The characteristic equation is \[\det\matrix{41-\lambda & -120 \\ 12 & -35-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 41-\lambda & -120 \\ 12 & -35-\lambda \end{array} \right\vert &= (41-\lambda)(-35-\lambda)+120\cdot12 \\
&= (41-\lambda)(-35-\lambda)+1440 \\ &= \lambda^2-6\,\lambda+5
\end{aligned}\] 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 \[A=\matrix{41 & -120 \\ 12 & -35}\] 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 \[A-I = \matrix{41 & -120 \\ 12 & -35} - \matrix{1 & 0 \\0& 1 }=\matrix{ 40 & -120 \\ 12 & -36}\] This can be done as follows:
\[\begin{aligned}
\matrix{40&-120\\12&-36\\}&\sim\matrix{1&-3\\12&-36\\}&{\blue{\begin{array}{c}{{1}\over{40}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 1\) equals \(\left\{ r \cv{3\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\1}\right\rangle\).

Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{41 & -120 \\ 12 & -35}\] 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 \[A-5 I = \matrix{41 & -120 \\ 12 & -35} - \matrix{5 & 0 \\0& 5 }=\matrix{ 36 & -120 \\ 12 & -40}\] This can be done as follows:
\[\begin{aligned}
\matrix{36&-120\\12&-40\\}&\sim\matrix{1&-{{10}\over{3}}\\12&-40\\}&{\blue{\begin{array}{c}{{1}\over{36}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{10}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{-10\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-10\\-3}\right\rangle\).
If possible, we avoided fractions in the solution.