Vector space/Basis and isomorphic to K^n/Remark

An isomorphism between an ${}n$-dimensional vector space ${}V$ and the standard space ${}K^{n}$ is essentially equivalent with the choice of a basis of ${}V$. For a basis

${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}\,,$

we associate the linear mapping

$\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,e_{i}\longmapsto v_{i},$

which maps from the standard space to the vector space by sending the ${}i$-th standard vector to the ${}i$-th basis vector of the given basis. This defines a unique linear mapping due to fact. Due to exercise, this mapping is bijective. It is just the mapping

$(a_{1},\ldots ,a_{n})\longmapsto \sum _{i=1}^{n}a_{i}v_{i}.$

The inverse mapping

$x=\Psi _{\mathfrak {v}}^{-1}\colon V\longrightarrow K^{n}$

is also linear, and it is called the coordinate mapping for this basis. The ${}i$-th component of this map, that is, the composed mapping

$x_{i}=p_{i}\circ x\colon V\longrightarrow K,v\longmapsto (\Psi _{\mathfrak {v}}^{-1}(v))_{i},$

is called the ${}i$-th coordinate function. It is denoted by ${}v_{i}^{*}$. It assigns to vector ${}v\in V$ with the unique representation

${}v=\sum _{i=1}^{n}\lambda _{i}v_{i}\,$

the coordinate ${}\lambda _{i}$. Note that the linear mapping ${}v_{i}^{*}$ depends on the basis, not only on the vector ${}v_{i}$.

If an isomorphism

$\Psi \colon K^{n}\longrightarrow V$

is given, then the images

$\Psi (e_{i}),i=1,\ldots ,n,$

form a basis of ${}V$.