The Orthogonal Projection of a Hilbert Space onto a Closed Subset

The Orthogonal Projection of a Hilbert Space onto a Closed Subset

Definition: Let $H$ be a Hilbert space and let $M \subseteq H$ be a closed subspace so that $H = M \oplus M^{\perp}$. The Orthogonal Projection is the bounded linear operator $P : H \to H$ such that for each $x = m + m' \in H$ ($m \in M$, $m' \in M^{\perp}$), $P(x) = m$.

We now describe some properties of the orthogonal projection operator.

Theorem 1: Let $H$ be a Hilbert space and let $M \subseteq H$ be a nonzero closed subspace. Let $P : H \to H$ be the orthogonal projection of $H$ onto $M$. Then: a) $\| x \|^2 = \| P(x) \|^2 + \| (I - P)(x) \|^2$ for all $x \in H$. b) $\| P \| = 1$. c) $\langle P(x), y \rangle = \langle x, P(y) \rangle$ for all $x, y \in H$.

• Proof of a) Let $x \in H$. Then:

(1)

\begin{align} \quad \| x \|^2 &= \| P(x) + (I - P)(x) \|^2 \\ &= \langle P(x) + (I - P)(x), P(x) + (I - P)(x) \rangle \\ &= \langle P(x), P(x) \rangle + 2 \mathrm{Re} (\langle P(x), (I - P)(x) \rangle ) + \langle (I - P)(x), (I - P)(x) \rangle \\ &= \| P(x) \|^2 + 2 \mathrm{Re} (\langle P(x), (I - P)(x) \rangle ) + \| (I - P)(x) \|^2 \end{align}

• But $P(x) \in M$ and $(I - P)(x) \in M^{\perp}$ and so $\langle P(x), (I - P)(x) \rangle = 0$. So $\mathrm{Re} (\langle P(x), (I - P)(x) \rangle ) = 0$, and thus:

(2)

\begin{align} \quad \| x \|^2 = \| P(x) \|^2 + \| (I - P)(x) \|^2 \quad \blacksquare \end{align}

• Proof of b) For each $x \in H$ we have that:

(3)

\begin{align} \quad \| P(x) \| = \| m \| \leq \| m + m' \| = \| x \| \end{align}

• Therefore:

(4)

\begin{align} \quad \| P \| \leq 1 \quad (*) \end{align}

• Now since $M \neq \{ 0 \}$ there must exist an $m \in M$ such that $\| m \| = 1$. So $P(m) = m$, and $\| P(m) \| = \| m \| = 1$. Hence:

(5)

\begin{align} \quad \| P \| \geq 1 \quad (**) \end{align}

• From $(*)$ and $(**)$ we conclude that:

(6)

\begin{align} \quad \| P \| = 1 \quad \blacksquare \end{align}

• Proof of c) Let $x, y \in H$. Then:

(7)

\begin{align} \quad \langle P(x), y \rangle &= \langle P(x), P(y) + (I - P)(y) \rangle \\ &= \langle P(x), P(y) \rangle + \langle P(x), (I - P)(y) \rangle \end{align}

• Since $P(x) \in M$ and $(I - P)(y) \in M^{\perp}$ we have that $\langle P(x), (I - P)(y) \rangle = 0$, and so:

(8)

\begin{align} \quad \langle P(x), y \rangle = \langle P(x), P(y) \rangle \quad (***) \end{align}

• Similarly we have that:

(9)

\begin{align} \quad \langle x, P(y) \rangle &= \langle P(x) + (I - P)(x), P(y) \rangle \\ &= \langle P(x), P(y) \rangle + \langle (I - P)(x), P(y) \rangle \end{align}

• Since $(I - P)(x) \in M^{\perp}$ and $P(y) \in M$ we have that $\langle (I - P)(x), P(y) \rangle = 0$, and so:

(10)

\begin{align} \quad \langle x, P(y) \rangle = \langle P(x), P(y) \rangle \quad (****) \end{align}

• From $(***)$ and $(****)$ we conclude that for all $x, y \in H$:

(11)

\begin{align} \quad \langle P(x), y \rangle = \langle x, P(y) \rangle \quad \blacksquare \end{align}