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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 ?

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  • $\begingroup$ So your thoughts please...? $\endgroup$
    – IAmNoOne
    Jun 15 '14 at 5:51
  • $\begingroup$ just updated in the question. $\endgroup$
    – GroundIns
    Jun 15 '14 at 5:55
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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.

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  • $\begingroup$ I mean to say that $\|x \| \leq \| x + ay \| \iff \left < x,y \right > = 0$ $\endgroup$
    – IAmNoOne
    Jun 15 '14 at 6:50
  • $\begingroup$ can you show me how to prove it without assuming $<x,y>=<y,x>$? Thanks! $\endgroup$
    – GroundIns
    Jun 15 '14 at 7:08
  • $\begingroup$ What do you mean NOT assuming $\left < x ,y \right > = \left < y,x \right >$? $\endgroup$
    – IAmNoOne
    Jun 15 '14 at 7:13
  • $\begingroup$ because in $\mathbb{C}$ $<x,y>= \overline{\ <y,x>}$ $\endgroup$
    – GroundIns
    Jun 15 '14 at 7:27
  • $\begingroup$ 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? $\endgroup$
    – IAmNoOne
    Jun 15 '14 at 8:02

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