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

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## Example

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])

## Proof

• 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 $\frac{\alpha}{\lambda}$ 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]