description/proof of that for linear surjection from finite-dimensional vectors space, if dimension of codomain is equal to or larger than that of domain, surjection is bijection
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: Note
- 4: Proof
Starting Context
- The reader knows a definition of linear map.
- The reader knows a definition of dimension of vectors space.
- The reader knows a definition of bijection.
- 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 from any finite-dimensional vectors space, if the dimension of the codomain is equal to or larger than that of the domain, the surjection is a bijection.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } d_1 \text{ -dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear surjections }\}\)
//
Statements:
\(dim V_1 \le dim V_2\)
\(\implies\)
\(f \in \{\text{ the bijections }\}\)
//
2: Natural Language Description
For any field, \(F\), any \(d_1\)-dimensional \(F\) vectors space, \(V_1\), any \(F\) vectors space, \(V_2\), and any linear surjection, \(f: V_1 \to V_2\), if \(dim V_1 \le dim V_2\), \(f\) is a bijection.
3: Note
\(V_2\) is not presupposed to be finite-dimensional, although \(V_2\) will be concluded to be finite-dimensional, in fact \(d_1\)-dimensional.
4: 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: suppose that there were some distinct elements of \(V_1\) that were mapped to the same element, and find a contradiction.
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:
Let us suppose that there were some \(v_1, v_2 \in V_1\) such that \(v_1 \neq v_2\) and \(f (v_1) = f (v_2)\).
\(v_1 = \sum_{j \in \{1, ..., d_1\}} {v_1}^{j} e_j\) and \(v_2 = \sum_{j \in \{1, ..., d_1\}} {v_2}^{j} e_j\). \(f (v_1) = \sum_{j \in \{1, ..., d_1\}} {v_1}^j f (e_j) = \sum_{j \in \{1, ..., d_1\}} {v_2}^j f (e_j) = f (v_2)\). As \(v_1 \neq v_2\), \({v_1}^k \neq {v_2}^k\) for a \(k\). Then, \(\sum_{j \in \{1, ..., d_1\}} ({v_1}^j - {v_2}^j) f (e_j) = 0\), but as \(v_1 - v_2 \neq 0\), \(\{f (e_1), ..., f (e_n)\}\) would be linearly dependent.
So, \(\{f (e_1), ..., f (e_n)\}\) would not be any basis of \(V_2\), but \(\{f (e_1), ..., f (e_n)\}\) could be reduced to be a basis of \(V_2\), by the proposition that for any vectors space, any finite generator can be reduced to be a basis, which would mean that \(dim V_2 \lt dim V_1\), a contradiction against \(dim V_1 \le dim V_2\).
So, \(f\) is injective, and \(f\) is bijective.
In fact, \(\{f (e_1), ..., f (e_n)\}\) is linearly independent and is a basis, because otherwise, \(dim V_2 \lt dim V_1\), so, \(dim V_2 = dim V_1\).
