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
Topics
About: vectors space
The table of contents of this article
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.
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.
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
\(V'\): \(\in \{\text{ the } d' \text{ -dimensional } F \text{ vectors spaces }\}\), with any inner product, \(\langle \bullet, \bullet \rangle\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional vectors subspaces of } V'\}\)
\(S\): \(= \{s \in V' \vert \forall v \in V (\langle s, v \rangle = 0)\}\)
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Statements:
\(S \in \{\text{ the } (d' - d) \text{ -dimensional vectors subspaces of } V'\}\)
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2: Proof
Whole Strategy: Step 1: take an orthonormal basis for \(V\) and expand it to an orthonormal basis for \(V'\), 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 \(V'\) to \(\widetilde{B} = \{\widetilde{b_1}, ..., \widetilde{b_d}\}\) by the Gram-Schmidt orthonormalization: \(\widetilde{b_1} = b_1 / \sqrt{\langle b_1, b_1 \rangle}\); \(\widetilde{b_2} = (b_2 - \langle b_2, \widetilde{b_1} \rangle \widetilde{b_1}) / \sqrt{\langle b_2 - \langle b_2, \widetilde{b_1} \rangle \widetilde{b_1}, b_2 - \langle b_2, \widetilde{b_1} \rangle \widetilde{b_1} \rangle}\); ....
\(\widetilde{B}\) is linearly independent on \(V'\): if \(v^j \widetilde{b_j} = 0\) had a not-all-0 \(\{v^1, ..., v^d\} \subseteq F\), \(\widetilde{B}\) would not be linearly independent on \(V\).
\(\widetilde{B}\) can be expanded to be a basis for \(V'\), \(B' = \{\widetilde{b_1}, ..., \widetilde{b_d}, b_{d + 1}, ..., b_{d'}\}\), 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.
\(B'\) can be orthonormalized to \(\widetilde{B'} = \{\widetilde{b_1}, ..., \widetilde{b_d}, \widetilde{b_{d + 1}}, ..., \widetilde{b_{d'}}\}\) by the Gram-Schmidt orthonormalization: the \(\widetilde{B}\) part is not changed by the orthonormalization.
Now, let us see that the \(F\) vectors space spanned by \(\{\widetilde{b_{d + 1}}, ..., \widetilde{b_{d'}}\}\), \(W\), is nothing but \(S\).
For each \(s \in S\), \(s = \sum_{j \in \{1, ..., d'\}} s^j \widetilde{b_j}\), because \(s \in V'\). But for each \(l \in \{1, ..., d\}\), \(\langle s, \widetilde{b_l}\rangle = 0\), where the left hand side is \(\langle \sum_{j \in \{1, ..., d'\}} s^j \widetilde{b_j}, \widetilde{b_l}\rangle = \sum_{j \in \{1, ..., d'\}} s^j \langle\widetilde{b_j}, \widetilde{b_l}\rangle = \sum_{j \in \{1, ..., d'\}} s^j \delta_{j, l} = s_l\). So, \(s^l = 0\), which means that \(s \in W\).
For each \(w \in W\), \(w = \sum_{j \in \{d + 1, ..., d'\}} w^j \widetilde{b_j}\). For each \(v \in V\), \(v = \sum_{l \in \{1, ..., d\}} v^l \widetilde{b_l}\), and \(\langle w, v \rangle = \langle \sum_{j \in \{d + 1, ..., d'\}} w^j \widetilde{b_j}, \sum_{l \in \{1, ..., d\}} v^l \widetilde{b_j} \rangle = \sum_{j \in \{d + 1, ..., d'\}} w^j \sum_{l \in \{1, ..., d\}} \overline{v^l} \langle \widetilde{b_j}, \widetilde{b_l} \rangle = 0\), which means that \(w \in S\).
So, \(S = W\), the \((d' - d)\)-dimensional vectors subspace of \(V'\) with the basis, \(\widetilde{B'} \setminus \widetilde{B}\).
