Question: 1. $\varphi: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}$ an endomorphism with the property $\varphi^2=\varphi$ and $x \in \mathbb{C}^3$ arbitrary where $\varphi(x) \neq 0$. Then $\varphi(x)$ is an eigenvector of $\varphi$
2. $\varphi: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}$ an endomorphism with the property $\varphi^2=\varphi$ and $x \in \mathbb{C}^3$ arbitrary where $x \neq 0$. Then $x$ is an eigenvector of $\varphi$.

It's clear that the two statements above are projections in $\mathbb{C}^{3}$ but I don't know how to continue, should I search for an algebraic or a geometric solution?

Thank you for your help in advance

Answer: 
1. is true because if you apply $\phi$ to $\phi(x)$ you get
$\phi(x)$.
2. Is true only if your map is $0$ or $1$. It is better to consider it geometrically, if the vector is not parallel to the subspace which is the range of your projection, its image is not a multiple of the vector unless the vector is in the kernel.
1. This is always true, because $\varphi(\varphi x)=\varphi^2x=\varphi x$, therefore $\varphi x$ is an eigenvector of $\varphi$ with eigenvalue $1$.
2. This is not always true (take $x$ from the kernel of $\varphi$).

Edit: I was thinking of non-zero eigenvalues, elements of the kernel can be considered eigenvectors with eigenvalue $0$.