A description/proof of that double dual of finite dimensional real vectors space is 'vectors spaces - linear morphisms' isomorphic to vectors space
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of vectors space.
- The reader knows a definition of dual of vectors space.
- The reader knows a definition of %category name% isomorphism.
- The reader admits the proposition that the dual space of any finite dimensional real vectors space constitutes a same dimensional vectors space.
- The reader admits the proposition that any linear surjection from any finite dimensional vectors space to any same dimensional vectors space is a 'vectors spaces - linear morphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that the double dual of any finite dimensional real vectors space is 'vectors spaces - linear morphisms' isomorphic to the original 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: Description
For any finite dimensional real vectors space, \(V\), the double dual, \(V^{**}\), is 'vectors spaces - linear morphisms' isomorphic to \(V\).
2: Proof
\(V^{**} = Hom (V^*, \mathbb{R})\), and for any \(f \in Hom (V^{*}, \mathbb{R})\), \(f: V^{*} \rightarrow \mathbb{R}, f' \mapsto f (f')\). \(V^{*} = Hom (V, \mathbb{R})\), and for any \(f' \in V^{*}\), \(f': V \rightarrow \mathbb{R}, v \mapsto f' (v)\), which means that with \(v\) fixed, there is a map, \(f'' (v): V^{*} \rightarrow \mathbb{R}, f' \mapsto f' (v)\). Is \(f'' (v)\) a vectors space homomorphism? For any \(r_1, r_2 \in \mathbb{R}\) and any \({f'}_1, {f'}_2 \in V^{*}\), \(f'' (v) (r_1 {f'}_1 + r_2 {f'}_2) = (r_1 {f'}_1 + r_2 {f'}_2) (v) = r_1 {f'}_1 (v) + r_2 {f'}_2 (v) = r_1 f'' (v) ({f'}_1) + r_2 f'' (v) ({f'}_2)\). So, yes, \(f'' (v)\) is a vectors space homomorphism. So, \(\{f'' (v)| v \in V\} \subseteq V^{**}\).
\(f'' (v)\) is linear with respect to \(v\), which means that \(f'' (r_1 v_1 + r_2 v_2) = r_1 f'' (v_1) + r_2 f'' (v_2)\), because \((f'' (r_1 v_1 + r_2 v_2)) (f') = f' (r_1 v_1 + r_2 v_2) = r_1 f' (v_1) + r_2 f' (v_2) = r_1 (f'' (v_1)) (f') + r_2 (f'' (v_2)) (f') = (r_1 f'' (v_1) + r_2 f'' (v_2)) (f')\).
As \(V\) is \(d\) dimensional, there is a basis, \(b_1, b_2, . . ., b_d\), and for any \(v \in V\), \(v = v^i b_i\). \(\{f'' (v)| v \in V\} = \{f'' (v^i b_i)| v^i \in \mathbb{R}\}\). \(\{f'' (v^i b_i)| v^i \in \mathbb{R}\}\) constitutes a vectors space, because \(r_1 f'' (v_1^i b_i) + r_2 f'' (v_2^i b_i) = r_1 v_1^i f'' (b_i) + r_2 v_2^i f'' (b_i) = (r_1 v_1^i + r_2 v_2^i) f'' (b_i) = f'' ((r_1 v_1^i + r_2 v_2^i) b_i)\). Is \(f'' (b_1), f'' (b_2), . . ., f'' (b_d)\) linearly independent? Suppose \(c^i f'' (b_i) = 0\) where \(c^i \in \mathbb{R}\). When \(b'^1, b'^2, . . ., b'^d\) is the dual basis for \(V^*\), \((c^i f'' (b_i)) (b'^j) = c^i (f'' (b_i)) (b'^j) = c^i b'^j (b_i) = c^i \delta^j_i = c^j = 0 (b'^j) = 0\). So, yes, \(f'' (b_1), f'' (b_2), . . ., f'' (b_d)\) is linearly independent. \(v^i f'' (b_i) = f'' (v^i b_i)\) covers \(\{f'' (v^i b_i)| v^i \in \mathbb{R}\}\). So, \(f'' (b_1), f'' (b_2), . . ., f'' (b_d)\) is a basis, and \(\{f'' (v^i b_i)| v^i \in \mathbb{R}\}\) is \(d\) dimensional. As \(V^{**}\) is a \(d\) dimensional vectors space by the proposition that the dual space of any finite dimensional real vectors space constitutes a same dimensional vectors space, \(\{f'' (v)| v \in V\} = (V^*)^*\).
For any \(v_1, v_2 \in V, v_1 \neq v_2\), \(f'' (v_1) \neq f'' (v_2)\), because \(f'' (v_1) = f'' (v_1^i b_i) = v_1^i f'' (b_i) \neq f'' (v_2) = f'' (v_2^i b_i) = v_2^i f'' (b_i)\), because \(f'' (b_1), f'' (b_2), . . ., f'' (b_d)\) is a basis.
So, there is the canonical bijection, \(f''': (V^*)^* = \{f'' (v)| v \in V\} \rightarrow V\). Is \(f'''\) linear? \(f''' (r_1 f'' (v_1) + r_2 f'' (v_2)) = f''' (f'' (r_1 v_1 + r_2 v_2)) = r_1 v_1 + r_2 v_2 = r_1 f''' (f'' (v_1)) + r_2 f''' (f'' (v_2))\). So, yes, \(f'''\) is linear. By the proposition that any linear surjection from any finite dimensional vectors space to any same dimensional vectors space is a 'vectors spaces - linear morphisms' isomorphism, \(f'''\) is a 'vectors spaces - linear morphisms' isomorphism.
3: Note
Although it is sometimes sloppily expressed like "The double dual of a finite dimensional vectors space is the original vectors space.", the double dual is not the same entity with the original vectors space, as the 2 have different meanings. They are just 'vectors spaces - linear morphisms' isomorphic, and having such a relation does not make 2 entities the same.
