Let \(\lambda = 1\) be an eigenvalue of the matrix \[A=\matrix{37 & 120 \\ -12 & -39}\] is. 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 \[A-I = \matrix{37 & 120 \\ -12 & -39} - \matrix{1 & 0 \\0& 1 }=\matrix{ 36 & 120 \\ -12 & -40}\] 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 \(\lambda = 1\) equals \(\left\{ r \cv{10\\-3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(1\) equals \(\left\langle\cv{10\\-3}\right\rangle\)