Let `B` = {`b`_{1}, `b`_{2}, ..., ` b_{n}`}
be a basis of a vector space

`c`_{1}`b`_{1} + `c`_{2}`b`_{2} + ... + `c _{n}b_{n}` =

The uniqueness of the coefficients make the following definition unambiguous.

The coordinates of ** v** with respect to the basis

[** v**]

Example The formula

(`x`_{1}, `x`_{2}, `x`_{3})
= `x`_{1}(1, 0, 0) + `x`_{2}(0, 1, 0) + `x`_{3}(0, 0, 1)
= `x`_{1}**e**_{1}
+ `x`_{2}**e**_{2}
+ `x`_{3}**e**_{3}

indicates that

[(`x`_{1}, `x`_{2}, `x`_{3})]_{{e1, e2, e3}}
= (`x`_{1}, `x`_{2}, `x`_{3}).

In other words, when we say `x`_{1} is the first coordinate of the vector
(`x`_{1}, `x`_{2}, `x`_{3}), we really mean it is the first
coordinate with regard to the standard basis {**e**_{1}, **e**_{2},
**e**_{3}}. To illustrate the subtlety here, the formula

(`x`_{1}, `x`_{2}, `x`_{3})
= `x`_{2}**e**_{2}
+ `x`_{1}**e**_{1}
+ `x`_{3}**e**_{3}

tells us

[(`x`_{1}, `x`_{2}, `x`_{3})]_{{e2, e1, e3}}
= (`x`_{2}, `x`_{1}, `x`_{3}),

so that `x`_{1} is the *second* coordinate of
(`x`_{1}, `x`_{2}, `x`_{3}) with regard to the basis
{**e**_{2}, **e**_{1}, **e**_{3}}.

By the definition of the coordinates, we also have

[`a`_{0} + `a`_{1}`t`
+ `a`_{2}`t`^{2} + ...
+ `a _{n}t^{n}`]

In other words, the coefficients of a polynomial are the coordinates with regard to the standard monomial basis.

Finally, the equality

[ | a |
b |
] = a[ |
1 | 0 | ] + b[ |
0 | 1 | ] + c[ |
0 | 0 | ] + d[ |
0 | 0 | ]
= aE_{11} + bE_{12}
+ cE_{21} + dE_{22} |

c |
d |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |

tells us

[ | a |
b |
]_{{E11,
E12, E21, E22}}
= (a, b, c, d). |

c |
d |

Example In an earlier example, the system

3x_{1} |
+ x_{2} |
- x_{3} |
= | 2 |

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 6 |

was shown to have a unique solution `x`_{1} = 1, `x`_{2} = 1,
`x`_{3} = 2. This means

1(3, 1, 2) + 1(1, -1, 2) + 2(-1, 1, 1) = (2, 2, 6),

Recall that in this case, the uniqueness of the solution actually
implies that {(3, 1, 2), (1, -1, 2), (-1, 1, 1)} is a basis of
**R**^{3}. Moreover, the equality above means that the coordinates of (2, 2, 6)
with regard to this basis is

[(2, 2, 6)]_{{(3, 1, 2), (1, -1, 2), (-1, 1, 1)}} = (1, 1, 2).

In general, the coordinates of a vector (`b`_{1}, `b`_{2}, `b`_{3}) with regard to the basis {(3, 1, 2), (1, -1, 2), (-1, 1, 1)} is obtained by solving the following system.

3x_{1} |
+ x_{2} |
- x_{3} |
= | b_{1} |

x_{1} |
- x_{2} |
+ x_{3} |
= | b_{2} |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | b_{3} |

The solution is

[ | b_{1} |
]_{{(3, 1, 2), (1, -1, 2), (-1, 1, 1)}}
= [ |
3 | 1 | -1 | ]^{-1}[ |
b_{1} |
] |

b_{1} |
1 | -1 | 1 | b_{2} |
||||

b_{3} |
2 | 2 | 1 | b_{3} |
||||

= [ | 1/4 | 1/4 | 0 | ] [ | b_{1} |
] | ||

-1/12 | -5/12 | 1/3 | b_{2} |
|||||

-1/3 | 1/3 | 1/3 | b_{3} |
|||||

= [ | (1/4)b_{1} + (1/4)b_{2} |
]. | ||||||

- (1/12)b_{1} - (5/12)b_{2} + (1/3)b_{3} |
||||||||

- (1/3)b_{1} + (1/3)b_{2} + (1/3)b_{3} |

In the example above, the coordinates of a vector ** v** with respect to
a basis of

Let the columns of ` B` = [

[** v**]