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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 $W$.

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+1 because I have fond memories of this problem. I met it as a student reading Axler's Linear Algebra Done Right. My memory is that it was the most difficult exercise in that book. – leslie townes Jan 7 '12 at 2:16
up vote 5 down vote accepted

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

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

  2. $P$ 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.

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thank You Sir for your help. I have a one question regarding this Answer. Are you using V is finite dimensional Some where other than in the existence of $W^{\bot}$. – Abcd J Jan 7 '12 at 12:33
@AbcdJ $W^\perp$ does exist always; but we want to decompose $V$ as a sum of $W$ and $W^\perp$ which is not possible in infinite-dimensional vector spaces. And on a related note: I do not know whether an orthogonal projection can be defined if I'm not given this. – Srivatsan Jan 7 '12 at 13:15
Actually, I want to ask if we are using finite dimensions of V just for the requirement that $V = W\oplus W^{\bot}.$ Then it should be true for closed subspace of Hilbert Space. Because for any closed subspace W of a Hilbert space V we have $V = W\oplus W^{\bot}.$ – Abcd J Jan 7 '12 at 19:09
@AbcdJ I think we need finite-dimensionality only for that decomposition to work. – Srivatsan Jan 10 '12 at 19:16

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