A basis of a vector space can be used to construct an isomorphism, via the coordinates, between the vector space and the euclidean space. The basis can also be used to identify the dual space with the euclidean space.

Let {**b**_{1}, **b**_{2}, ..., ` b_{n}`} be a basis of

`f` → (`f`(**b**_{1}), `f`(**b**_{2}), ..., `f`(` b_{n}`)),

is an isomorphism. In particular, we have `dim V` =

Proof Denote the transformation by ` E`(

` E`(

= ((

= (

= (

=

We may verify ` E`(

Next, we construct an inverse for ` E`. For any

`f _{a}`(

Strictly speaking, we need to explain that `f _{a}` is well-defined and linear. This can be proved directly, or can be seen from the fact that

` v` =

where the first → is the coordinates with respect to the basis {**b**_{1}, **b**_{2}, ..., ` b_{n}`}, and the second → is a well-known linear function on the euclidean space.

By the way `f _{a}` is constructed, it is easy to verify that the transformation

In the special case ** V** =

Under the isomorphism `f` → (`f`(**b**_{1}), `f`(**b**_{2}), ..., `f`(` b_{n}`)),

The dual basis {`β`_{1}, `β`_{2}, ... , `β _{n}`} of a basis {

β(_{i}) = { b_{j} |
1, | if i = j |

0, | if i ≠ j |

Example The linear functions

`ε`_{1}(`x`_{1}, `x`_{2}, `x`_{3}) = `x`_{1},

`ε`_{2}(`x`_{1}, `x`_{2}, `x`_{3}) = `x`_{2},

`ε`_{3}(`x`_{1}, `x`_{2}, `x`_{3}) = `x`_{3},

form a basis of (**R**^{3})^{*}, which is in fact the dual basis of the standard basis of **R**^{3}.

The linear functions

`α`(`a` + `bt` + `ct`^{2} + `dt`^{3}) = `a`,

`β`(`a` + `bt` + `ct`^{2} + `dt`^{3}) = `b`,

`γ`(`a` + `bt` + `ct`^{2} + `dt`^{3}) = `c`,

`δ`(`a` + `bt` + `ct`^{2} + `dt`^{3}) = `d`,

form the dual basis of the basis {1, `t`, `t`^{2}, `t`^{3}} of `P`_{3}.

Example In an earlier example, we had a basis `B` = {(3, 1, 2), (1, -1, 2), (-1, 1, 1)} of **R**^{3}. Let `B`^{*} = {`a`_{11}`x`_{1} + `a`_{12}`x`_{2}
+ `a`_{13}`x`_{3}, `a`_{21}`x`_{1} + `a`_{22}`x`_{2}
+ `a`_{23}`x`_{3}, `a`_{31}`x`_{1} + `a`_{32}`x`_{2}
+ `a`_{33}`x`_{3}} be its dual basis. Then the definition of the dual basis can be reformulated exactly as

[ | a_{11} |
a_{12} |
a_{13} |
][ | 3 | 1 | -1 | ] = [ | 1 | 0 | 0 | ]. |

a_{21} |
a_{22} |
a_{23} |
1 | -1 | 1 | 0 | 1 | 0 | ||||

a_{31} |
a_{32} |
a_{33} |
2 | 2 | 1 | 0 | 0 | 1 |

Then by this example, we get

[ | a_{11} |
a_{12} |
a_{13} |
] = [ | 3 | 1 | -1 | ]^{-1} = [ |
1/4 | 1/4 | 0 | ]. |

a_{21} |
a_{22} |
a_{23} |
1 | -1 | 1 | -1/12 | -5/12 | 1/3 | ||||

a_{31} |
a_{32} |
a_{33} |
2 | 2 | 1 | -1/3 | 1/3 | 1/3 |

We conclude that the dual basis is `B`^{*} = {(1/4)`x`_{1} + (1/4)`x`_{2}, -(1/12)`x`_{1} - (5/12)`x`_{2} + (1/3)`x`_{3}, -(1/3)`x`_{1} + (1/3)`x`_{2} + (1/3)`x`_{3}}.