Range of Orthogonal Projection
Jump to navigation
Jump to search
Theorem
It has been suggested that this page be renamed. In particular: Image of Orthogonal Projection To discuss this page in more detail, feel free to use the talk page. |
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:
- $P_K \sqbrk H \subseteq K$
$\Box$
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:
- $K \subseteq P_K \sqbrk H$
$\Box$
Hence, by definition of set equality:
- $K = P_K \sqbrk H$
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text{(d)}$