description/proof of that linear range of finite-dimensional vectors space is vectors space
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of %field name% 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 range 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f\): \(V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
//
Statements:
\(f (V_1) \in \{\text{ the } F \text{ vectors spaces }\}\)
//
2: Natural Language Description
For any field, \(F\), any finite-dimensional \(F\) vectors space, \(V_1\), any \(F\) vectors space, \(V_2\), and any linear map, \(f: V_1 \to V_2\), the range of \(f\), \(f (V_1)\), is a \(F\) vectors space.
3: Proof
Whole Strategy: Step 1: take any basis for \(V_1\), \(\{b_1, b_2, . . ., b_d\}\); Step 2: take any maximal linearly independent subset of \(\{f (b_1), f (b_2), . . ., f (b_d)\}\), \(\{f (b_1), f (b_2), . . ., f (b_r)\}\), with the basis for \(V_1\) re-indexed; Step 3: see that the space spanned by \(\{f (b_1), f (b_2), . . ., f (b_r)\}\) is \(f (V_1)\).
Step 1:
Let us take any basis for \(V_1\), \(\{b_1, b_2, . . ., b_d\}\).
Step 2:
Let us take any maximal linearly independent subset of \(\{f (b_1), f (b_2), . . ., f (b_d)\}\).
What that means is this: if \(\{f (b_1), f (b_2), . . ., f (b_d)\}\) is linearly independent, it is the thing; if not, an \(f (b_j)\) is a linear combination of the rest, remove \(f (b_j)\), and if the leftover is linearly independent, it is the thing; if not, an \(f (b_k)\) is a linear combination of the rest, remove \(f (b_k)\), and if the leftover is linearly independent, it is the thing; and so on. If nothing is left, \(f (V_1)\) is the 0 vectors space, which is a kind of vectors space, and we are done. Let us suppose otherwise hereafter.
We have a linearly independent subset, but by re-indexing \(\{b_1, b_2, . . ., b_d\}\), we can say that \(\{f (b_1), f (b_2), . . ., f (b_r)\}\) is the linearly independent subset.
Step 3:
Let us see that the space spanned by \(\{f (b_1), f (b_2), . . ., f (b_r)\}\), \(V\), is \(f (V_1)\).
Let \(f (\sum_{j \in \{1, ..., d\}} c^j b_j) \in f (V_1)\) be any. \(f (\sum_{j \in \{1, ..., d\}} c^j b_j) = \sum_{j \in \{1, ..., d\}}c^j f (b_j)\), but as \(f (b_j)\) for each \(r \lt j\) is a linear combination of \(\{f (b_1), f (b_2), . . ., f (b_r)\}\), it is \({\sum_{j \in \{1, ..., r\}}c'^j f (b_j)} \in V\). So, \(f (V_1) \subseteq V\).
Let \(\sum_{j \in \{1, ..., r\}} c'^j f (b_j) \in V\) be any. Let us take \(\sum_{j \in \{1, ..., r\}} c'^j b_j \in V_1\). Then, \(f (\sum_{j \in \{1, ..., r\}} c'^j b_j) = \sum_{j \in \{1, ..., r\}} c'^j f (b_j)\). So, \(V \subseteq f (V_1)\).
So, \(f (V_1) = V\).
