description/proof of that for finite-dimensional vectors space, symmetric or antisymmetric non-degenerate \((0, 2)\)-tensor, and subspace, tensor complement of tensor complement of subspace is contained in subspace
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of non-degenerate \((0, 2)\)-tensor of vectors space.
- The reader knows a definition of \((0, 2)\)-tensor complement of subset of vectors space.
- The reader knows a definition of vectors subspace generated by subset of vectors space.
- The reader admits the proposition that for any finite-dimensional vectors space, any non-degenerate \((0, 2)\)-tensor, and any vectors subspace, the dimension of the space is the dimension of the subspace plus the dimension of the tensor complement of the subspace.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space, any symmetric or antisymmetric non-degenerate \((0, 2)\)-tensor, and any subspace, the tensor complement of the tensor complement of the subspace is contained in the 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(t\): \(\in \{\text{ the non-degenerate } (0, 2) \text{ -tensors of } V'\} \cap \{\text{ the symmetric or antisymmetric tensors }\}\)
\(V\): \(\in \{\text{ the vectors subspaces of } V'\}\)
\(V^\perp_t\): \(= \text{ the } t \text{ complement of } V\)
\({V^\perp_t}^\perp_t\): \(= \text{ the } t \text{ complement of } V^\perp_t\)
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Statements:
\({V^\perp_t}^\perp_t \subseteq V\)
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2: Note
For each \(v \in V\), for each \(v' \in V^\perp_t\), \(t (v', v) = 0\), directly by the definition of \(V^\perp_t\), but this proposition is claiming that for any \(v' \in V'\), if for each \(v'' \in V^\perp_t\), \(t (v', v'') = 0\), \(v' \in V\).
3: Proof
Whole Strategy: Step 1: suppose that there was a \(v' \in {V^\perp_t}^\perp_t \setminus V\); Step 2: take \(W := Span (V \cup \{v'\})\), and see that \(V^\perp_t = W^\perp_t\); Step 3: apply the proposition that for any finite-dimensional vectors space, any non-degenerate \((0, 2)\)-tensor, and any vectors subspace, the dimension of the space is the dimension of the subspace plus the dimension of the tensor complement of the subspace to see that \(Dim (V') = Dim (V) + Dim (V^\perp_t) = Dim (W) + Dim (W^\perp_t)\), and find a contradiction.
Step 1:
Let us suppose that there was a \(v' \in {V^\perp_t}^\perp_t \setminus V\).
Step 2:
Let us take \(W := (V \cup \{v'\})\), the vectors space generated by \(V \cup \{v'\}\), which is a vectors subspace of \(V'\), \(= Span (V \cup \{v'\})\): refer to Note for the definition of vectors subspace generated by subset of vectors space.
\(Dim (W) = Dim (V) + 1\), because for any basis for \(V\), \(B = \{b_1, ..., b_d\}\), \(\{b_1, ..., b_d, v'\}\) would be linearly independent, because for \(r^1 b_1 + ... + r^d b_d + r' v' = 0\), \(r' = 0\), because otherwise, \(v' = 1 / r' (r^1 b_1 + ... + r^d b_d) \in V\), a contradiction, so, \(r^1 b_1 + ... + r^d b_d = 0\), which would imply all the \(r^j\) s are \(0\).
Let us see that \(V^\perp_t = W^\perp_t\).
For each \(p \in V^\perp_t\), for each \(v \in V\), \(t (p, v) = 0\), but for each \(w \in W\), \(w = v + r' v'\) for a \(v \in V\) and an \(r' \in F\), and \(t (p, w) = t (p, v + r' v') = t (p, v) + r' t (p, v') = t (p, v) + \text{ or }- r' t (v', p) = 0 + \text{ or } - r' 0\), because \(v' \in {V^\perp_t}^\perp_t\) and \(p \in V^\perp_t\), \(= 0\), so, \(p \in W^\perp_t\).
For each \(p \in W^\perp_t\), for each \(w \in W\), \(t (p, w) = 0\), but for each \(v \in V\), as \(v \in W\), \(t (p, v) = 0\), so, \(p \in V^\perp_t\).
Step 3:
By the proposition that for any finite-dimensional vectors space, any non-degenerate \((0, 2)\)-tensor, and any vectors subspace, the dimension of the space is the dimension of the subspace plus the dimension of the tensor complement of the subspace, \(Dim (V') = Dim (V) + Dim (V^\perp_t) = Dim (W) + Dim (W^\perp_t) = Dim (V) + 1 + Dim (V^\perp_t)\), a contradiction.
So, there is no \(v' \in {V^\perp_t}^\perp_t \setminus V\).
That means that \({V^\perp_t}^\perp_t \subseteq V\).
