1089: For Finite-Dimensional Vectors Space with Inner Product and Vectors Subspace, Set of Vectors Normal to Subspace Is Complementary-Dimensional Vectors Subspace

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description/proof of that for finite-dimensional vectors space with inner product and vectors subspace, set of vectors normal to subspace is complementary-dimensional vectors subspace

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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: Proof

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Starting Context

• The reader knows a definition of inner product on real or complex vectors space.
• The reader knows a definition of basis of module.
• The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space with any inner product and any vectors subspace, the set of the vectors normal to the subspace is a complementary-dimensional vectors subspace.

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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
$V^{\prime }$: $\in \{\text{ the }{d}^{\prime }\text{ -dimensional }F\text{ vectors spaces }\}$, with any inner product, $⟨\bullet ,\bullet ⟩$
$V$: $\in \{\text{ the }d\text{ -dimensional vectors subspaces of }{V}^{\prime }\}$
$S$: $=\{s\in V^{\prime }|\mathrm{\forall }v\in V(⟨s,v⟩=0)\}$
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Statements:
$S\in \{\text{ the }(d^{\prime }-d)\text{ -dimensional vectors subspaces of }{V}^{\prime }\}$
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2: Proof

Whole Strategy: Step 1: take an orthonormal basis for $V$ and expand it to an orthonormal basis for $V^{\prime }$, and see that $S$ is nothing but the vectors subspace spanned by the complementary elements of the $V$ basis.

Step 1:

As $V$ is a $d$-dimensional $F$ vectors space, there is a basis for $V$, $B=\{b_1,...,b_d\}$.

$B$ can be orthonormalized with respect to the inner product for $V^{\prime }$ to $\tilde{B}=\{\widetilde{b_1},...,\widetilde{b_d}\}$ by the Gram-Schmidt orthonormalization: $\widetilde{b_1}=b_1/\sqrt{⟨b_1,b_1⟩}$; $\widetilde{b_2}=(b_2-⟨b_2,\widetilde{b_1}⟩\widetilde{b_1})/\sqrt{⟨b_2-⟨b_2,\widetilde{b_1}⟩\widetilde{b_1},b_2-⟨b_2,\widetilde{b_1}⟩\widetilde{b_1}⟩}$; ....

$\tilde{B}$ is linearly independent on $V^{\prime }$: if $v^j\widetilde{b_j}=0$ had a not-all-0 $\{v^1,...,v^d\}\subseteq F$, $\tilde{B}$ would not be linearly independent on $V$.

$\tilde{B}$ can be expanded to be a basis for $V^{\prime }$, $B^{\prime }=\{\widetilde{b_1},...,\widetilde{b_d},b_{d+1},...,b_{d^{\prime }}\}$, by the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements.

$B^{\prime }$ can be orthonormalized to $\widetilde{B^{\prime }}=\{\widetilde{b_1},...,\widetilde{b_d},\widetilde{b_{d+1}},...,\widetilde{b_{d^{\prime }}}\}$ by the Gram-Schmidt orthonormalization: the $\tilde{B}$ part is not changed by the orthonormalization.

Now, let us see that the $F$ vectors space spanned by $\{\widetilde{b_{d+1}},...,\widetilde{b_{d^{\prime }}}\}$, $W$, is nothing but $S$.

For each $s\in S$, $s=\sum _{j\in \{1,...,d^{\prime }\}}{s}^j\widetilde{b_j}$, because $s\in V^{\prime }$. But for each $l\in \{1,...,d\}$, $⟨s,\widetilde{b_l}⟩=0$, where the left hand side is $⟨\sum _{j\in \{1,...,d^{\prime }\}}{s}^j\widetilde{b_j},\widetilde{b_l}⟩=\sum _{j\in \{1,...,d^{\prime }\}}{s}^j⟨\widetilde{b_j},\widetilde{b_l}⟩=\sum _{j\in \{1,...,d^{\prime }\}}{s}^j{\delta }_{j,l}=s_l$. So, $s^l=0$, which means that $s\in W$.

For each $w\in W$, $w=\sum _{j\in \{d+1,...,d^{\prime }\}}{w}^j\widetilde{b_j}$. For each $v\in V$, $v=\sum _{l\in \{1,...,d\}}{v}^l\widetilde{b_l}$, and $⟨w,v⟩=⟨\sum _{j\in \{d+1,...,d^{\prime }\}}{w}^j\widetilde{b_j},\sum _{l\in \{1,...,d\}}{v}^l\widetilde{b_j}⟩=\sum _{j\in \{d+1,...,d^{\prime }\}}{w}^j\sum _{l\in \{1,...,d\}}\overline{v^l}⟨\widetilde{b_j},\widetilde{b_l}⟩=0$, which means that $w\in S$.

So, $S=W$, the $(d^{\prime }-d)$-dimensional vectors subspace of $V^{\prime }$ with the basis, $\widetilde{B^{\prime }}\setminus \tilde{B}$.

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References

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