### Change of Basis

##### 3. Matrix for linear transformation

Let `T`: `V` → `W` be a linear transformation. Let `B` = {`b`_{1}, `b`_{2}, ..., **b**_{n}} and `C` = {`c`_{1}, `c`_{2}, ..., **c**_{m}} be bases of `V` and `W`. Then by isomorphisms [?]_{B}: `V` → **R**^{n}, [?]_{C}: `W` → **R**^{m}, `T` is translated to a linear transformation [?]_{C} **T** [?]_{B}^{-1}: **R**^{n} → **R**^{m}. The linear transformation [?]_{C} **T** [?]_{B}^{-1} is given by a matrix `A`, which we call matrix of the linear transformation with respect to the bases `B` and `C`.

Since

`i`-th column of `A`

= [?]_{C} **T** [?]_{B}^{-1}(**e**_{i})
(this formula)

= [?]_{C} **T**(**b**_{i})
([**b**_{i}]_{B} = **e**_{i})

= [**T**(**b**_{i})]_{C},

we conclude the following formula for **A**.

The matrix of a linear transformation `T`: `V` → `W` with respect to bases `B` = {`b`_{1}, `b`_{2}, ..., **b**_{n}} and `C` = {`c`_{1}, `c`_{2}, ..., **c**_{m}} of `V` and `W` is

[**T**(`B`)]_{C} = [ [**T**(**b**_{1})]_{C} [**T**(**b**_{2})]_{C} ... [**T**(**b**_{n})]_{C} ].

The notation [**T**(`B`)]_{C} in introduced to suggest that the matrix is obtained by the following steps:

- Apply
`T` to vectors in `B`.
- Take the coordinates with respect to
`C`.

For an explicit example, in case `m` = 3, `n` = 2 and

**T**(**b**_{1}) = `a`_{11}`c`_{1} + `a`_{12}`c`_{2} + `a`_{13}`c`_{3},

**T**(**b**_{2}) = `a`_{21}`c`_{1} + `a`_{22}`c`_{2} + `a`_{23}`c`_{3},

the matrix of **T** is

[ |
`a`_{11} |
`a`_{21} |
]. |

`a`_{12} |
`a`_{22} |

`a`_{13} |
`a`_{23} |

More concrete examples are given here, in which `V` = `W` = `P`_{2} and `B` = `C` = {1, `t`, `t`^{2}}, and given in this exercise and this exercise, in which `V` = `W` = **P**_{n} and `B` = `C` = {1, `t`, `t`^{2}, ..., `t`^{n}}.

Example Consider the linear transformation

**T**(`p`(`t`)) = [ |
`p`(0) |
`p`(1) |
]: `P`_{3} → `M`(2, 2). |

`p`'(0) |
`p`'(1) |

We use the bases

`B` = `Hermite` = {1, `t`, `t`^{2} - 1, `t`^{3} - 3`t`}, `C` = { [ |
1 |
0 |
], [ |
0 |
1 |
], [ |
0 |
0 |
], [ |
0 |
0 |
] } |

0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |

for `P`_{3} and `M`(2, 2). Then

**T**(1) = [ |
1 |
1 |
], **T**(`t`) = [ |
0 |
1 |
], **T**(`t`^{2} - 1) = [ |
-1 |
0 |
], **T**(`t`^{3} - 3`t`) = [ |
0 |
-2 |
] |

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

tell us the matrix for `T` with respect to the bases is

[**T**(`Hermite`)]_{C} = [ [**T**(1)]_{C} [**T**(`t`)]_{C} [**T**(2`t`^{2})]_{C} [**T**(4`t`^{3} - 3`t`)]_{C} ] = [ |
1 |
0 |
-1 |
0 |
]. |

1 |
1 |
0 |
-2 |

0 |
1 |
0 |
-3 |

0 |
1 |
0 |
-3 |

We may conclude, for example, that `rank`**T** = 3 from the matrix.

The following lists some properties of the matrix for linear transformation.

- linearity
- [(
**T** + **S**)(`B`)]_{C} = [**T**(`B`)]_{C} + [**S**(`B`)]_{C}, [`c`**T**(`B`)]_{C} = `c`[**T**(`B`)]_{C}
- transitivity
- [
**T**(`C`)]_{D} [**T**(`B`)]_{C} = [**T**(`B`)]_{D}
- inverse
- [
**T**(`B`)]_{C}^{-1} = [**T**^{-1}(`C`)]_{B}
- identity
- [
**id**(`B`)]_{B} = `I`