Prove that linear operator T is the projection Operator

Question

I got question prove or disprove

Let be $T:\mathbb R^n \to \mathbb R^n$ linear oprator and let be $U\subset\mathbb R^n$ hyperplane.

If known that $Image(T)=U$ and also that for every $u\in U$, $T(u)=u$.

So $T$ is projection operator on $U$.

Thank you for the help :)

Answer 1

I assume, the projection operator is defined as $T^2=T$.

Consider $v\in \mathbb R^n$. Can you think of a way, why $T^2v=Tv$ holds? It is maybe easier to see if you write $$ T^2v=Tv \Leftrightarrow T(T(v)) = T(v). $$ You don't to it normally this way, but let us make it one step more obvious:

We define $w=Tv$. Now we get $$ T(T(v))=T(v) \Leftrightarrow T(w)=w. $$ Consider, that $w$ is an element of the image of $T$. And that this holds, is writen in your condition.