description/proof of that for Hilbert space and its closed vectors subspace, space is vectors space as direct sum of subspace and its orthogonal complement
Topics
About: Hilbert space
The table of contents of this article
Starting Context
- The reader knows a definition of Hilbert space.
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of orthogonal complement of subset of vectors space with inner product.
- The reader knows a definition of module as direct sum of finite number of submodules.
- The reader admits the proposition that for any vectors space with the topology induced by the metric induced by the norm induced by any inner product, the orthogonal complement of any subset of the vectors space is a closed vectors subspace.
- The reader admits the local criterion for openness.
- The reader admits the proposition that for any Hilbert space, any nonempty closed convex subset, and any point on the Hilbert space, there is the unique point on the subset whose distance to the point is the minimum.
Target Context
- The reader will have a description and a proof of the proposition that for any Hilbert space and its any closed vectors subspace, the space is the vectors space as direct sum of the subspace and its orthogonal complement.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V'\): \(\in \{\text{ the } F \text{ Hilbert spaces }\}\)
\(V\): \(\in \{\text{ the closed vectors subspaces of } V'\}\)
\(V^\perp\): \(= \text{ the orthogonal complement of } V\)
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Statements:
\(V' = \text{ the direct sum of } V, V^\perp\)
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2: Proof
Whole Strategy: Step 1: see that \(V^\perp\) is a closed vectors subspace of \(V'\); Step 2: see that \(V \cap V^\perp = \{0\}\); Step 3: when \(\{0\} \subset V \subset V'\), take any \(v' \in V' \setminus V\), take \(C := v' + V\), see that \(C\) is a nonempty convex closed subset of \(V'\), take the unique \(c \in C\) such that \(\Vert c \Vert = inf_{c' \in C} \Vert c' \Vert\), and see that \(c \in V^\perp\); Step 4: conclude the proposition.
Step 1:
\(V^\perp\) is a closed vectors subspace of \(V'\), by the proposition that for any vectors space with the topology induced by the metric induced by the norm induced by any inner product, the orthogonal complement of any subset of the vectors space is a closed vectors subspace.
Step 2:
Let us see that \(V \cap V^\perp = \{0\}\).
Let \(v \in V \cap V^\perp\) be any.
\(\langle v, v \rangle = 0\).
So, \(v = 0\).
Step 3:
When \(V = V'\), whatever \(V^\perp\) is, \(V' = V + V^\perp\).
When \(V = \{0\}\), \(V^\perp = V'\), so, \(V' = V + V^\perp\).
Let us suppose that \(\{0\} \subset V \subset V'\) hereafter.
When \(v' \in V\), \(v' = v' + 0 \in V + V^\perp\).
Let \(v' \in V' \setminus V\) be any.
Let us take \(C := v' + V\).
\(C\) is nonempty, because \(v' + 0 = v' \in C\).
\(C\) is convex, because for each \(c_1, c_2 \in C\), \(c_1 + t (c_2 - c_1) \in C\), because \(c_1 = v' + v_1\) and \(c_2 = v' + v_2\) where \(v_1, v_2 \in V\), and \(c_1 + t (c_2 - c_1) = v' + v_1 + t (v' + v_2 - (v' + v_1)) = v' + v_1 + t (v_2 - v_1)\), but \(v_1 + t (v_2 - v_1) \in V\).
\(C\) is a closed subset of \(V'\), because for each \(p \in V' \setminus C\), \(p - v' \in V' \setminus V\), but as \(V' \setminus V\) is open, there is an open ball around \(p - v'\), \(B_{p - v', \epsilon}\), such that \(B_{p - v', \epsilon} \subseteq V' \setminus V\), and \(v' + B_{p - v', \epsilon} \subseteq V' \setminus C\), because if \(v' + p' \in C\) for a \(p' \in B_{p - v', \epsilon}\), \(v' + p' = v' + v\) for a \(v \in V\), and \(p' = v \in V\), a contradiction against \(p' \in B_{p - v', \epsilon} \subseteq V' \setminus V\), but \(v' + B_{p - v', \epsilon} = B_{p, \epsilon}\) obviously, so, \(V' \setminus C\) is open, by the local criterion for openness.
By the proposition that for any Hilbert space, any nonempty closed convex subset, and any point on the Hilbert space, there is the unique point on the subset whose distance to the point is the minimum, there is the unique \(c \in C\) such that \(\Vert c \Vert = \Vert c - 0 \Vert = inf_{c' \in C} \Vert c' - 0 \Vert = inf_{c' \in C} \Vert c' \Vert\).
For each \(v \in V\) such that \(v \neq 0\) and each \(\lambda \in F\), \(c' = c + \lambda v \in C\).
\(\Vert c' \Vert^2 = \Vert c + \lambda v \Vert^2 = \langle c + \lambda v, c + \lambda v \rangle = \langle c, c \rangle + \langle c, \lambda v \rangle + \langle \lambda v, c \rangle + \langle \lambda v, \lambda v \rangle = \langle c, c \rangle + \langle c, \lambda v \rangle + \overline{\langle c, \lambda v \rangle} + \langle \lambda v, \lambda v \rangle = \langle c, c \rangle + 2 Re (\langle c, \lambda v \rangle) + \langle \lambda v, \lambda v \rangle = \langle c, c \rangle + 2 Re (\overline{\lambda} \langle c, v \rangle) + \vert \lambda \vert^2 \langle v, v \rangle\).
When \(F = \mathbb{R}\), take \(\langle c, v \rangle = r\) and \(\lambda = s\), then, \(= \langle c, c \rangle + 2 s r + s^2 \langle v, v \rangle\).
When \(F = \mathbb{C}\), take \(\langle c, v \rangle = r e^{i \theta}\) and \(\lambda = s e^{i \theta}\), then, \(= \langle c, c \rangle + 2 Re (s e^{- i \theta} r e^{i \theta}) + s^2 \langle v, v \rangle = \langle c, c \rangle + 2 Re (s r) + s^2 \langle v, v \rangle = \langle c, c \rangle + 2 s r + s^2 \langle v, v \rangle\).
Anyway, \(= \langle c, c \rangle + \langle v, v \rangle (s^2 + 2 s r / \langle v, v \rangle) = \langle c, c \rangle + \langle v, v \rangle ((s + r / \langle v, v \rangle)^2 - (r / \langle v, v \rangle)^2)\).
If \(r \neq 0\), by nonzero \(s\) such that \(s + r / \langle v, v \rangle = 0\), \(= \langle c, c \rangle + \langle v, v \rangle 0 - (r / \langle v, v \rangle)^2 = \langle c, c \rangle - (r / \langle v, v \rangle)^2 = \Vert c' \Vert^2 \lt \langle c, c \rangle = \Vert c \Vert^2\), a contradiction against \(\Vert c \Vert = inf_{c' \in C} \Vert c' \Vert\).
So, \(r = 0\), so, \(\langle c, v \rangle = 0\).
So, \(c \in V^\perp\).
As \(c \in C = v' + V\), \(c = v' + v\) for a \(v \in V\).
So, \(v' = - v + c \in V + V^\perp\).
As \(v' \in V' \setminus V\) is arbitrary, \(V' = V + V^\perp\).
Step 4:
By Step 1 and Step 3, \(V'\) is the vectors space as direct sum of \(V\) and \(V^\perp\).
