2025-10-05

1332: For Vectors Space with Inner Product, Orthonormal Subset Is Linearly Independent

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description/proof of that for vectors space with inner product, orthonormal subset is linearly independent

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 any inner product, any orthonormal subset is linearly independent.

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
\(S\): \(\in \{\text{ the orthonormal subsets of } V\}\)
//

Statements:
\(S \in \{\text{ the linearly independent subsets of } V\}\)
//


2: Proof


Whole Strategy: Step 1: take any finite subset of \(S\), \(S^` = \{s_1, ..., s_n\} \subseteq S\), and take \(r^1 s_1 + ... + r^n s_n = 0\), and see that \(r^j = 0\).

Step 1:

Let \(S^` = \{s_1, ..., s_n\} \subseteq S\) be any finite subset.

Let \(r^1 s_1 + ... + r^n s_n = 0\), where \(r^j \in F\).

Let \(j \in \{1, ..., n\}\) be any.

\(r^j = r^j \langle s_j, s_j \rangle = \langle r^1 s_1 + ... + r^n s_n, s_j \rangle = \langle 0, s_j \rangle = 0\).

So, \(S\) is linearly independent.


References


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