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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.

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  • $\begingroup$ No. Try $P(v)=-v$. $\endgroup$
    – Alex Fish
    Nov 5 '15 at 0:44
  • $\begingroup$ Oh yeah, I will rewrite the problem in another way. $\endgroup$
    – peter
    Nov 5 '15 at 0:46
  • $\begingroup$ If $\mathcal{P}$ is just a projection (not necessarily orthogonal) then we can NOT conclude that $v\in \mathcal{S}$. $\endgroup$
    – Ramiro
    Nov 5 '15 at 1:31
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    $\begingroup$ Let $\mathcal{S}= \{(x,y)\in \mathbf{R}^2 : y=0\}$. Let $\mathcal{P}$: $\mathbf{R}^2\rightarrow \mathcal{S}$ defined by $\mathcal{P}(x,y)=(x+y,0)$. Then we have that $\mathcal{P}$ is linear and $\mathcal{P}\circ \mathcal{P}= \mathcal{P}$, so $\mathcal{P}$ is a projection. However, $\mathcal{P}$ is NOT orthogonal and if we take $v=(0,1)$ we have $\mathcal{P}(v)=(1,0)$ and so $||\mathcal{P}(v)||_2=||v||_2$, but $v\notin \mathcal{S}$. $\endgroup$
    – Ramiro
    Nov 5 '15 at 1:32
  • $\begingroup$ Thank you guys. I think the orthogonality is necessary. I just set the space as a Hilbert space, thus it is nature to consider one orthogonal projection. Thank you for your comments! $\endgroup$
    – peter
    Nov 5 '15 at 1:42
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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 $$ w\text{ not an element of }\mathcal{S}=\operatorname{span}\begin{pmatrix}1 \\ 0\end{pmatrix} $$

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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 $v \in \text{im } P$.

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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 $$v=\sum_{j=1}^n<v,v_j>v_j=\sum_{j=1}^k<v,v_j>v_j=P(v)$$

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    $\begingroup$ Sorry, but when you write $P(v)=\sum_{j=1}^k<v,v_j>v_j$, you are assuming that $P$ is an ORTHOGONAL projection. However, we only know that $P$ is projection (not necessarily orthogonal). $\endgroup$
    – Ramiro
    Nov 5 '15 at 1:50

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