The characteristic equation is \[\det\matrix{37-\lambda & 126 \\ -12 & -41-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 37-\lambda & 126 \\ -12 & -41-\lambda \end{array} \right\vert &= (37-\lambda)(-41-\lambda)-126\cdot-12 \\
&= (37-\lambda)(-41-\lambda)+1512 \\ &= \lambda^2+4\,\lambda-5
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+4\,\lambda-5 = (\lambda+5)(\lambda-1) \] So, the eigenvalues are \(\lambda_1 = -5\) and \(\lambda_2 = 1\).

Let \(\lambda = -5\) be an eigenvalue of the matrix \[A=\matrix{37 & 126 \\ -12 & -41}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-5\vec{v}\), this 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 \[A+5 I = \matrix{37 & 126 \\ -12 & -41} - \matrix{-5 & 0 \\0& -5 }=\matrix{ 42 & 126 \\ -12 & -36}\] This can be done as follows:
\[\begin{aligned}
\matrix{42&126\\-12&-36\\}&\sim\matrix{1&3\\-12&-36\\}&{\blue{\begin{array}{c}{{1}\over{42}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{3\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\-1}\right\rangle\).

Let \(\lambda = 1\) be an eigenvalue of the matrix \[A=\matrix{37 & 126 \\ -12 & -41}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=1\vec{v}\), that 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 \[A-I = \matrix{37 & 126 \\ -12 & -41} - \matrix{1 & 0 \\0& 1 }=\matrix{ 36 & 126 \\ -12 & -42}\] This can be done as follows:
\[\begin{aligned}
\matrix{36&126\\-12&-42\\}&\sim\matrix{1&{{7}\over{2}}\\-12&-42\\}&{\blue{\begin{array}{c}{{1}\over{36}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{7}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 1\) equals \(\left\{ r \cv{7\\-2} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{7\\-2}\right\rangle\).
If possible, we avoided fractions in the solution.