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

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Example

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

• $\text{End}(V)\equiv \mathbb{F}$ (where the $\mathbb{F}$ here is considered as a vector space, not a field)
  • $\equiv$ denotes linear isomorphism with a canonical linear map for the isomorphism itself.

Proof

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

The isomorphism

We define: $A:\text{End}(V)\rightarrow \mathbb{F}$ as the map:

• $A:f\mapsto \dfrac{\alpha}{\lambda}$ where:
  • For any basis, $e$, of $\text{End}(V)$ $\lambda$ and $\alpha$ are such that:
    1. $f\eq\alpha e$ and
    2. $\text{Id}\eq \lambda e$

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