1069: For Vectors Space with Norm Induced by Inner Product, Finite-Dimensional Subspace, and Vector on Superspace, There Is Unique Vector on Subspace Whose Distance to Vector Is Minimum

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description/proof of that for vectors space with norm induced by inner product, finite-dimensional subspace, and vector on superspace, there is unique vector on subspace whose distance to vector is minimum

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of norm induced by inner product on real or complex vectors space.
• The reader admits 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.

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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 finite-dimensional subspace, and any vector on the superspace, there is the unique vector on the subspace whose distance to the vector is the minimum.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$V^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$, with any inner product, $⟨\bullet ,\bullet ⟩:V^{\prime }\times V^{\prime }\to F$, and the norm induced by $⟨\bullet ,\bullet ⟩$
$V$: $\in \{\text{ the }n\text{ -dimensional subspaces of }{V}^{\prime }\}$
$v^{\prime }$: $\in V^{\prime }$
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Statements:
$!\mathrm{\exists }{v}_0\in V(\mathrm{\forall }v\in V(\Vert v^{\prime }-v_0\Vert \le \Vert v^{\prime }-v\Vert ))$
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2: Note

By 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, $\mathrm{\forall }v\in V(⟨v^{\prime }-v_0,v⟩=0)$, and $v_0$ is such the unique one.

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.

$V^{\prime }$ does not need to be finite-dimensional.

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3: Proof

Whole Strategy: find a $v_0\in V$ such that $\mathrm{\forall }v\in V(⟨v^{\prime }-v_0,v⟩=0)$ and apply 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; Step 1: take any orthonormal basis of $V$, $B=\{b_1,...,b_n\}$; Step 2: define $v_0:=\sum _{j\in \{1,...,n\}}⟨v^{\prime },b_j⟩b_j$; Step 3: see that $\mathrm{\forall }v\in V(⟨v^{\prime }-v_0,v⟩=0)$ holds; Step 4: conclude the proposition.

Step 1:

Let us take any orthonormal basis of $V$, $B=\{b_1,...,b_n\}$, which is possible by Gram - Schmidt orthonormalization: for any basis, $B^{\prime }=\{b_1^{\prime },...,b_n^{\prime }\}$, $b_1:=1/\sqrt{⟨b_1^{\prime },b_1^{\prime }⟩}{b}_1^{\prime }$; $b_2:=1/\sqrt{⟨b_2^{\prime }-⟨b_2^{\prime },b_1⟩b_1,b_2^{\prime }-⟨b_2^{\prime },b_1⟩b_1⟩}(b_2^{\prime }-⟨b_2^{\prime },b_1⟩b_1)$; $b_3=1/\sqrt{⟨b_3^{\prime }-⟨b_3^{\prime },b_1⟩b_1-⟨b_3^{\prime },b_2⟩b_2,b_3^{\prime }-⟨b_3^{\prime },b_1⟩b_1-⟨b_3^{\prime },b_2⟩b_2⟩}(b_3^{\prime }-⟨b_3^{\prime },b_1⟩b_1-⟨b_3^{\prime },b_2⟩b_2)$; ....

Step 2:

Let us define $v_0:=\sum _{j\in \{1,...,n\}}⟨v^{\prime },b_j⟩b_j$.

Indeed, $v_0\in V$.

Step 3:

Let us see that $\mathrm{\forall }v\in V(⟨v^{\prime }-v_0,v⟩=0)$ holds.

$⟨v^{\prime }-v_0,b_l⟩=⟨v^{\prime }-\sum _{j\in \{1,...,n\}}⟨v^{\prime },b_j⟩b_j,b_l⟩=⟨v^{\prime },b_l⟩-\sum _{j\in \{1,...,n\}}⟨v^{\prime },b_j⟩⟨b_j,b_l⟩=⟨v^{\prime },b_l⟩-\sum _{j\in \{1,...,n\}}⟨v^{\prime },b_j⟩{\delta }_{j,l}=⟨v^{\prime },b_l⟩-⟨v^{\prime },b_l⟩=0$.

As any $v\in V$ is $v^j{b}_j$, $⟨v^{\prime }-v_0,v⟩=⟨v^{\prime }-v_0,v^j{b}_j⟩=\overline{v^j}⟨v^{\prime }-v_0,b_j⟩=0$.

Step 4:

By 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, $v_0$ is the unique one that satisfies the requirement of this proposition.

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References

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