Direct Sum of Subspace and Orthocomplement

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Theorem

Let $H$ be a Hilbert space.

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

Denote by $M^\perp$ its orthocomplement.


Then the direct sum $M \oplus M^\perp$ is isomorphic to $H$.


Proof

Assert that $U: M \oplus M^\perp \to H: \left({m, m^\perp}\right) \mapsto m + m^\perp$ is an isomorphism.

According to the definition of isomorphism, it is sufficient to check that $U$ is surjective and that:

$\left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H = \left\langle{ \left({m, m^\perp}\right), \left({n, n^\perp}\right) }\right\rangle_{M \oplus M^\perp}$


First the surjectivity:

Denote by $P$ the orthogonal projection on $M$.

Then for any $h \in H$, have $h = Ph + \left({h - Ph}\right)$.

By definition of $P$, $Ph \in M$.

Furthermore, by Properties of Orthogonal Projection, $h - Ph \perp M$; that is, $h - Ph \in M^\perp$.

It follows that $h = U \left({Ph, h - Ph}\right)$, showing that $U$ is a surjection.


It remains to check that $U$ preserves the inner product:

\(\ds \left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H\) \(=\) \(\ds \left\langle{ m + m^\perp, n + n^\perp}\right\rangle_H\)
\(\ds \) \(=\) \(\ds \left\langle{m, n}\right\rangle_H + \left\langle{m^\perp, n}\right\rangle_H + \left\langle{m, n^\perp}\right\rangle_H + \left\langle{m^\perp, n^\perp}\right\rangle_H\) Linearity of $\left\langle{\cdot, \cdot}\right\rangle_H$
\(\ds \) \(=\) \(\ds \left\langle{m, n}\right\rangle_H + \left\langle{m^\perp, n^\perp}\right\rangle_H\) Defining property of orthocomplement
\(\ds \) \(=\) \(\ds \left\langle{ \left({m, m^\perp}\right), \left({n, n^\perp}\right) }\right\rangle_{M \oplus M^\perp}\) Definition of $\left\langle{\cdot, \cdot}\right\rangle_{M \oplus M^\perp}$


Hence $U$ is an isomorphism, and so $M \oplus M^\perp$ is isomorphic to $H$.

$\blacksquare$