Let $\lambda = 3$ be an eigenvalue of the matrix $$A=\matrix{3 & 4 \\ 0 & -1}$$ is. Then there must be a vector $\vec{v}$ such that $A\vec{v}=3\vec{v}$, that is, $$(A-3 I)\vec{v}=\vec{0}\text.$$ In other words, we must find the kernel of the matrix $A-3 I$.
We can do this through row reduction of the matrix $$A-3 I = \matrix{3 & 4 \\ 0 & -1} - \matrix{3 & 0 \\0& 3 }=\matrix{ 0 & 4 \\ 0 & -4}$$ This can be done as follows:
$$\begin{aligned} \matrix{0&4\\0&-4\\}&\sim\matrix{0&1\\0&-4\\}&{\blue{\begin{array}{c}{{1}\over{4}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{0&1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+4R_1\end{array}}} \end{aligned}$$ So the eigenspace for $\lambda = 3$ equals $\left\{ r \cv{1\\0} \middle|\;r\in\mathbb R\right\}$.
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue $3$ equals $\left\langle\cv{1\\0}\right\rangle$