Why is the following linear transformation not the identity map?

Question

A question that I attempted stated the following:

"Let $T:V\to V$ be a linear transformation such that $T^2=T $ ie. $T \circ T=T $

I thought that this was in fact the identity map in disguise but it turns out it isn't.

Why exactly is it not the identity map?

(My answer to my question is: V is the codomain and the domain in this case so one might be tempted to think that $T(v)=T(T(v))\implies$ T is the identity map(as the definition of the identity map is the map $T:V \to V$ such that $T(v)=v$ $ \forall v\in V$)( my mistake was that I equated the situation that we have in the question where T(v)=T(T(v)) for all T(v) in V to the definition the identity map which is incorrect because the linear transformation in the question is not necessarily surjective hence the T(v)'s do not constitute the whole of V the codomain, and as the codomain and the domain are the same in this question the T(v)'s do not constitute the whole domain hence the fact that $T(v)=T(T(v))$ for all $T(v) \in V$ does not imply that the linear transformation in the question is the identity map.

This means that were the map surjective then the linear transformation would be the identity map)

Answer 1

You have to find an example which disproves that claim, like $T(x)=e_1=0, T(e_2)=e_2$ in $Vect(e_1,e_2)$

Answer 2

For example, the map

$$T:\Bbb R^2\to\Bbb R^2\;,\;\;\;T\binom xy:=\binom x0$$

fulfills $\;T^2=T\;$ ...and clearly $\;T\neq Id\;$ . Such a map is called projection.