Orthogonal Projection is Linear Transformation

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Theorem

Let $\GF \in \set {\R, \C}$.


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Let $H$ be a Hilbert space over $\mathbb F$ with inner product $\innerprod \cdot \cdot$.

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

Let $P_K$ denote the orthogonal projection on $K$.


Then $P_K$ is a linear transformation on $H$.


Proof

Let $x, y \in H$.

Let $\alpha, \beta \in \GF$.

Let $k \in \KK$.


Since the inner product is linear in its first argument, we have:

$\innerprod {\paren {\alpha x + \beta y} - \paren {\alpha \map {P_K} x + \beta \map {P_K} y} } k = \alpha \innerprod {x - \map {P_K} x} k + \beta \innerprod {y - \map {P_K} y} k$

From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have:

$\alpha \innerprod {x - \map {P_K} x} k = 0$

and:

$\beta \innerprod {y - \map {P_K} y} k = 0$

Since $k$ was arbitrary, we have:

$\innerprod {\paren {\alpha x + \beta y} - \paren {\alpha \map {P_K} x + \beta \map {P_K} y} } k = 0$

for all $k \in K$.

Note that from Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space:

$\map {P_K} {\alpha x + \beta y}$ is the unique point in $H$ such that $\innerprod {\paren {\alpha x + \beta y} - \map {P_K} {\alpha x + \beta y} } k = 0$ for each $k \in K$.

So we therefore have:

$\alpha \map {P_K} x + \beta \map {P_K} y = \map {P_K} {\alpha x + \beta y}$

for each $x, y \in H$ and $\alpha, \beta \in \mathbb F$.

So $P_K$ is a linear transformation, as required.

$\blacksquare$


Sources

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