Example:Canonical linear isomorphism between a one dimensional vector space and its field

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Let [ilmath]\mathbb{F} [/ilmath] be a field and let [ilmath](V,\mathbb{F})[/ilmath] be a vector space. If the dimension of [ilmath]V[/ilmath] is [ilmath]1[/ilmath] then:

  • [ilmath]\text{End}(V)\equiv \mathbb{F} [/ilmath][1] (where the [ilmath]\mathbb{F} [/ilmath] here is considered as a vector space, not a field[Note 1])


  • Let [ilmath]e[/ilmath], [ilmath]e'[/ilmath] be different bases of [ilmath]\text{End}(V)[/ilmath] - this means [ilmath]e,e':V\rightarrow V[/ilmath] are linear maps
    • Notice that:
      1. [ilmath]\text{Id}\in\text{End}(V)[/ilmath] (the identity map) can be expressed as: [ilmath]\text{Id}\eq\lambda e\eq\lambda e'[/ilmath]
      2. [ilmath]f\in\text{End}(V)[/ilmath] (any member) can be expressed as: [ilmath]f\eq \alpha e\eq \alpha'e'[/ilmath]
    • But [ilmath] f\eq\mu\text{Id} [/ilmath] also (as [ilmath]\text{Id} [/ilmath] can also be a basis of [ilmath]\text{End}(V)[/ilmath])
    • We see:
      • [ilmath]f\eq\mu\lambda e\eq\mu\lambda'e'\eq\alpha e\eq\alpha'e'[/ilmath]
      • Explicitly: [ilmath]\mu\lambda\eq \alpha[/ilmath] and [ilmath]\mu\lambda'\eq \alpha'[/ilmath]
        • Rearranging we see: [ilmath]\mu\eq\frac{\alpha}{\lambda}\eq\frac{\alpha'}{\lambda'} [/ilmath]
  • Thus regardless of basis [math]\frac{\alpha}{\lambda} [/math] is the same value ([ilmath]\mu[/ilmath] above)

The isomorphism

We define: [ilmath]A:\text{End}(V)\rightarrow \mathbb{F} [/ilmath] as the map:

  • [ilmath]A:f\mapsto \dfrac{\alpha}{\lambda} [/ilmath] where:
    • For any basis, [ilmath]e[/ilmath], of [ilmath]\text{End}(V)[/ilmath] [ilmath]\lambda[/ilmath] and [ilmath]\alpha[/ilmath] are such that:
      1. [ilmath]f\eq\alpha e[/ilmath] and
      2. [ilmath]\text{Id}\eq \lambda e[/ilmath]
Grade: A*
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These were rushed notes


  1. Remember every field is a vector space in its own right


  1. Linear Algebra via Exterior Products - Sergei Winitzki

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