Let V be $n$-Dimensional ($n\ge1$) inner product space .
Let $T:V \rightarrow V$ be a linear map which maintains $ T^2=T$ , $\forall v \in V\ ||Tv||\le||v||$.
Prove that there is exists a subspace $U\subseteq V$ ,$0 \le dimU\le n $, such that T is the Orthogonal projection on U.
I tried to prove that $ImT \perp KerT $.But I don't know how to use the fact that $||Tv||\le||v||$.Maybe Cauchy–Schwarz inequality can helps somehow ?
Your idea is good, let's start defining those quantities
$$Im(T) = \{ x \in V : x = Tv, v \in V \}$$
$$Ker(T) = \{ y \in V : Ty = 0 \}.$$
We need to show that for any $x \in Im(T)$ and $y \in Ker(T)$ we must have $\left < x, y\right > = 0.$
As $x \in Im(T)$, there is a $v \in V$ such that $$x = Tv.$$
By the idempotent property, we also have
$$T(x) = T(Tv) = Tv = x. $$
and since $y \in Ker(T)$ we get $T(x + ay) = T(x) + aT(y) = T(x) = x$ for any scalar $a$.
Using the inequality we get $\| x \| = \| T(x+ ay) \| \leq \|x + ay \|.$
As the above is true for an arbitrary chosen $a$, the result now follows.