & -5}\] 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 \[A+5 I = \matrix{-1 & -12 \\ 0 & -5} - \matrix{-5 & 0 \\0& -5 }=\matrix{ 4 & -12 \\ 0 & 0}\] This can be done as follows:
\[\begin{aligned}
\matrix{4&-12\\0&0\\}&\sim\matrix{1&-3\\0&0\\}&{\blue{\begin{array}{c}{{1}\over{4}}R_1\\\phantom{x}\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{-3\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-3\\-1}\right\rangle\).

Let \(\lambda = -1\) be an eigenvalue of the matrix \[A=\matrix{-1 & -12 \\ 0 & -5}\] 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 \[A+I = \matrix{-1 & -12 \\ 0 & -5} - \matrix{-1 & 0 \\0& -1 }=\matrix{ 0 & -12 \\ 0 & -4}\] This can be done as follows:
\[\begin{aligned}
\matrix{0&-12\\0&-4\\}&\sim\matrix{0&1\\0&-4\\}&{\blue{\begin{array}{c}-{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{0&1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+4R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -1\) equals \(\left\{ r \cv{1\\0} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\0}\right\rangle\).
If possible, we avoided fractions in the solution.