definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product
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.
Target Context
- The reader will have a definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
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, \(\langle \bullet, \bullet \rangle: V \times V \to F\)
\( S\): \(\in \{\text{ the countable subsets of } V\}\), \(= \{s_1, s_2, ...\}\)
\(*\widetilde{S}\): \(\in \{\text{ the countable subsets of } V\}\), \(= \{b_1, b_2, ...\}\), as specified below
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Conditions:
\(b_n\) is defined iteratively:
\(\{\}\) has been already determined from \(\{\}\)
Suppose that \(\{b_1, ..., b_{n - 1}\}\) has been already determined from \(\{s_1, ..., s_{n'}\}\)
When there is the smallest \(\{s_1, ..., s_{n''}\}\) such that \(\{b_1, ..., b_{n - 1}, s_{n''}\}\) is linearly independent, \(b_n := (s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j) / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert\), and \(\{b_1, ..., b_n\}\) has been determined from \(\{s_1, ..., s_{n''}\}\)
Otherwise, the iteration is finished, and \(\widetilde{S} := \{b_1, ..., b_{n - 1}\}\)
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"countable" does not mean that it is infinite: any finite set is countable.
2: Note
Often, \(S\) is supposed to be linearly independent, but this definition does not require \(S\) to be linearly independent.
Let us see that \(\widetilde{S}\) is orthonormal, iteratively.
For \(\{b_1\}\), \(\langle b_1, b_1 \rangle = \langle s_{n''} / \Vert s_{n''} \Vert, s_{n''} / \Vert s_{n''} \Vert \rangle = 1 / \Vert s_{n''} \Vert^2 \langle s_{n''}, s_{n''} \rangle = 1 / \Vert s_{n''} \Vert^2 \Vert s_{n''} \Vert^2 = 1\).
So, \(\{b_1\}\) is orthonormal.
For \(\{b_1, b_2\}\), \(\langle b_1, b_2 \rangle = \langle b_1, (s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j) / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert \rangle = 1 / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert \langle b_1, (s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j) \rangle = 1 / \Vert s_{n''} - \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_1, s_{n''} \rangle - \overline{\langle s_{n''}, b_1 \rangle} \langle b_1, b_1\rangle) = 1 / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_1, s_{n''} \rangle - \overline{\langle s_{n''}, b_1 \rangle} 1) = 1 / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_1, s_{n''} \rangle - \overline{\langle s_{n''}, b_1 \rangle}) = 1 / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_1, s_{n''} \rangle - \langle b_1, s_{n''} \rangle) = 0\).
\(\langle b_2, b_2 \rangle = \langle (s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j) / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert, (s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j) / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert \rangle = 1 / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert^2 \langle s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j, s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \rangle = 1 / \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert^2 \Vert s_{n''} - \sum^1_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert^2 = 1\).
So, \(\{b_1, b_2\}\) is orthonormal.
Let us suppose that \(\{b_1, ..., b_{n - 1}\}\) is orthonormal.
For \(\{b_1, ..., b_n\}\), for each \(1 \le l \le n - 1\), \(\langle b_l, b_n \rangle = \langle b_l, (s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j) / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert \rangle = 1 / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert \langle b_l, s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \rangle = 1 / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_l, s_{n''} \rangle - \sum^{n - 1}_{j = 1} \overline{\langle s_{n''}, b_j \rangle} \langle b_l, b_j \rangle) = 1 / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_l, s_{n''} \rangle - \sum^{n - 1}_{j = 1} \overline{\langle s_{n''}, b_j \rangle} \delta_{l, j}) = 1 / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_l, s_{n''} \rangle - \overline{\langle s_{n''}, b_l \rangle}) = 1 / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert (\langle b_l, s_{n''} \rangle - \langle b_l, s_{n''} \rangle) = 0\).
\(\langle b_n, b_n \rangle = \langle (s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j) / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert, (s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j) / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert \rangle = 1 / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert^2 \langle s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j, s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \rangle = 1 / \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert^2 \Vert s_{n''} - \sum^{n - 1}_{j = 1} \langle s_{n''}, b_j \rangle b_j \Vert^2 = 1\).
So, \(\{b_1, ..., b_n\}\) is orthonormal.
Then, let \(b_l, b_m \in \widetilde{S}\) be any such that \(l \neq m\).
Let \(l \lt m\) without loss of generality.
So, \(b_l, b_m \in \{b_1, ..., b_m\}\).
As \(\{b_1, ..., b_m\}\) is orthonormal, \(\langle b_l, b_m \rangle = 0\).
\(\langle b_l, b_l \rangle = 1\).
