A description/proof of that linear image of finite dimensional vectors space is vectors space
Topics
About: vectors space
About: map
The table of contents of this article
Starting Context
- The reader knows a definition of vectors space.
- The reader knows a definition of linear map.
Target Context
- The reader will have a description and a proof of the proposition that the image of any finite dimensional vectors space under any linear map is a 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 vectors space, \(V_1\), any vectors space, \(V_2\), and any linear map, \(f: V_1 \rightarrow V_2\), the image of the map, \(f (V_1)\) is a vectors space.
2: Proof
For any basis for \(V_1\), \(b_1, b_2, . . ., b_d\), \(f (b_1), f (b_2), . . ., f (b_d)\) may be or not be linearly independent on \(V_2\), but if not, an \(f (b_i)\) is a linear combination of the rest, which may be or not be linearly independent, but if not, an \(f (b_i)\) in the rest is a linear combination of the new rest, and so on, and after all, \(f (b_1), f (b_2), . . ., f (b_r) = (b'_1, b'_2, . . ., b'_r)\) is linearly independent, with the indexes of the basis for \(V_1\) renumbered if necessary. \(f (V_1) = {f (c^i b_i)} = {c^i f (b_i)} = {c'^i b'_i}\). The issue is whether whatever combination \((c'^1, c'^2, . . ., c'^r)\) can be realized by a combination \((c^1, c^2, . . ., c^d)\). In fact, it can, because \((c^1, c^2, . . ., c^d) = (c'^1, c'^2, . . ., c'^r, 0, . . ., 0)\) suffices.
