Proving that $P\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=\left(x_{1}, x_{2}, 0,0,0\right)$ is linear.

Question

Consider the vector space $\mathbb{C}^5$ and the map $$P\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=\left(x_{1}, x_{2}, 0,0,0\right)$$ I want to prove that it is linear. I argue that from the property of vector addition that it can be proven by: $$P(\alpha x+\beta y)=P\left(\alpha\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right]+\beta\left[\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \\ y_{5} \end{array}\right]\right)=P\left(\left[\begin{array}{l} \alpha x_{1}+\beta y_{1} \\ \alpha x_{2}+\beta y_{2} \\ \alpha x_{3}+\beta y_{3} \\ \alpha x_{4}+\beta y_{4} \\ \alpha x_{5}+\beta y_{5} \end{array}\right]\right)\\ =\left[\begin{array}{c} \alpha x_{1}+\beta y_{1} \\ \alpha x_{2}+\beta y_{2} \\ 0 \\ 0 \\ 0 \end{array}\right]=\alpha\left[\begin{array}{c} x_{1} \\ x_{2} \\ 0 \\ 0 \\ 0 \end{array}\right]+\beta\left[\begin{array}{c} y_{2} \\ y_{2} \\ 0 \\ 0 \\ 0 \end{array}\right]=\alpha P(x)+\beta P(y)$$ Did I 'break any laws', or is this the correct way to show that it is linear?