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This question is a follow-up/extension to this question. Suppose $P$ is an orthogonal operator on a finite dimensional inner product space $V$. By definition, this means that $$ \langle Pu, Pv \rangle = \langle u, v \rangle $$ for all $u,v \in V$. I want to show that $P^TP = I$.

By definition of the transpose,

$$ \langle u, v \rangle = \langle Pu, Pv \rangle = \langle u, P^T(Pv) \rangle $$

and by bilinearity of the inner product it follows that $$ \langle u , (I - P^TP)v \rangle = 0 $$ If $v \neq 0 \neq u$, I want to use the fact that the inner product is nondegenerate to conclude that $I - P^TP = 0$ which would prove the claim. However, I'm not so sure that this applies here for the nondegenerate criterion is a statement about all vectors and not just a particular vector. That is, I don't believe it necessarily true that $\langle x, y \rangle = 0$ and $x \neq 0 \implies y = 0$.

On the other hand, without invoking that the inner product is nondegenerate one can see from inspection that $P^TP = I$ satisfies $$ \langle u, v \rangle = \langle u , (P^TP)v \rangle $$
Is this enough?

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3 Answers

up vote 5 down vote accepted

You're almost there. Notice that in fact, given $v \in V$ $$\langle u, (I - P^TP)v \rangle = 0$$ for all $u \in V$. In particular, setting $u = (I - P^TP)v$ gives

$$\|(I - P^TP)v\|^2 = 0$$ which implies $$(I - P^TP)v = 0$$

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Beaten by 2 seconds... :-) –  Henning Makholm Jul 2 '12 at 1:26
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Since you know $\langle u,(I-P^tP)v\rangle=0$, for arbitrary $u$ and $v$, you can fix any $v$ and then chose $u$ to be $(I-P^tP)v$. Then you know $\langle (I-P^tP)v,(I-P^tP)v\rangle=0$ for every $v$, and because the inner product is positive definite, that must mean $(I-P^tP)v=0$ for all $v$...

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In general it is indeed true that if $\langle u, w \rangle=\langle v,w\rangle$ for all $w$ then $u$ must equal $v$. The other answers show one way of seeing this, and here is another: take $w$ to be each element of an orthonormal basis in turn, say $e_1\dots e_n$, then $\langle u, e_i \rangle$ is the $i^\mathrm{th}$ component of $u$ with respect to this basis. Hence, corresponding components of $u$ and $v$ are equal, so of course $u$ and $v$ are equal.

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