Let `B` = {**b**_{1}, `b`_{2}, `b`_{3}} and `C` = {`c`_{1}, `c`_{2}, `c`_{3}} be two bases of the euclidean space **R**^{3} (or a subspace of **R**^{3}). Then they induce matrices [`B`] = [`b`_{1} `b`_{2} `b`_{3}], [`C`] = [`c`_{1} `c`_{2} `c`_{3}]. As we have seen in the proof of the formula [** v**]

[`B`] = [`C`][`B`]_{C}.

In particular, if we take `B` to be the standard basis of **R**^{3}, so that [`B`] = ** I** and [

The equality [`B`] = [`C`][`B`]_{C} shows that the columns of the matrix [`B`]_{C} are the solutions of the system [`C`]** x** =

Example Let

= [A |
1 | 3 | 2 | 0 | 1 | ]. |

-1 | -1 | -1 | 1 | 0 | ||

0 | 4 | 2 | 4 | 3 | ||

1 | 3 | 2 | -2 | 0 |

In this example, we found a basis `B` = {(1, -1, 0, 1), (3, -1, 4, 3), (0, 1, 4, -2)} of `col A`. In this example, we found another basis

[ [C] [B] ] = [ |
1 | 0 | 0 | 1 | 3 | 0 | ] |

-1 | 1 | 0 | -1 | -1 | 1 | ||

0 | 2 | 1 | 0 | 4 | 4 | ||

1 | 0 | -1 | 1 | 3 | -2 |

Row operations change this into

[ | 1 | 0 | 0 | 1 | 3 | 0 | ]. |

0 | 1 | 0 | 0 | 2 | 1 | ||

0 | 0 | 1 | 0 | 0 | 2 | ||

0 | 0 | 0 | 0 | 0 | 0 |

Then it is easy to see that the columns in heavily shaded submatrix are the solutions of [`C`]** x** =

[B]_{C} = [ |
1 | 3 | 0 | ]. |

0 | 2 | 1 | ||

0 | 0 | 2 |

Now by an earlier computation, we have

[(0, 1, 2, 0)]_{B} = (-3/2, 1/2, 0).

Then

[(0, 1, 2, 0)]_{C} = [ |
1 | 3 | 0 | ] [ | -3/2 | ] = [ | 0 | ]. |

0 | 2 | 1 | 1/2 | 1 | ||||

0 | 0 | 2 | 0 | 0 |

This is consistent with the fact that the second vector in `C` is (0, 1, 2, 0).

Example The following are two bases for the vector space **P**_{3} .

`Chebyshev` = {1, `t`, 2`t`^{2}, 4`t`^{3} - 3`t`}, `Hermite` = {1, `t`, `t`^{2} - 1, `t`^{3} - 3`t`}

To find the matrix for changing the Chebyshev coordinates to Hermite coordinates, we use the translation [`a` + `bt` + `ct`^{2} + `dt`^{3}] = (`a`, `b`, `c`, `d`) and the method of the previous example. Thus we form the matrix

[ [Hermite] [Chebychev] ] = [ |
1 | 0 | -1 | 0 | 1 | 0 | -1 | 0 | ]. |

0 | 1 | 0 | -3 | 0 | 1 | 0 | -3 | ||

0 | 0 | 1 | 0 | 0 | 0 | 2 | 0 | ||

0 | 0 | 0 | 1 | 0 | 0 | 0 | 4 |

Row operations give us

[ | 1 | 0 | 0 | 0 | 1 | 0 | 2 | 0 | ]. |

0 | 1 | 0 | 0 | 0 | 1 | 0 | 9 | ||

0 | 0 | 1 | 0 | 0 | 0 | 2 | 0 | ||

0 | 0 | 0 | 1 | 0 | 0 | 0 | 4 |

Thus we conclude that the Hermite coordinates may be obtained by multiplying

[Chebychev] = [_{Hermite} |
1 | 0 | 2 | 0 | ] |

0 | 1 | 0 | 9 | ||

0 | 0 | 2 | 0 | ||

0 | 0 | 0 | 4 |

to (the left of) the Chebychev coordinates.