Orthogonal Projection is Linear Transformation

Theorem
Let $\mathbb F \in \set {\R, \C}$.

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

Let $\KK$ be a closed linear subspace of $\HH$.

Let $P_\KK$ denote the orthogonal projection on $\KK$.

Then $P_\KK$ is a linear transformation on $\HH$.

Proof
Let $x, y \in \HH$.

Let $\alpha, \beta \in \mathbb F$.

Let $k \in \KK$.

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


 * $\innerprod {\paren {\alpha x + \beta y} - \paren {\alpha \map {P_\KK} x + \beta \map {P_\KK} y} } k = \alpha \innerprod {x - \map {P_\KK} x} k + \beta \innerprod {y - \map {P_\KK} y} k$

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


 * $\alpha \innerprod {x - \map {P_\KK} x} k = 0$

and:


 * $\beta \innerprod {y - \map {P_\KK} y} k = 0$

Since $k$ was arbitrary, we have:


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

for all $k \in \KK$.

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


 * $\map {P_\KK} {\alpha x + \beta y}$ is the unique point in $\HH$ such that $\innerprod {\paren {\alpha x + \beta y} - \map {P_\KK} {\alpha x + \beta y} } k = 0$ for each $k \in \KK$.

So we therefore have:


 * $\alpha \map {P_\KK} x + \beta \map {P_\KK} y = \map {P_\KK} {\alpha x + \beta y}$

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

So $P_\KK$ is a linear transformation, as required.