# 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 $v \in \mathbf R^n$ and $a_1,\ldots,a_n \in \mathbf R$, then it does make sense to ask whether $v = (a_1,\ldots,a_n)$.
If $V$ is an arbitrary finite-dimensional real vector space, $\beta = (v_1, \ldots, v_n)$ is a basis for $V$, and $a_1,\ldots,a_n \in \mathbf R$, then it does make sense to ask whether $[v]^\beta = (a_1,\ldots,a_n)$, or equivalently, to ask whether $v = a_1v_1 + \cdots + a_nv_n$.
If $v \in \mathbf R^n$ and $\beta = (v_1, \ldots, v_n)$ is a basis for $\mathbf R^n$, then $v$ and $[v]^\beta$ have the same type, so it makes sense to ask whether $v = [v]^\beta$. When are the two equal? If $\beta = (e_1, \ldots, e_n)$ is the standard basis and $v = (a_1, \ldots, a_n)$, then $v = a_1e_1 + \cdots + a_ne_n$ so $[v]^\beta = (a_1, \ldots, a_n) = v$. But the converse is not true: given $v = [v]^\beta$, there can be many bases that give the same coordinates. For instance if $\beta = (e_1, e_2)$ and $\beta' = (e_2, e_1)$, and $v=(1,1)$, then $v = [v]^\beta = [v]^{\beta'}$.