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Projection onto a finite-dimensional subspace of an arbitrary real inner-product space

Statement

Let $E$ be a real inner-product space, let $M \subseteq E$ be a finite-dimensional subspace, and let $f \in E$ . There is a unique $p \in M$ such that $f - p$ is orthogonal to every vector of $M$ . This vector is the unique nearest point in $M$ to $f$ , and it satisfies $∥ p ∥ \leq ∥ f ∥$ . The conclusion does not require $E$ to be complete.

Statement references

Sheldon Axler, Linear Algebra Done Right, 4th ed., Result 6.61

Facts & Assumptions

Given: A real inner-product space $E$ , a finite-dimensional subspace $M \subseteq E$ , and $f \in E$ .

[F1]

In a finite-dimensional inner-product space $V$ , the orthogonal projection of a vector onto a subspace is its unique nearest point (The orthogonal projection is the unique nearest point in the subspace).

[F2]

In a finite-dimensional inner-product space $V$ , the orthogonal projection $P_{W} f$ belongs to $W$ and $f - P_{W} f \in W^{⊥}$ (The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ ).

[F3]

If $u ⊥ v$ , then $∥ u + v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2}$ (Pythagoras, the parallelogram identity, and the real and complex polarisation identities).

Proof

1.1givenF1construct

Set $V : = span (M \cup {f})$ . Since $M$ is finite-dimensional and adjoining one vector increases dimension by at most one, $V$ is a finite-dimensional real inner-product space with the inherited inner product; $M$ is a subspace of $V$ and $f \in V$ . Apply [F1] inside $V$ and set $p : = P_{M} f$ .

2.1step 1.1F2

By [F2], $p \in M$ and $f - p \in M^{⊥}$ .

3.1step 2.1F1F3algebra∎

Since $p \in M$ and $f - p \in M^{⊥}$ , Pythagoras [F3] gives $∥ f ∥^{2} = ∥ p ∥^{2} + ∥ f - p ∥^{2} \geq ∥ p ∥^{2}$ , so $∥ p ∥ \leq ∥ f ∥$ . If $q \in M$ also has $f - q \in M^{⊥}$ , then for every $r \in M$ , [F3] gives $∥ f - r ∥^{2} = ∥ f - q ∥^{2} + ∥ q - r ∥^{2} \geq ∥ f - q ∥^{2}$ . Thus $q$ is a nearest point to $f$ , and [F1] forces $q = p$ .

Depends on

Used by

Dependency tree · two levels

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Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., Definition 6.55 and Result 6.61 (standard reference, not scraped)