description/proof of that for finite-dimensional vectors space, non-degenerate \((0, 2)\)-tensor, and vectors subspace, dimension of space is dimension of subspace plus dimension of tensor complement of 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 dimension of vectors space.
- The reader knows a definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.
- The reader admits the proposition that for any vectors space and any linearly independent subset, the subset can be expanded to be a basis.
- The reader admits the rank-nullity law for linear map between finite-dimensional vectors spaces.
Target Context
- The reader will have a description and a proof of 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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(t\): \(\in \{\text{ the non-degenerate } (0, 2) \text{ -tensors of } V'\}\)
\(V\): \(\in \{\text{ the vectors subspaces of } V'\}\)
\(V^\perp_t\): \(= \text{ the } t \text{ complement of } V\)
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Statements:
\(Dim (V') = Dim (V) + Dim (V^\perp_t)\)
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2: Note
Typically, \(t\) is symmetric or antisymmetric, but this proposition does not require so, as will be seen in Proof.
3: Proof
Whole Strategy: Step 1: take the 'vectors spaces - linear morphisms' isomorphism, \(\widehat{t}: V' \to V'^*, v' \mapsto i_{v'} (t)\); Step 2: take \(f: V' \to V^*, v' \mapsto i_{v'} (t) \vert_V\), and see that \(f\) is a surjective linear map and \(V^\perp_t = Ker (f)\); Step 3: apply the rank-nullity law for linear map between finite-dimensional vectors spaces to see that \(Dim (V') = Dim (V^*) + Dim (V^\perp_t) = Dim (V) + Dim (V^\perp_t)\).
Step 1:
Let us take \(\widehat{t}: V' \to V'^*, v' \mapsto i_{v'} (t)\), which is a 'vectors spaces - linear morphisms' isomorphism: refer to the definition of non-degenerate \((0, 2)\)-tensor of vectors space.
Step 2:
Let us take \(f: V' \to V^*, v' \mapsto i_{v'} (t) \vert_V\).
That is well-defined, because \(i_{v'} (t) \vert_V \in V^*\), because \(i_{v'} (t) \vert_V = t (v', \bullet) \vert_V \in V^*\).
\(f\) is linear, because for each \(v'_1, v'_2 \in V'\) and \(r_1, r_2 \in F\), \(f (r_1 v'_1 + r_2 v'_2) = t (r_1 v'_1 + r_2 v'_2, \bullet) \vert_V = r_1 t (v'_1, \bullet) \vert_V + r_2 t (v'_2, \bullet) \vert_V = r_1 f (v'_1) + r_2 f (v'_2)\).
Let us see that \(f\) is surjective.
Let \(v^* \in V^*\) be any.
\(v^*\) can be extended to \(v'^* \in V'^*\) like this: let \(B = \{b_1, ..., b_d\}\) be any basis for \(V\); \(B\) is obviously linearly-independent on \(V'\) and can be expanded to a basis for \(V'\), \(B' = \{b_1, ..., b_d, b'_{d + 1}, ..., b'_{d'}\}\), by the proposition that for any vectors space and any linearly independent subset, the subset can be expanded to be a basis; let \(v'^*: V' \to F, r^1 b_1 + ... + r^d b_d + r^{d + 1} b'_{d + 1} + ... + r^{d'} b'_{d'} \mapsto v^* (r^1 b_1 + ... + r^d b_d)\), which is indeed in \(V'^*\), because it is linear, because for each \(v'_1 = r^1_1 b_1 + ... + r^d_1 b_d + r^{d + 1}_1 b'_{d + 1} + ... + r^{d'}_1 b'_{d'}, v'_2 = r^1_2 b_1 + ... + r^d_2 b_d + r^{d + 1}_2 b'_{d + 1} + ... + r^{d'}_2 b'_{d'} \in V'\) and \(s^1, s^2 \in F\), \(v'^* (s^1 v'_1 + s^2 v'_2) = v'^* (s^1 (r^1_1 b_1 + ... + r^d_1 b_d + r^{d + 1}_1 b'_{d + 1} + ... + r^{d'}_1 b'_{d'}) + s^2 (r^1_2 b_1 + ... + r^d_2 b_d + r^{d + 1}_2 b'_{d + 1} + ... + r^{d'}_2 b'_{d'})) = v'^* ((s^1 r^1_1 + s^2 r^1_2) b_1 + ... + (s^1 r^d_1 + s^2 r^d_2) b_d + (s^1 r^{d + 1}_1 + s^2 r^{d + 1}_2) b'_{d + 1} + ... + (s^1 r^{d'}_1 + s^2 r^{d'}_2) b'_{d'}) = v^* ((s^1 r^1_1 + s^2 r^1_2) b_1 + ... + (s^1 r^d_1 + s^2 r^d_2) b_d) = v^* (s^1 (r^1_1 b_1 + ... + r^d_1 b_d) + s^2 (r^1_2 b_1 + ... + r^d_2 b_d)) = s^1 v^* (r^1_1 b_1 + ... + r^d_1 b_d) + s^2 v^* (r^1_2 b_1 + ... + r^d_2 b_d) = s^1 v'^* (r^1_1 b_1 + ... + r^d_1 b_d + r^{d + 1}_1 b'_{d + 1} + ... + r^{d'}_1 b'_{d'}) + s^2 v'^* (r^1_2 b_1 + ... + r^d_2 b_d + r^{d + 1}_2 b'_{d + 1} + ... + r^{d'}_2 b'_{d'}) = s^1 v'^* (v'_1) + s^2 v'^* (v'_2)\).
\(v'^* \vert_V = v^*\), because for each \(v = r^1 b_1 + ... + r^d b_d \in V\), \(v'^* \vert_V (v) = v'^* (r^1 b_1 + ... + r^d b_d) = v^* (r^1 b_1 + ... + r^d b_d) = v^* (v)\).
As \(\widehat{t}: V' \to V'^*\) is bijective, there is a \(v' \in V'\) such that \(\widehat{t} (v') = v'^*\).
\(f (v') = i_{v'} (t) \vert_V = \widehat{t} (v') \vert_V = v'^* \vert_V = v^*\).
So, \(f\) is surjective.
\(V^\perp_t = Ker (f)\), because for each \(v' \in V^\perp_t\), for each \(v \in V\), \(t (v', v) = 0\), but \(t (v', v) = f (v') (v)\), so, \(f (v') = 0\); for each \(v' \in Ker (f)\), \(f (v') = 0\), which means that for each \(v \in V\), \(f (v') (v) = 0\), but \(f (v') (v) = t (v', v)\), so, \(v' \in V^\perp_t\).
Step 3:
By applying the rank-nullity law for linear map between finite-dimensional vectors spaces to \(f\), \(Dim (V') = Dim (Ker (f)) + Dim (f (V')) = Dim (V^\perp_t) + Dim (V^*)\).
But as \(Dim (V^*) = Dim (V)\) (refer to Note for the definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space), \(= Dim (V) + Dim (V^\perp_t)\).
