Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2009 / Algebra / 2

The orthogonal complement is $W^{\perp }=\{v\in V:⟨v,w⟩=0\text{ for every }w\in W\}$. Put $Pv={\sum }_i⟨v,e_i⟩e_i$, so $T=I-P$. Orthonormality gives $P^2=P$, and therefore $T^2=T$. Also $⟨Tv,e_j⟩=0$ for every $j$, so $Tv\in W^{\perp }$. Every $v=Pv+Tv$ is thus a sum from $W$ and $W^{\perp }$, and their intersection is zero, since a vector orthogonal to itself has squared norm zero. This proves $V=W\oplus W^{\perp }$. The decomposition is unique, so its two projection maps depend only on $W$, not on the chosen orthonormal basis.

Finite dimensionality of $W$ cannot simply be dropped. Take the real Hilbert space $V={\ell }^2(\mathbb{N};\mathbb{R})$ and $W=c_{00}$, the finitely supported sequences. This is an infinite-dimensional proper subspace. A vector orthogonal to every coordinate unit vector is zero, so $W^{\perp }=\{0\}$. Yet, for example, $(2^{-j}{)}_{j\ge 1}$ belongs to $V\setminus W$. Hence $W\oplus W^{\perp }=W\ne V$. The failure occurs even though $V$ itself is complete; the proper dense subspace $W$ is not closed.

For any $w\in W$, the decomposition $v-w=Tv+(Pv-w)$ is orthogonal. The Pythagorean identity givesEquality is attained at $w=Pv$. Thus $\boxed{\underset{w\in W}{\inf }\Vert v-w\Vert =\Vert Tv\Vert }$, and the minimizer in $W$ is uniquely $Pv$.

The assumption $v\notin W^{\perp }$ means $p=Pv\ne 0$. For every unit vector $w\in W$, orthogonality and Cauchy–Schwarz give $⟨v,w⟩=⟨p,w⟩\le \Vert p\Vert$. Equality is attained atEquality in the real Cauchy–Schwarz inequality requires $w$ to be a scalar multiple of $p$, and attaining the positive supremum fixes its sign; the unit norm fixes its magnitude. This proves uniqueness.