# Orthogonal projection in Inner product space

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 ?

• So your thoughts please...? Jun 15 '14 at 5:51
• just updated in the question. Jun 15 '14 at 5:55

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.

• I mean to say that $\|x \| \leq \| x + ay \| \iff \left < x,y \right > = 0$ Jun 15 '14 at 6:50
• can you show me how to prove it without assuming $<x,y>=<y,x>$? Thanks! Jun 15 '14 at 7:08
• What do you mean NOT assuming $\left < x ,y \right > = \left < y,x \right >$? Jun 15 '14 at 7:13
• because in $\mathbb{C}$ $<x,y>= \overline{\ <y,x>}$ Jun 15 '14 at 7:27
• If I am understanding you correctly, $\| x + ay \|^2 = \|x \|^2 + \left < x , y \right > + \left <y,x \right > + \|a y \|^2 = \|x \|^2 + \bar{a}\left < x , y \right > + \bar{a\left <x,y \right >} + \| ay \|^2 = \|x \|^2 + 2Re(\bar{a}\left < x , y \right >) + \| ay \|^2$ is this where you were stuck? Jun 15 '14 at 8:02