Kernel 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:

$\ker P_K = K^\bot$

where:

$\ker P_K$ denotes the kernel of $P_K$

$K^\bot$ denotes the orthocomplement of $K$.

Proof

We first prove that:

$\ker P_K \subseteq K^\bot$

Let $h \in \ker P_K$.

Then:

$\map {P_K} h = 0$

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

$h - \map {P_K} h \in K^\bot$

That is:

$h \in K^\bot$

So:

$\ker P_K \subseteq K^\bot$

We now prove that:

$K^\bot \subseteq \ker P_K$

Let $h \in K^\bot$.

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

$\map {P_K} h$ is the unique point such that $h - \map {P_K} h \in K^\bot$

So, since $h - 0 \in K^\bot$, we have $\map {P_K} h = 0$.

So, we have $h \in \ker P_K$, and we obtain:

$K^\bot = \ker P_K$

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text {(d)}$