Show that $T(v)$ and $T(v)-v$ are eigenvectors of $T$

Question

Let $V$ be a vector space, $T:V\to V$ be linear s.t. $T\circ T=T$, $v\in V$ s.t. $T(v)\neq 0$ and $T(v)\neq v$.
Show that $T(v)$ and $T(v)-v$ are eigenvectors of $T$. What are their eigenvalues?

Attempt:
It's clear that since $T(T(v))=T(v)$ we derive that $T(v)$ is an eigenvector with a corresponding eigenvalue of $1$. However, the struggle is with $T(v)-v$ since $T(T(v)-v)=T(T(v))-T(v)$ and the fact that $T(v)\neq v$ apparently hints that a certain trick is required.

Answer 1

Note that$$T\bigl(T(v)\bigr)-T(v)=T(v)-T(v)=0.$$Since $T(v)\neq v$, this proves that $T(v)-v$ is an eigenvector with eigenvalue $0$.

Answer 2

We have $T(T(v)) = (T \circ T)(v) = T(v)$ therefore $T(T(v) - v) = 0$.