Orthogonal Projection in Hilbert Spaces 📂Hilbert Space

Orthogonal Projection in Hilbert Spaces

Definition¹

Let’s assume a closed subspace $V$ of a Hilbert space $H$ is given.

When $\mathbf{v} \in H$ is represented as $\mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2}$ with respect to $\mathbf{v}_{1} \in V$ and $\mathbf{v}_{2} \in V^{\perp}$ , a surjection $P :H \to V$ that satisfies the following is called an orthogonal projection.

$P \mathbf{v} = \mathbf{v}_{1}$

Explanation

Orthogonal projection has the following properties:

$P$ is linear, bounded, and $| P | = 1$ .
$P$ is a self-adjoint operator, i.e., $P^{ \ast } = P$ .
$P$ is idempotent, which means $P^{2} = P$ .

Orthogonal projection extended to the Hilbert space naturally covers orthogonal projection in matrix algebra, and it feels a bit more abstract in its definition.

Theorem

Let’s assume a [normalized orthogonal basis] $\left\{ \mathbf{e}_{k} \right\}_{k \in \mathbb{N}}$ of the closed subspace $V$ of the Hilbert space $H$ is given. For every $\mathbf{v} \in H$ , orthogonal projection $P : H \to V$ can be represented as follows.

$P \mathbf{v} = \sum_{k=1}^{\infty} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \mathbf{e}_{k}$

Proof

Since $\left\{ \mathbf{e}_{k} \right\}_{k=1}^{\infty}$ is a basis of $V$ , $P \mathbf{v} \in V$ is represented as follows for $\left\{ a_{k} \right\}_{k=1}^{\infty} \subset \mathbb{C}$ not equal to $a_{1} = \cdots = 0$ .

$P \mathbf{v} = \sum_{k =1}^{\infty} a_{k} \mathbf{e}_{k}$

Because of the orthogonality of $\left\{ \mathbf{e}_{k} \right\}_{k=1}^{\infty}$ , $\left\langle \mathbf{e}_{i} , \mathbf{e}_{i} \right\rangle = 1$ and therefore $\left\langle \mathbf{e}_{i} , \mathbf{e}_{j} \right\rangle = 0$ for $i \ne j$ .

$\left\langle P \mathbf{v} ,P \mathbf{v} \right\rangle = \sum_{k =1}^{\infty} a_{k}^{2}$

Meanwhile, due to the property $P^{ \ast }=P$ and $P^{2} = P$ from $P \mathbf{v} = \sum_{k =1}^{\infty} a_{k} \mathbf{e}_{k}$ ,

$\left\langle P \mathbf{v} ,P \mathbf{v} \right\rangle = \left\langle \mathbf{v} ,P^{ \ast }P \mathbf{v} \right\rangle = \left\langle \mathbf{v} ,P \mathbf{v} \right\rangle = \sum_{k =1}^{\infty} a_{k} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle$

Hence

$\sum_{k =1}^{\infty} a_{k}^{2} = \sum_{k =1}^{\infty} a_{k} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle$

In summary,

$\sum_{k =1}^{\infty} a_{k} \left( a_{k} - \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \right) = 0$

Therefore, for every $k \in \mathbb{N}$ , $a_{k} = \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle$ must hold.

$P \mathbf{v} = \sum_{k=1}^{\infty} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \mathbf{e}_{k} \qquad , \mathbf{v} \in H$

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