341: Finite Dimensional Vectors Spaces Related by Linear Bijection Are of Same Dimension|T.B.P.

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A description/proof of that finite dimensional vectors spaces related by linear bijection are of same dimension

Topics

About: vectors space

Starting Context

The reader knows a definition of vectors space.
The reader knows a definition of bijection.
The reader admits the Clamer theorem.

Target Context

The reader will have a description and a proof of the proposition that any 2 finite dimensional vectors spaces such that there is a linear bijection from one of them to the other are of the same dimension.

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: Description

For any finite dimensional vectors spaces,

V_{1}

and

V_{2}

, such that there is any linear bijection,

f : V_{1} \to V_{2}

, the dimensions of

V_{1}

and

V_{2}

are the same.

2: Proof

There are a basis for

V_{1}

(b_{11}, b_{12}, . . ., b_{1 d_{1}})

, and a basis for

V_{2}

(b_{21}, b_{22}, . . ., b_{2 d_{2}})

. Any

v_{1} \in V_{1}

v_{1} = {v_{1}}^{i} b_{1 i}

f (v_{1}) = {v_{1}}^{i} f (b_{1 i})

f (b_{1 i}) = {f^{j}}_{i} b_{2 j}

f (v_{1}) = {v_{1}}^{i} {f^{j}}_{i} b_{2 j} = {v_{2}}^{i} b_{2 i}

(\begin{matrix} {f^{1}}_{1} & . . . & {f^{1}}_{d_{1}} \\ . . . & . . . & . . . \\ {f^{d_{2}}}_{1} & . . . & {f^{d_{2}}}_{d_{1}} \end{matrix}) (\begin{matrix} {v_{1}}^{1} \\ . . . \\ {v_{1}}^{d_{1}} \end{matrix}) = (\begin{matrix} {v_{2}}^{1} \\ . . . \\ {v_{2}}^{d_{2}} \end{matrix})

. By the Clamer theorem, if the rank of

(\begin{matrix} {f^{j}}_{i} \end{matrix})

, k, was

k < d_{1}

, supposing that the theorem's condition for existence of solutions for

v_{1}

was satisfied (it had to be satisfied indeed), there would be solutions for

v_{1}

with

{v_{1}}^{k + 1}, . . ., {v_{1}}^{d_{1}}

arbitrary, which would mean that

f

was not injective, so,

k = d_{1}

, so,

d_{1} \leq d_{2}

. By the Clamer theorem, if

d_{1} < d_{2}

, supposing that some rows and some columns had been already reordered according to the theorem,

d e t (\begin{matrix} {f^{1}}_{1} & . . . & {f^{1}}_{d_{1}} & {v_{2}}^{1} \\ . . . & . . . & . . . & . . . \\ {f^{d_{1}}}_{1} & . . . & {f^{d_{1}}}_{d_{1}} & {v_{2}}^{d_{1}} \\ {f^{d_{1} + 1}}_{1} & . . . & {f^{d_{1} + 1}}_{d_{1}} & {v_{2}}^{d_{1} + 1} \end{matrix}) = 0

in order for any solution for

v_{1}

to exist, but it would not be possible for every possible

v_{2}

, because the algebraic complement with respect to

{v_{2}}^{d_{1} + 1}

was not 0, so, the determinant could be deviated from 0 by choosing a

{v_{2}}^{d_{1} + 1}

, which would mean that

f

was not surjective, so,

d_{1} = d_{2}

References

The Bias Planet
Definitions and Propositions
School Mathematics from Higher Viewpoints
To Disentangle Confusing Terms or Discourses
Information Tables
Exploiting LibreOffice or OpenOffice
Exposing the Java Programming Language
Exposing C++
Exposing Python
Exposing C#
Exposing Gradle
Exposing Git
UNO Dispatch Commands

2022-08-28