How to show $x \in R(P)$ when if $P$ is orthogonal projector?

Question

For an orthogonal projector $P$, if $\|Px\|_2=\|x\|_2$, show that $x \in R(P)$, where $R(P)$ is the range of $P$, $x \in \mathbb{R}^n$, and $P$ is an $n \times n$ matrix.

We need to show there exist $y \in \mathbb{R}^n$ such that $x=Py$.

My try:

Since $\|Px\|_2=\|x\|_2$, we get $\|Px\|_2^2=\|x\|_2^2$, i.e.,

$$ x^TP^TPx=x^Tx $$

How would you come up with $x=Py$?

Answer 1

$\|x\|^{2}=\|Px+(x-Px)\|^{2}=\|Px\|^{2}+\|x-Px\|^{2}$ because $Px$ is orthogonal to $(I-P)(x)$. So $x-Px=0$.