A description/proof of that dual of finite dimensional real vectors space constitutes same dimensional 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 admits the proposition that for any 2 vectors spaces over any same field, the set of vectors space homomorphisms constitutes a vectors space.
Target Context
- The reader will have a description and a proof of the proposition that the dual of any finite dimensional real vectors space constitutes a same dimensional 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 dual, \(V^*\), constitutes a same dimensional vectors space.
2: Proof
\(V^*\) is \(Hom (V, \mathbb{R})\), and by the proposition that for any 2 vectors spaces over any same field, the set of vectors space homomorphisms constitutes a vectors space, \(V^*\) constitutes a vectors space.
As \(V\) is \(d\) dimensional, there is a basis, \(b_1, b_2, . . ., b_d\), and for any \(v \in V\), \(v = v^i b_i\). Let us define an element, \(b'^i \in V^*\), as \(b'^i (v^j b_j) = v^i\) where \(i = 1, 2, ..., d\), which is indeed linear, because \(b'^i (r_1 v_1 + r_2 v_2) = b'^i (r_1 {v_1}^j b_j + r_2 {v_2}^j b_j) = r_1 {v_1}^i + r_2 {v_2}^i = r_1 b'^i (v_1) + r_2 b'^i (v_2)\). Is \(b'^1, b'^2, . . ., b'^d\) linearly independent? For \(c_i b'^i = 0\) where \(c_i \in \mathbb{R}\), \(c_i b'^i (b_j) = c_j = 0 (b_j) = 0\). So, yes, \(b'^1, b'^2, . . ., b'^d\) is linearly independent. Does \({v'}_i b'^i\) cover \(V^*\)? For any \(f \in V^*\), let us define \(f' = f (b_i) b'^i\). For any \(v \in V\), \(v = v^i b_i\), and \(f (v) = f (v^i b_i) = v^i f(b_i) = v^i f' (b_i) = f' (v^i b_i) = f' (v)\), so, \(f\) and \(f'\) are the same. So, yes, \({v'}_i b'^i\) covers \(V^*\). So, \(b'^1, b'^2, . . ., b'^d\) is a basis for \(V^*\), and \(V^*\) is a \(d\) dimensional vectors space.
