description/proof of that for vectors space with norm induced by inner product, subspace, and vector on superspace, vector on subspace whose distance to vector is minimum is unique and difference is perpendicular to subspace, and vector on subspace s.t. difference is perpendicular to subspace is unique and distance is minimum
Topics
About: vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space with the norm induced by any inner product, any subspace, and any vector on the superspace, if there is a vector on the subspace whose distance to the vector is the minimum, it is unique and the difference of the vectors is perpendicular to the subspace, and if there is a vector on the subspace such that the difference of the vectors is perpendicular to the subspace, it is unique and the distance is the minimum.
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
\(V'\): \(\in \{\text{ the } F \text{ vectors spaces } \}\), with any inner product, \(\langle \bullet, \bullet \rangle: V' \times V' \to F\), and the norm induced by \(\langle \bullet, \bullet \rangle\), \(\Vert \bullet \Vert\)
\(V\): \(\in \{\text{ the subspaces of } V'\}\)
\(v'\): \(\in V'\)
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Statements:
(
\(\exists v_0 \in V (\forall v \in V (\Vert v' - v_0 \Vert \le \Vert v' - v \Vert))\)
\(\implies\)
(
\(v_0 \text{ is unique }\)
\(\land\)
\(\forall v \in V (\langle v' - v_0, v \rangle = 0)\)
)
)
\(\land\)
(
\(\exists v_0 \in V (\forall v \in V (\langle v' - v_0, v \rangle = 0))\)
\(\implies\)
(
\(v_0 \text{ is unique }\)
\(\land\)
\(\forall v \in V (\Vert v' - v_0 \Vert \le \Vert v' - v \Vert)\)
)
)
//
2: Note
This proposition does not claim that there is such a \(v_0\).
Compare with 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.
3: Proof
Whole Strategy: Step 1: suppose that there is a \(v_0 \in V\) such that for each \(v \in V\), \(\Vert v' - v_0 \Vert \le \Vert v' - v \Vert\); Step 2: take any \(v \in V\) such that \(v \neq v_0\) and define \(r := \Vert v - v_0 \Vert\) and \(w := 1 / r (v - v_0)\), and take \(\widetilde{v} := \widetilde{r} e^{\widetilde{\theta} i} w + v_0\) for arbitrary \(\widetilde{r}\) and \(\widetilde{\theta}\) and see that \(\widetilde{v} \in V\); Step 3: expand \(\langle v' - \widetilde{v}, v' - \widetilde{v} \rangle = \langle v' - v_0 + v_0 - \widetilde{v}, v' - v_0 + v_0 - \widetilde{v} \rangle\), and see that \(\langle v' - v_0, w \rangle = 0\); Step 4: see that \(\langle v' - v_0, v_0 \rangle = 0\) and \(\langle v' - v_0, v \rangle = 0\); Step 5: see that \(v_0\) is unique; Step 6: suppose that there is a \(v_0 \in V\) such that for each \(v \in V\), \(\langle v' - v_0, v \rangle = 0\); Step 7: expand \(\langle v' - v, v' - v \rangle = \langle v' - v_0 + v_0 - v, v' - v_0 + v_0 - v \rangle\) and see that \(\langle v' - v_0, v' - v_0 \rangle \le \langle v' - v, v' - v \rangle\); Step 8: see that \(v_0\) is unique.
Step 1:
Let us suppose that there is a \(v_0 \in V\) such that for each \(v \in V\), \(\Vert v' - v_0 \Vert \le \Vert v' - v \Vert\).
Step 2:
Let \(v \in V\) be any such that \(v \neq v_0\).
Let us define \(r := \Vert v - v_0 \Vert\) and \(w := 1 / r (v - v_0) \in V\). Certainly, \(w \in V\), because \(1 / r \in F\) and \(v, v_0 \in V\).
Inevitably, \(\langle w, w \rangle = \langle 1 / r (v - v_0), 1 / r (v - v_0) \rangle = 1 / r^2 \langle v - v_0, v - v_0 \rangle = 1 / r^2 r^2 = 1\).
Let us take \(\widetilde{v} := \widetilde{r} e^{\widetilde{\theta} i} w + v_0 \in V\) for any \(\widetilde{r} \in \mathbb{R}\) such that \(0 \lt \widetilde{r}\) and any \(\widetilde{\theta} \in \mathbb{R}\) such that \(\widetilde{\theta} \in \{0, \pi\}\) when \(F = \mathbb{R}\) and \(0 \le \widetilde{\theta} \lt 2 \pi\) when \(F = \mathbb{C}\). Certainly, \(\widetilde{v} \in V\), because \(\widetilde{r} e^{\widetilde{\theta} i} \in F\) and \(w, v_0 \in V\).
Step 3:
Let us expand \(\langle v' - \widetilde{v}, v' - \widetilde{v} \rangle = \langle v' - v_0 + v_0 - \widetilde{v}, v' - v_0 + v_0 - \widetilde{v} \rangle\).
\(= \langle v' - v_0, v' - v_0 \rangle + \langle v' - v_0, v_0 - \widetilde{v} \rangle + \langle v_0 - \widetilde{v}, v' - v_0 \rangle + \langle v_0 - \widetilde{v}, v_0 - \widetilde{v} \rangle = \langle v' - v_0, v' - v_0 \rangle + \langle v' - v_0, - \widetilde{r} e^{\widetilde{\theta} i} w \rangle + \langle - \widetilde{r} e^{\widetilde{\theta} i} w, v' - v_0 \rangle + \langle - \widetilde{r} e^{\widetilde{\theta} i} w, - \widetilde{r} e^{\widetilde{\theta} i} w \rangle = \langle v' - v_0, v' - v_0 \rangle + \widetilde{r} e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle + \widetilde{r} e^{\widetilde{\theta} i} \langle - w, v' - v_0 \rangle + (- \widetilde{r} e^{\widetilde{\theta} i}) (- \widetilde{r} e^{- \widetilde{\theta} i}) \langle w, w \rangle = \langle v' - v_0, v' - v_0 \rangle + \widetilde{r} e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle + \widetilde{r} e^{\widetilde{\theta} i} \langle - w, v' - v_0 \rangle + \widetilde{r}^2\).
\(\widetilde{r} e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle\) can be made real non-positive by choosing a \(\widetilde{\theta}\).
\(\widetilde{r} e^{\widetilde{\theta} i} \langle - w, v' - v_0 \rangle = \overline{\widetilde{r} e^{- \widetilde{\theta} i}} \overline{\langle v' - v_0, - w \rangle} = \overline{\widetilde{r} e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle} = \widetilde{r} e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle\), because it has been made real above.
So, \(\langle v' - \widetilde{v}, v' - \widetilde{v} \rangle = \langle v' - v_0, v' - v_0 \rangle + 2 \widetilde{r} e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle + \widetilde{r}^2\).
As \(\langle v' - v_0, v' - v_0 \rangle \le \langle v' - \widetilde{v}, v' - \widetilde{v} \rangle\), \(0 \le 2 \widetilde{r} e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle + \widetilde{r}^2\), so, \(0 \le 2 e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle + \widetilde{r}\). While \(2 e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle\) is non-positive, \(\widetilde{r}\) can be chosen any smaller, and \(2 e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle = 0\) is implied: otherwise, by choosing a small enough \(\widetilde{r}\), \(2 e^{- \widetilde{\theta} i} \langle v' - v_0, - w \rangle + \widetilde{r} \lt 0\), a contradiction.
So, \(\langle v' - v_0, - w \rangle = 0\), so, \(\langle v' - v_0, w \rangle = 0\).
Step 4:
So, \(\langle v' - v_0, 1 / r (v - v_0) \rangle = 0\), so, \(\langle v' - v_0, v - v_0 \rangle = 0\).
Let us suppose that \(v_0 \neq 0\).
As \(v\) can be taken to be \(2 v_0\) (\(2 v_0 \neq v_0\)), \(\langle v' - v_0, v - v_0 \rangle = \langle v' - v_0, v_0 \rangle = 0\).
So, \(\langle v' - v_0, v \rangle = \langle v' - v_0, v - v_0 + v_0 \rangle = \langle v' - v_0, v - v_0 \rangle + \langle v' - v_0, v_0 \rangle = 0 + 0 = 0\).
Let us suppose that \(v_0 = 0\).
\(\langle v' - v_0, v \rangle = \langle v' - v_0, v - v_0 \rangle = 0\).
So, anyway, \(\langle v' - v_0, v \rangle = 0\).
When \(v = v_0\), \(\langle v' - v_0, v \rangle = \langle v' - v_0, v_0 \rangle = 0\) is seen above.
So, for any \(v \in V\), \(\langle v' - v_0, v \rangle = 0\).
Step 5:
Let us see that \(v_0\) is unique.
Let us suppose that there was another \(v_1 \neq v_0\).
As \(v_1\) satisfies the conditions for \(v_0\), we already know that \(v' - v_1\) is perpendicular to \(V\).
\(\langle v_1 - v_0, v_1 - v_0 \rangle = \langle v_1 - v' + v' - v_0, v_1 - v' + v' - v_0 \rangle = \langle v_1 - v', v_1 - v' \rangle + \langle v_1 - v', v' - v_0 \rangle + \langle v' - v_0, v_1 - v' \rangle + \langle v' - v_0, v' - v_0 \rangle = \langle v_1 - v', - v' \rangle + \langle v_1 - v', v' \rangle + \langle v' - v_0, - v' \rangle + \langle v' - v_0, v' \rangle\), because \(v_0, v_1 \in V\), \(= 0\), which implies that \(v_1 - v_0 = 0\), which implies that \(v_1 = v_0\).
So, there is not any other \(v_1\).
Step 6:
Let us suppose that there is a \(v_0 \in V\) such that for each \(v \in V\), \(\langle v' - v_0, v \rangle = 0\).
Step 7:
\(\langle v' - v, v' - v \rangle = \langle v' - v_0 + v_0 - v, v' - v_0 + v_0 - v \rangle = \langle v' - v_0, v' - v_0 \rangle + \langle v' - v_0, v_0 - v \rangle + \langle v_0 - v, v' - v_0 \rangle + \langle v_0 - v, v_0 - v \rangle = \langle v' - v_0, v' - v_0 \rangle + 0 + 0 + \langle v_0 - v, v_0 - v \rangle\), because \(v_0 - v \in V\), so, \(\langle v' - v_0, v' - v_0 \rangle \le \langle v' - v, v' - v \rangle\), because \(0 \le \langle v_0 - v, v_0 - v \rangle\).
Step 8:
\(v_0\) is unique by Step 5.
