Orthogonal Projection onto Orthocomplement

Theorem

Let $H$ be a Hilbert space.

Let $A$ be a closed linear subspace of $H$.

Let $P_A: H \to H$ be the orthogonal projection onto $A$.

Let $P_{A^\perp}: H \to H$ be the orthogonal projection onto $A^\perp$, the orthocomplement of $A$.

Let $I: H \to H$ be the identity operator on $H$.

Then:

$P_{A^\perp} = I - P_A$

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Proof

Let $h \in H$.

By definition of orthogonal projection, $P_A h = a \in A$ such that $h - a \in A^\perp$.

Then the claim is that for all $h \in H$:

$\paren {I - P_A } h = h - P_A h$

is in $A^\perp$, and:

$h - \paren {h - P_A h} \in \paren {A^\perp}^\perp$

That $h - P_A h \in A^\perp$ is part of the definition of $P_A$.

For the other criterion, note:

$h - \paren {h - P_A h} = P_A h \in A$

By Double Orthocomplement is Closed Linear Span: Corollary:

$\paren{ A^\perp }^\perp = A$

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Exercise $2$