The characteristic equation is \[\det\matrix{9-\lambda & 8 \\ -4 & -3-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 9-\lambda & 8 \\ -4 & -3-\lambda \end{array} \right\vert &= (9-\lambda)(-3-\lambda)-8\cdot-4 \\
&= (9-\lambda)(-3-\lambda)+32 \\ &= \lambda^2-6\,\lambda+5
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-6\,\lambda+5 = (\lambda-1)(\lambda-5) \] So, the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 5\).

Let \(\lambda = 1\) be an eigenvalue of the matrix \[A=\matrix{9 & 8 \\ -4 & -3}\] 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 \[A-I = \matrix{9 & 8 \\ -4 & -3} - \matrix{1 & 0 \\0& 1 }=\matrix{ 8 & 8 \\ -4 & -4}\] This can be done as follows:
\[\begin{aligned}
\matrix{8&8\\-4&-4\\}&\sim\matrix{1&1\\-4&-4\\}&{\blue{\begin{array}{c}{{1}\over{8}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+4R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 1\) equals \(\left\{ r \cv{1\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-1}\right\rangle\).

Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{9 & 8 \\ -4 & -3}\] 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 \[A-5 I = \matrix{9 & 8 \\ -4 & -3} - \matrix{5 & 0 \\0& 5 }=\matrix{ 4 & 8 \\ -4 & -8}\] This can be done as follows:
\[\begin{aligned}
\matrix{4&8\\-4&-8\\}&\sim\matrix{1&2\\-4&-8\\}&{\blue{\begin{array}{c}{{1}\over{4}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+4R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{2\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{2\\-1}\right\rangle\).
If possible, we avoided fractions in the solution.