User:IssaRice/Linear algebra/Type checking vector spaces

If $V$ is an arbitrary vector space, it does not in general make sense to ask whether $v \in V$ is a string of numbers. This is because we have not chosen a coordinate system.

If $v \in R^{n}$ and $a_{1}, \dots, a_{n} \in R$ , then it does make sense to ask whether $v = (a_{1}, \dots, a_{n})$ .

If $V$ is an arbitrary finite-dimensional real vector space, $β = (v_{1}, \dots, v_{n})$ is a basis for $V$ , and $a_{1}, \dots, a_{n} \in R$ , then it does make sense to ask whether $[v]^{β} = (a_{1}, \dots, a_{n})$ , or equivalently, to ask whether $v = a_{1} v_{1} + \dots + a_{n} v_{n}$ .

If $v \in R^{n}$ and $β = (v_{1}, \dots, v_{n})$ is a basis for $R^{n}$ , then $v$ and $[v]^{β}$ have the same type, so it makes sense to ask whether $v = [v]^{β}$ . When are the two equal? If $β = (e_{1}, \dots, e_{n})$ is the standard basis and $v = (a_{1}, \dots, a_{n})$ , then $v = a_{1} e_{1} + \dots + a_{n} e_{n}$ so $[v]^{β} = (a_{1}, \dots, a_{n}) = v$ . But the converse is not true: given $v = [v]^{β}$ , there can be many bases that give the same coordinates. For instance if $β = (e_{1}, e_{2})$ and $β^{'} = (e_{2}, e_{1})$ , and $v = (1, 1)$ , then $v = [v]^{β} = [v]^{β^{'}}$ .