description/proof of that for real or complex vectors space with inner product, linear combination of finite vectors cannot be perpendicular to each constituent without being 0
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 description and a proof of the proposition that for any real or complex vectors space with any inner product, any linear combination of any finite vectors cannot be perpendicular to each constituent without being 0.
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:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(\langle \bullet, \bullet\rangle\): \(: V \times V \to F\), \(\in \{\text{ the inner products on } V\}\)
\(\{v_1, ..., v_k\}\): \(\subseteq V\)
\(v\): \(= r^1 v_1 + ... + r^k v_k \in V\), where \(r^j \in F\)
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Statements:
\(\forall j \in \{1, ..., k\} (\langle v, v_j \rangle = 0)\)
\(\implies\)
\(v = 0\)
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2: Proof
Let us suppose that \(\forall j \in \{1, ..., k\} (\langle v, v_j \rangle = 0)\).
\(\langle v_j, v \rangle = \overline{\langle v, v_j \rangle} = 0\).
\(\langle v, v \rangle = \langle r^1 v_1 + ... + r^k v_k, v \rangle = r^1 \langle v_1, v \rangle + ... + r^k \langle v_k, v \rangle = r^1 0 + ... + r^k 0 = 0\), which implies that \(v = 0\), by the definition of inner product.
