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
- 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 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the d-dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the d-dimensional } F \text{ vectors spaces }\}\)
\(f\): \(V_1 \to V_2\), \(\in \{\text{ the linear surjections }\}\)
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Statements:
\(f \in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
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2: Natural Language Description
For any field, \(F\), any d-dimensional \(F\) vectors space, \(V_1\), and any d-dimensional \(F\) vectors space, \(V_2\), any linear surjection, \(f: V_1 \to V_2\), is a 'vectors spaces - linear morphisms' isomorphism.
3: Proof
Whole Strategy: Step 1: take any basis for \(V_1\), \(\{b_1, b_2, . . ., b_d\}\); Step 2: see that \(\{f (b_1), f (b_2), . . ., f (b_d)\}\) is a basis for \(V_2\); Step 3: see that \(f\) is injective; Step 4: see that \(f^{-1}\) is linear.
Step 1:
Let us take any basis for \(V_1\), \(\{b_1, b_2, . . ., b_d\}\).
Step 2:
Let us see that \(\{f (b_1), f (b_2), . . ., f (b_d)\}\) is a basis for \(V_2\).
\(\{f (b_1), f (b_2), . . ., f (b_d)\}\) is linearly independent on \(V_2\), because otherwise, \(V_2 = f (V_1) = \{f (\sum_{j \in \{1, ..., d\}} c^j b_j)\} = \{\sum_{j \in \{1, ..., d\}} c^j f (b_j)\}\) would not be d-dimensional.
\(\{f (b_1), f (b_2), . . ., f (b_d)\}\) spans \(V_2\), because \(V_2 = \{\sum_{j \in \{1, ..., d\}} c^j f (b_j)\}\).
Step 3:
Let us see that \(f\) is injective.
For each \(p = \sum_{j \in \{1, ..., d\}} c^j b_j, p' = \sum_{j \in \{1, ..., d\}} c'^j b_j \in V_1\) such that \(p \neq p'\), \(c^j \neq c'^j\) for a \(j\), so, \(f (\sum_{j \in \{1, ..., d\}} c^j b_j) = \sum_{j \in \{1, ..., d\}} c^j f (b_j) \neq \sum_{j \in \{1, ..., d\}} c'^j f (b_j) = f (\sum_{j \in \{1, ..., d\}} c'^j b_j)\), because \(\{f (b_1), f (b_2), . . ., f (b_d)\}\) is a basis.
So, \(f\) is bijective, and the inverse, \(f^{-1}\), exists.
Step 4:
Let us see that \(f^{-1}\) is linear.
\(f^{-1} (\sum_{j \in \{1, ..., d\}} c^j f (b_j)) = \sum_{j \in \{1, ..., d\}} c^j b_j\), because \(f (\sum_{j \in \{1, ..., d\}} c^j b_j) = \sum_{j \in \{1, ..., d\}} c^j f (b_j)\), \(= \sum_{j \in \{1, ..., d\}} c^j f^{-1} (f (b_j))\).
