Is one projection which preserves norm is identity? Given an orthogonal projection $\mathcal{P}$: $\mathbf{R}^{n}\rightarrow \mathcal{S}$, where $\mathcal{S}\text{ is one subspace of  } \mathbf{R}^{n}$. And let $||\cdot||_2$ denote $l_2$ norm. Then for one vector $v\in \mathbf{R}^{n}$, if $||\mathcal{P}(v)||_2=||v||_2$, can we conclude that: $v\in \mathcal{S}$ ?
Thank you.
 A: Ramiro is right in his comment when he says that 

if $\mathcal{P}$ is just a projection (not necessarily orthogonal)
  then we can NOT conclude that $v\in \mathcal{S}$

for this to see, we construct a simple counterexample. We take as a projection
$$
\mathcal{P}=\begin{pmatrix}1 & -1 \\ 0 & 0\end{pmatrix}
$$
operating on $\mathbb{R}^2$ and subspace generated by $$v=\begin{pmatrix}1 \\ 0\end{pmatrix}$$
so $\mathcal{S}=\operatorname{span}(v)$. Now we take 
$$
w=\begin{pmatrix}0 \\ -1\end{pmatrix}, ||w||_2=1
$$
but we also have 
$$
\mathcal{P}w=\begin{pmatrix}1 & -1 \\ 0 & 0\end{pmatrix}w=\begin{pmatrix}1\\ 0\end{pmatrix}\implies ||\mathcal{P}w||_2= ||w||_2=1
$$
but we have 
$$
w\text{ not an element of }\mathcal{S}=\operatorname{span}\begin{pmatrix}1 \\ 0\end{pmatrix}
$$
A: A projection is a map of the form $P^2=P$. Every projection splits $\mathbb{R}^n$ in the following way:
$$\mathbb{R}^n \cong \ker P\oplus \text{im } P,$$
which is easily seen from the fact that $x=(I-P)x+Px$.
An orthogonal projection is a projection for which $\ker P$ and $\text{im }P$ are orthogonal. Hence, if you are in an orthogonal projection,
$$\|P(v)\|=\|P(v_k+v_i)\|=\|P(v_k)+P(v_i)\|=\|P(v_i)\|=\|v_i\|.$$
Therefore, if $\|P(v)\|=\|v\|$, then $\|v\|=\|v_i\|$. Since $\|v\|=\|v_i\|+\|v_k\|$ (here we use the orthogonal assumption), then $\|v_k\|=0$, which implies $v=v_i$, and then that $v \in \text{im } P$.
A: Sorry, I missed that $P$ is a projection. In that case it's correct. Let $\{v_1,v_2,...,v_k\}$ be an orthonormal basis for $S$. Let us complete it to an orthonormal basis for $\mathbb{R}^n$ $\{v_1,v_2,...,v_k,v_{k+1},...,v_n\}$. We know $v=\sum_{j=1}^n<v,v_j>v_j$ and $P(v)=\sum_{j=1}^k<v,v_j>v_j$. Now $||v||_2^2=||\sum_{j=1}^n<v,v_j>v_j||_2^2=\sum_{j=1}^n|<v,v_j>|^2$ and $||P(v)||_2^2=||\sum_{j=1}^k<v,v_j>v_j||_2^2=\sum_{j=1}^k|<v,v_j>|^2$. With these represantations it is easy to see that $||P(v)||_2=||v||_2 \iff \forall j\in \{k+1,...,n\}$ $|<v,v_j>|=0 \iff \forall j\in \{k+1,...,n\}$ $<v,v_j>=0$ which implies $$v=\sum_{j=1}^n<v,v_j>v_j=\sum_{j=1}^k<v,v_j>v_j=P(v)$$
