### Change of Basis

##### 5. Similar matrix

To build the matrix for a linear transformation **T**: **V** → **V** from a vector space to itself, we naturally want to choose one basis `B` = {**b**_{1}, **b**_{2}, ..., **b**_{n}} for both copies of **V**. Then by this formula, the matrix for **T** with respect to `B` is [**T**(`B`)]_{B}.

Let `B'` be another basis of **V**. Then by this formula, the matrices of **T** with respect to `B` and `B'` are related by the equality

[**T**(`B'`)]_{B'} = [`B`]_{B'} [**T**(`B`)]_{B} [`B`]_{B'}^{-1}.

The relation leads to the following definition.

Two square matrices **A** and **B** are said to be similar if there is an invertible **P**, such that **B** = `PAP`^{-1}.

Thus the matrix of a *self* linear transformation is changed by similarity when the basis is changed.

Example Consider the linear transformation

**T**(`x`, `y`) = (13`x` - 4`y`, - 4`x` + 7`y`): **R**^{2} → **R**^{2}

from **R**^{2} to itself. Since `x` and `y` are the coordinates with respect to the standard basis `B` = {**e**_{1}, **e**_{2}} = {(1, 0), (0, 1)}, the matrix

**A** = [**T**(`B`)]_{B} = [ |
13 |
-4 |
] |

-4 |
7 |

that we usually associate to **T** is really the matrix of **T** with respect to the standard basis. If we choose a different basis `B'` = {**u**_{1}, **u**_{2}} = {(1, 2), (2, -1)}. Then

**T**(**u**_{1}) = (13×1 - 4×2, - 4×1 + 7×2) = (5, 10) = 5**u**_{1}

**T**(**u**_{2}) = (13×2 - 4×(-1), - 4×2 + 7×(-1)) = (30, -15) = 15**u**_{2}

tell us the matrix for **T** with respect to `B'` is

**B** = [**T**(`B'`)]_{B'} = [ |
5 |
0 |
]. |

0 |
15 |

Denote

**P** = [`B'`]_{B} = [**u**_{1} **u**_{2}] = [ |
1 |
2 |
]. |

2 |
-1 |

It is easy to directly verify that **A** = `PBP`^{-1}.

We note that the original formula for **T** (with respect to {**e**_{1}, **e**_{2}}) is too complicated for us to have a good understanding of what **T** is doing on the plane **R**^{2}. However, if we look through the basis {**u**_{1}, **u**_{2}} instead, then we have a very good understanding of the transformation. Basically **T** rescale the **u**_{1} direction by 5 times and the **u**_{2} direction by 15 times.

The example shows the importance of similarity. To gain a good undertanding of a self linear transformation, we find a basis (a suitable invertible matrix **P**) so that the the transformation behaves particularly nice with respect to the basis (the similarity change via **P** gives a diagonal matrix).

Example Consider the linear transformation

**T**(`p`) = (-1/3)(1 - `t` )^{2}`p'''` + (- 1 + `t`)`p''` - 2(1 + `t`)`p'` + 6`p`: **P**_{3} → **P**_{3}.

From

**T**(1) = 6

**T**(`t`) = - 2 + 4`t`

**T**(`t`^{2}) = - 2 - 2`t` + 2`t`^{2}

**T**(`t`^{3}) = - 2 - 2`t` - 2`t`^{2}

the linear transformation has the following matrix with respect to the usual choice {1, `t`, `t`^{2}, `t`^{3}^{}} of basis.

**A** = [ |
6 |
-2 |
-2 |
-2 |
] |

0 |
4 |
-2 |
-2 |

0 |
0 |
2 |
-2 |

0 |
0 |
0 |
0 |

To see more clearly what the transformation does to polynomials, we consider a new basis {1, 1 + `t`, 1 + `t` + `t`^{2}, 1 + `t` + `t`^{2} + `t`^{3}^{}}. From

**T**(1) = 6 = 6(1)

**T**(1 + `t`) = 4(1 + `t`)

**T**(1 + `t` + `t`^{2}) = 2(1 + `t` + `t`^{2})

**T**(`t`^{3}) = 0 = 0(1 + `t` + `t`^{2} + `t`^{3})

the matrix with respect to the new basis is

**B** = [ |
6 |
0 |
0 |
0 |
]. |

0 |
4 |
0 |
0 |

0 |
0 |
2 |
0 |

0 |
0 |
0 |
0 |

Geometrically, this means that **T** is rescaling by factors 6, 4, 2, 0 in the new basis directions.

The matrix for changing the coordinates in the new basis to the coordinates in the usual basis is

**Q** = [{1, 1 + `t`, 1 + `t` + `t`^{2}, 1 + `t` + `t`^{2} + `t`^{3}^{}}]_{{1, t, t2, t3}} = [ |
1 |
1 |
1 |
1 |
]. |

0 |
1 |
1 |
1 |

0 |
0 |
1 |
1 |

0 |
0 |
0 |
1 |

Direct verification shows **B** = **Q**^{-1}`AQ` (by [`B`]_{B'} = [`B`']_{B}^{-1}, this **Q** is the inverse of **P** above).