The Hilbert orthogonal projection onto a closed subspace

Assume the Axiom of Countable Choice. Let $M$ be a closed linear subspace of a real or complex Hilbert space $H$ (Linear subspace of a vector space). By the orthogonal-decomposition theorem (Orthogonal decomposition by a closed subspace) every $x\in H$ has a unique representation

$x=P_{M}x+(x-P_{M}x),P_{M}x\in M,x-P_{M}x\in M^{\perp },$

and the Hilbert orthogonal projection onto $M$ is the map

$P_M:H⟶M\subseteq H,x⟼P_{M}x,$

assigning to $x$ its unique $M$-component. Its defining properties are therefore

$P_{M}x\in M,x-P_{M}x\in M^{\perp }(x\in H),$

which characterise $P_M$ uniquely: a map with these defining properties must agree with the $M$-component of the unique decomposition of each $x$.

Agreement with the finite-dimensional projection. If $V$ is a finite-dimensional inner-product space and $W\subseteq V$ a subspace, then For a subspace $W$ of a finite-dimensional inner product space, $V=W\oplus W^{\perp }$ writes $V=W\oplus W^{\perp }$ and The orthogonal projection $P_{W}v$ is the $W$-component in $V=W\oplus W^{\perp }$ defines $P_{W}v$ as the unique $W$-component of $v$; the defining properties displayed above are the same, so they define the same map on a finite-dimensional Hilbert space.

Range and kernel. $P_{M}x\in M$ for every $x$, and $P_{M}m=m$ for $m\in M$ because $m=m+0$ with $0\in M^{\perp }$; conversely $P_{M}x=0$ says exactly that $x=x-0\in M^{\perp }$. Hence the range of $P_M$ is $M$ and its kernel is $M^{\perp }$, and $P_M$ is the identity on $M$ and zero on $M^{\perp }$.