description/proof of that for finite-dimensional vectors space and basis, vectors space is 'vectors spaces - linear morphisms' isomorphic to components 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: Note
- 4: Proof
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of basis of module.
- The reader knows a definition of %category name% isomorphism.
- The reader admits the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the basis.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(B\): \(= \{b_1, ..., b_d\} \subseteq V\), \(\in \{\text{ the bases of } V\}\)
\(F^d\): \(= \text{ the vectors space }\)
\(f\): \(: V \to F^d, v \to (v^1, ..., v^d)\), where \(v = v^1 b_1 + ... + v^d b_d\)
//
Statements:
\(f \in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
//
2: Natural Language Description
For any field, \(F\), any \(d\)-dimensional \(F\) vectors space, \(V\), any basis, \(B = \{b_1, ..., b_d\} \subseteq V\), the vectors space, \(F^d\), and \(f: V \to F^d, v \to (v^1, ..., v^d)\), where \(v = v^1 b_1 + ... + v^d b_d\), \(f\) is a 'vectors spaces - linear morphisms' isomorphism.
3: Note
\(F\) does not need to be \(\mathbb{R}\) or \(\mathbb{C}\).
This proposition is usually regarded to be obvious intuitively, but let us be at ease with conscience that we have indeed proved it once and for all.
4: Proof
Whole Strategy: Step 1: see that \(F^d\) is indeed an \(F\) vectors space; Step 2: see that \(f\) is linear; Step 3: see that \(f\) is bijective; Step 4: conclude the proposition.
Step 1:
Let us see that \(F^d\) is indeed an \(F\) vectors space.
1) for any elements, \(v_1, v_2 \in F^d\), \(v_1 + v_2 \in F^d\) (closed-ness under addition); 2) for any elements, \(v_1, v_2 \in F^d\), \(v_1 + v_2 = v_2 + v_1\) (commutativity of addition); 3) for any elements, \(v_1, v_2, v_3 \in F^d\), \((v_1 + v_2) + v_3 = v_1 + (v_2 + v_3)\) (associativity of additions); 4) there is a 0 element, \(0 \in F^d\), such that for any \(v \in F^d\), \(v + 0 = v\) (existence of 0 vector); 5) for any element, \(v \in F^d\), there is an inverse element, \(v' \in F^d\), such that \(v' + v = 0\) (existence of inverse vector); 6) for any element, \(v \in F^d\), and any scalar, \(r \in F\), \(r . v \in F^d\) (closed-ness under scalar multiplication); 7) for any element, \(v \in F^d\), and any scalars, \(r_1, r_2 \in F\), \((r_1 + r_2) . v = r_1 . v + r_2 . v\) (scalar multiplication distributability for scalars addition); 8) for any elements, \(v_1, v_2 \in F^d\), and any scalar, \(r \in F\), \(r . (v_1 + v_2) = r . v_1 + r . v_2\) (scalar multiplication distributability for vectors addition); 9) for any element, \(v \in F^d\), and any scalars, \(r_1, r_2 \in F\), \((r_1 r_2) . v = r_1 . (r_2 . v)\) (associativity of scalar multiplications); 10) for any element, \(v \in F^d\), \(1 . v = v\) (identity of 1 multiplication).
Step 2:
\(f\) is well-defined, because the decomposition of each vector with respect to any basis is unique: if \(v = v^1 b_1 + ... + v^d b_d = v'^1 b_1 + ... + v'^d b_d\), \((v^1 - v'^1) b_1 + ... + (v^d - v'^d) b_d = 0\), which implies that \(v^j = v'^j\).
Let us see that \(f\) is linear.
Let \(v, v' \in V\) and \(r, r' \in F\) be any.
Let \(v = v^1 b_1 + ... + v^d b_d\) and \(v' = v'^1 b_1 + ... + v'^d b_d\).
\(f (r v + r' v') = f (r (v^1 b_1 + ... + v^d b_d) + r' (v'^1 b_1 + ... + v'^d b_d)) = f ((r v^1 + r' v'^1)b_1 + ... + (r v^d + r' v'^d) b_d) = (r v^1 + r' v'^1, ..., r v^d + r' v'^d) = (r v^1, ..., r v^d) + (r' v'^1, ..., r' v'^d) = r (v^1, ..., v^d) + r' (v'^1, ..., v'^d) = r f (v) + r' f (v')\).
Step 3:
Let us see that \(f\) is bijective.
Let \(v \neq v' \in V\) be any.
\(f (v) \neq f (v')\), because the decomposition of each vector with respect to any basis is unique: mentioned before.
So, \(f\) is injective.
Let \((v^1, ..., v^d) \in F^d\) be any.
For \(v := v^1 b_1 + ... + v_d b_d \in V\), \(f (v) = f (v^1 b_1 + ... + v_d b_d) = (v^1, ..., v^d)\).
So, \(f\) is surjective.
Step 4:
By the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism, \(f\) is a 'vectors spaces - linear morphisms' isomorphism.
