### Properties

##### 2. About several matrices

In this part, we consider the eigenvalues and eigenvectors of several matrices.

Let **v** be an eigenvector of **A** and **B** with eigenvalues `λ` and `μ`. Then **v** is an eigenvector of `a`**A** + `b`**B** with eigenvalue `aλ` + `bμ`, and an eigenvector **AB** with eigenvalues `λμ`.

Proof From `Av` = `λ`**v**, `Bv` = `μ`**v**, we have

(`a`**A** + `b`**B**)**v** = `a`**Av** + `b`**Bv** = `aλ`**v** + `bμ`**v** = (`aλ` + `bμ`)**v**,

(`AB`)**v** = **A**(`Bv`) = **A**(`μ`**v**) = `λμ`**v**.

Example Consider the matrices

**A** = [ |
13 |
-4 |
], **B** = [ |
6 |
2 |
]. |

-4 |
7 |
2 |
9 |

From an earlier example, we know **v**_{1} = (1, 2), **v**_{2} = (-2, 1) are eigenvectors of **A** with eigenvalues 5 and 15. It is also easy to verify that they are eigenvectors of **B** with eigenvalues 10 and 5. We may then conclude that **v**_{1} = (1, 2), **v**_{2} = (-2, 1) are eigenvectors of the following matrices with the following eigenvalues:

**A** + **B** = [ |
19 |
-2 |
], `λ`_{1} = 5 + 10 = 15, `λ`_{2} = 15 + 5 = 20. |

-2 |
16 |

**A** + 2**B** = [ |
25 |
0 |
], `λ`_{1} = 5 + 20 = 15, `λ`_{2} = 15 + 10 = 25. |

0 |
25 |

**AB** = [ |
70 |
-10 |
], `λ`_{1} = 5×10 = 50, `λ`_{2} = 15×5 = 75. |

-10 |
55 |

**AB** - **B**^{2} = [ |
30 |
-40 |
], `λ`_{1} = 50 - 100 = -50, `λ`_{2} = 75 - 25 = 50. |

-40 |
-30 |

Let `AB` = `BA`. Let **v** be an eigenvector of **A** with eigenvalue `λ` and `Bv` ≠ **0**. Then `Bv` is again an eigenvector of **A** with eigenvalue `λ`.

Proof **A**(`Bv`) = (`AB`)**v** = (`BA`)**v** = **B**(`Av`) = **B**(`λ`**v**) = `λ`**Bv**.

Example Let **A** be a 3 by 3 matrix commuting with

**B** = [ |
0 |
2 |
0 |
]. |

1 |
-1 |
1 |

3 |
-5 |
3 |

Suppose **v**_{1} = (1, 1, 0), **v**_{2} = (1, 1, 2) are eigenvectors of **A** with eigenvalues 4 and -2. Then `Bv`_{1} = (2, 0, -2) and `Bv`_{2} = (2, 2, 4) are also eigenvectors of **A** with eigenvalues 4 and -2. Note that **v**_{1}, **v**_{2}, `Bv`_{1} form a basis of eigenvectors of **A** with eigenvalues 4, -2, 4. Therefore for

**P** = [**v**_{1} **v**_{2} `Bv`_{1}] = [ |
1 |
1 |
2 |
], **D** = [ |
4 |
0 |
0 |
], |

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

0 |
2 |
-2 |
0 |
0 |
4 |

we have

**A** = **PDP**^{-1} = [ |
1 |
3 |
-3 |
]. |

-3 |
7 |
-3 |

-6 |
6 |
-2 |

The matrix appeared in this example. You may also directly verify that `AB` = `BA`.

Finally, we remark that the properties also holds for linear transformations **T**, **S**: **V** → **V**. For example, if **v** is an engenvector for both **T** and **S**, then **v** is an engenvector for `a`**T** + `b`**S** and **TS**.