The characteristic equation is \[\det\matrix{23-\lambda & 9 \\ -54 & -22-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 23-\lambda & 9 \\ -54 & -22-\lambda \end{array} \right\vert &= (23-\lambda)(-22-\lambda)-9\cdot-54 \\
&= (23-\lambda)(-22-\lambda)+486 \\ &= \lambda^2-\lambda-20
\end{aligned}\] 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 \[A=\matrix{23 & 9 \\ -54 & -22}\] 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 \[A+4 I = \matrix{23 & 9 \\ -54 & -22} - \matrix{-4 & 0 \\0& -4 }=\matrix{ 27 & 9 \\ -54 & -18}\] This can be done as follows:
\[\begin{aligned}
\matrix{27&9\\-54&-18\\}&\sim\matrix{1&{{1}\over{3}}\\-54&-18\\}&{\blue{\begin{array}{c}{{1}\over{27}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+54R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -4\) equals \(\left\{ r \cv{1\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-3}\right\rangle\).

Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{23 & 9 \\ -54 & -22}\] 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 \[A-5 I = \matrix{23 & 9 \\ -54 & -22} - \matrix{5 & 0 \\0& 5 }=\matrix{ 18 & 9 \\ -54 & -27}\] This can be done as follows:
\[\begin{aligned}
\matrix{18&9\\-54&-27\\}&\sim\matrix{1&{{1}\over{2}}\\-54&-27\\}&{\blue{\begin{array}{c}{{1}\over{18}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+54R_1\end{array}}} \end{aligned}\] 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.