A description/proof of that linear surjection from finite dimensional vectors space to same dimensional vectors space is 'vectors spaces - linear morphisms' isomorphism
Topics
About: vectors space
About: category
The table of contents of this article
Starting Context
- The reader knows a definition of vectors space.
- The reader knows a definition of surjection.
- The reader knows a definition of %category name% isomorphism.
Target Context
- The reader will have a description and a proof of 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.
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 vectors space, \(V_1\), and any same dimensional vectors space, \(V_2\), any linear surjection, \(f: V_1 \rightarrow V_2\), is a 'vectors spaces - linear morphisms' isomorphism.
2: Proof
For any basis for \(V_1\), \((b_1, b_2, . . ., b_d)\), \(f (b_1), f (b_2), . . ., f (b_d)\) are linearly independent on \(V_2\), because otherwise, \(V_2 = {f (c^i b_i)} = {c^i f (b_i)}\) would not be d dimensional, and \(f (b_1), f (b_2), . . ., f (b_d)\) spans \(V_2\), because \(V_2 = {c^i f (b_i)}\), so, \(f (b_1), f (b_2), . . ., f (b_d)\) is a basis for \(V_2\). \(f\) is injective, because for different vectors on \(V_1\), \(c^i b_i\) and \(c'^i b_i\), \(c^i \neq c'^i\) for an \(i\), so, \(f (c^i b_i) = c^i f (b_i) \neq c'^i f (b_i) = f (c'^i b_i)\), so, \(f\) is a bijection. The inverse, \(f^{-1}\), is linear, because \(f^{-1} (c^i f (b_i)) = c^i b_i = c^i f^{-1} (f (b_i))\). So, the pair of \(f\) and \(f^{-1}\) constitutes an isomorphism of the category of 'all the vectors spaces and all the linear morphisms'.
