A projection satisfying $\| Px \| \leq \|x\|$ for all $x$ is an orthogonal projection How to prove that if $V$ is a finite dimensional inner product space and $W$ a subspace of $V$, if $P$ is projection map ($P^2=P$) having $W$ as its range and is such that $\|Px\| \leq \|x\|$ for all $x \in V$, then $P$ is orthogonal projection of $V$ onto $W$.
 A: Any linear transformation $V \to V$ is uniquely specified by its restrictions on $W$ and $W^\perp$. Therefore, to establish that $P$ is the orthogonal projection on to $W$, it suffices to check that


*

*$P$ fixes all vectors in its range $W$; i.e., $Px = x$ for all $x \in W$.

*$P$ kills all vectors in $W^\perp$; i.e., $Px = 0$ for all $x \in W^\perp$.
Now (1.) is a general fact about any projection map. The proof is easy and I will skip it here, focusing on showing (2.) in complete detail. 
Fix $x \in V$ and let $\lambda \in \mathbb R$ be arbitrary. Observe that $P(x + \lambda Px) = (1+\lambda) Px$. Therefore, applying the given hypothesis on $x + \lambda Px$, we get $\| x + \lambda Px \|^2 \geqslant (1+\lambda)^2 \| P x \|^2$.
Expanding out both sides and cancelling one term, we get
$$ \| x \|^2 + 2\lambda \langle x, Px \rangle \geqslant \| P x \|^2 + 2 \lambda \| P x \
\|^2 . \tag{$\dagger$}$$ 
Since $\lambda$ is a free variable, $(\dagger)$ will hold for all $\lambda \in \mathbb R$ if and only if
$$
\begin{cases}
\| x\|^2 &\geqslant& \|Px\|^2,
\\ \langle x, Px \rangle &=& \| Px \|^2. 
\end{cases}
$$
We are interested in the second conclusion $\| Px \|^2 = \langle x, Px \rangle$; notice that we have an exact equality here, not an inequality. 
Finally, if $x \in W^\perp$, we have $\| Px \|^2 = 0$ since $x$ is orthogonal to $Px \in W$. This implies that $Px = 0$ for all $x \in W^\perp$, and we are done. 
