definition of \((0, 2)\)-tensor complement of subset of vectors space
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of \((p, q)\)-tensors space of vectors space.
Target Context
- The reader will have a definition of \((0, 2)\)-tensor complement of subset of vectors space.
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:
\( F\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\( t\): \(\in T^0_2 (V)\)
\( S\): \(\subseteq V\)
\(*S^\perp_t\): \(= \{v \in V \vert \forall s \in S (t (v, s) = 0)\}\), \(\in \{\text{ the vectors subspaces of } V\}\)
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Conditions:
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2: Note
\(S\) does not need to be any vectors subspace.
But when \(S \neq \emptyset\), \(S^\perp_t = (S)^\perp_t\), where \((S) = Span (S)\) (see Note for the definition of vectors subspace generated by subset of vectors space).
That is because for each \(v \in S^\perp_t\), for each \(s \in (S)\), \(t (v, s) = 0\), because \(s = r^1 s_1 + ... + r^n s_n\) where \(s_j \in S\), and \(t (v, s) = t (v, r^1 s_1 + ... + r^n s_n) = r^1 t (v, s_1) + ... + r^n t (v, s_n) = 0 + ... + 0 = 0\); for each \(v \in (S)^\perp_t\), for each \(s \in S\), \(t (v, s) = 0\), because \(s \in (S)\).
Let us see that \(S^\perp_t\) is indeed a vectors subspace of \(V\).
Let \(v_1, v_2 \in S^\perp_t\) and \(r_1, r_2 \in F\) be any.
\(r_1 v_1 + r_2 v_2 \in S^\perp_t\), because for each \(s \in S\), \(t (r_1 v_1 + r_2 v_2, s) = r_1 t (v_1, s) + r_2 t (v_2, s) = 0 + 0 = 0\).
So, \(S^\perp_t\) is a vectors subspace of \(V\), by the proposition that for any vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination.
Typically, \(t\) is non-degenerate and symmetric or antisymmetric, but this definition does not require so.
