description/proof of that for linear surjection between finite-dimensional vectors spaces, dimension of codomain is equal to or smaller than that of domain
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 linear map.
- The reader knows a definition of dimension of vectors space.
- The reader admits the proposition that for any vectors space, any finite generator can be reduced to be a basis.
Target Context
- The reader will have a description and a proof of the proposition that for any linear surjection between any finite-dimensional vectors spaces, the dimension of the codomain is equal to or smaller than that of the domain.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } d_1 \text{ -dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } d_2 \text{ -dimensional } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear surjections }\}\)
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Statements:
\(dim V_2 \le dim V_1\)
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2: Natural Language Description
For any field, \(F\), any \(d_1\)-dimensional \(F\) vectors space, \(V_1\), any \(d_2\)-dimensional \(F\) vectors space, \(V_2\), and any linear surjection, \(f: V_1 \to V_2\), \(dim V_2 \le dim V_1\).
3: Proof
Whole Strategy: Step 1: choose any basis for \(V_1\); Step 2: express the range of \(f\) with the image of the basis and see that \(V_2\) is spanned by the image of the basis; Step 3: conclude the proposition.
Step 1:
Let us choose any basis for \(V_1\), \(\{e_1, ..., e_{d_1}\}\).
Step 2:
For each \(v \in V_1\), \(v = \sum_{j \in \{1, ..., d_1\}} v^j e_j\), where \(v^j \in F\), \(f (v) = \sum_{j \in \{1, ..., d_1\}} v^j f (e_j)\), and \(f (V_1) = V_2 = \{\sum_{j \in \{1, ..., d_1\}} v^j f (e_j) \vert v^j \in F\}\), which means that \(V_2\) is spanned by \(\{f (e_1), ..., f (e_{d_1})\}\).
Step 3:
A basis of \(V_2\) is a subset of \(\{f (e_1), ..., f (e_{d_1})\}\), by the proposition that for any vectors space, any finite generator can be reduced to be a basis, which implies that \(dim V_2 \le dim V_1\).
