The characteristic equation is \[\det\matrix{-15-\lambda & -16 \\ 12 & 13-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -15-\lambda & -16 \\ 12 & 13-\lambda \end{array} \right\vert &= (-15-\lambda)(13-\lambda)+16\cdot12 \\
&= (-15-\lambda)(13-\lambda)+192 \\ &= \lambda^2+2\,\lambda-3
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+2\,\lambda-3 = (\lambda-1)(\lambda+3) \] So, the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = -3\).

Let \(\lambda = 1\) be an eigenvalue of the matrix \[A=\matrix{-15 & -16 \\ 12 & 13}\] 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 \[A-I = \matrix{-15 & -16 \\ 12 & 13} - \matrix{1 & 0 \\0& 1 }=\matrix{ -16 & -16 \\ 12 & 12}\] This can be done as follows:
\[\begin{aligned}
\matrix{-16&-16\\12&12\\}&\sim\matrix{1&1\\12&12\\}&{\blue{\begin{array}{c}-{{1}\over{16}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-12R_1\end{array}}} \end{aligned}\] 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 \[A=\matrix{-15 & -16 \\ 12 & 13}\] 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 \[A+3 I = \matrix{-15 & -16 \\ 12 & 13} - \matrix{-3 & 0 \\0& -3 }=\matrix{ -12 & -16 \\ 12 & 16}\] This can be done as follows:
\[\begin{aligned}
\matrix{-12&-16\\12&16\\}&\sim\matrix{1&{{4}\over{3}}\\12&16\\}&{\blue{\begin{array}{c}-{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{4}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -3\) equals \(\left\{ r \cv{4\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{4\\-3}\right\rangle\).
If possible, we avoided fractions in the solution.