Let \(\lambda = 1\) be an eigenvalue of the matrix \[A=\matrix{13 & -16 \\ 12 & -15}\] 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 \[A-I = \matrix{13 & -16 \\ 12 & -15} - \matrix{1 & 0 \\0& 1 }=\matrix{ 12 & -16 \\ 12 & -16}\] This can be done as follows:
\[\begin{aligned}
\matrix{12&-16\\12&-16\\}&\sim\matrix{1&-{{4}\over{3}}\\12&-16\\}&{\blue{\begin{array}{c}{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{4}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 1\) equals \(\left\{ r \cv{-4\\-3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(1\) equals \(\left\langle\cv{-4\\-3}\right\rangle\)