Let \(\lambda = 4\) be an eigenvalue of the matrix \[A=\matrix{11 & -7 \\ 14 & -10}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=4\vec{v}\), that is, \[(A-4 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-4 I\).
We can do this through row reduction of the matrix \[A-4 I = \matrix{11 & -7 \\ 14 & -10} - \matrix{4 & 0 \\0& 4 }=\matrix{ 7 & -7 \\ 14 & -14}\] This can be done as follows:
\[\begin{aligned}
\matrix{7&-7\\14&-14\\}&\sim\matrix{1&-1\\14&-14\\}&{\blue{\begin{array}{c}{{1}\over{7}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-14R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 4\) equals \(\left\{ r \cv{-1\\-1} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

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