Range of Orthogonal Projection

Theorem
Let $H$ be a Hilbert space.

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

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

Then:


 * $P_K \sqbrk H = K$

where $P_K \sqbrk H$ denotes the image of $H$ under $P_K$.

Proof
We first show that $P_K \sqbrk H \subseteq K$.

Let $k \in P_K \sqbrk H$.

Then there exists $h \in H$ such that:


 * $\map {P_K} h = k$

From the definition of the orthogonal projection, we have:


 * $\map {P_K} h \in K$

so:


 * $h \in K$

giving:


 * $P_K \sqbrk H \subseteq K$

We now show that:


 * $K \subseteq P_K \sqbrk H$

Let $k \in K$.

Then, from Fixed Points of Orthogonal Projection:


 * $\map {P_K} k = k$

So:


 * $k \in P_K \sqbrk H$

and so:


 * $K \subseteq P_K \sqbrk H$

Hence, by definition of set equality:


 * $K = P_K \sqbrk H$