The characteristic equation is \[\det\matrix{13-\lambda & -9 \\ 18 & -14-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 13-\lambda & -9 \\ 18 & -14-\lambda \end{array} \right\vert &= (13-\lambda)(-14-\lambda)+9\cdot18 \\
&= (13-\lambda)(-14-\lambda)+162 \\ &= \lambda^2+\lambda-20
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+\lambda-20 = (\lambda-4)(\lambda+5) \] So, the eigenvalues are \(\lambda_1 = 4\) and \(\lambda_2 = -5\).

Let \(\lambda = 4\) be an eigenvalue of the matrix \[A=\matrix{13 & -9 \\ 18 & -14}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=4\vec{v}\), this 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 \[A-4 I = \matrix{13 & -9 \\ 18 & -14} - \matrix{4 & 0 \\0& 4 }=\matrix{ 9 & -9 \\ 18 & -18}\] This can be done as follows:
\[\begin{aligned}
\matrix{9&-9\\18&-18\\}&\sim\matrix{1&-1\\18&-18\\}&{\blue{\begin{array}{c}{{1}\over{9}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-18R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 4\) equals \(\left\{ r \cv{-1\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-1\\-1}\right\rangle\).

Let \(\lambda = -5\) be an eigenvalue of the matrix \[A=\matrix{13 & -9 \\ 18 & -14}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-5\vec{v}\), that 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 \[A+5 I = \matrix{13 & -9 \\ 18 & -14} - \matrix{-5 & 0 \\0& -5 }=\matrix{ 18 & -9 \\ 18 & -9}\] This can be done as follows:
\[\begin{aligned}
\matrix{18&-9\\18&-9\\}&\sim\matrix{1&-{{1}\over{2}}\\18&-9\\}&{\blue{\begin{array}{c}{{1}\over{18}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-18R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{-1\\-2} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-1\\-2}\right\rangle\).
If possible, we avoided fractions in the solution.