### Eigenvalue and Eigenvector

##### 2. Computation

A linear transformation **T**: **R**^{n} → **R**^{n} is given by an `n` by `n` matrix **A**. The eigenvalue `λ` and the eigenvector **v** of **T** are defined by

**Av** = `λ`**v**.

Equivalently, **v** is a nonzero vector in the null space `nul`(**A** - `λ`**I**). The number `λ` and the vector **v** are also called the eigenvalue and the eignevctor of **A**.

The following suggests how to find eigenvalues.

`λ` is an eigenvalue of **A**

⇔ **Av** = `λ`**v** for some **v** ≠ **0** (definition of eigenvalue)

⇔ (**A** - `λ`**I**)**x** = **0** has **x** = **v** as a nontrivial solution

⇔ **A** - `λ`**I** is not invertible (invertibility criterion for square matrices)

⇔ `det`(**A** - `λ`**I**) = 0. (this relation between `det` and invertiblity)

The characteristic polynomial of a square matrix **A** is `det`(**A** - `λ`**I**).

Thus the eigenvalues and the eigenvectors can be computed as follows.

Step 1: Solve the characteristic equation `det`(**A** - `λ`**I**) = 0 and get the eigenvalues `λ`_{1}, `λ`_{2}, ... .

Step 2: For each eigenvalue `λ`_{i}, solve the homogeneuous system (**A** - `λ`_{i}**I**)**x** = **0** and get the eigenvectors with `λ`_{i} as the eigenvalue.

In the second step, the answer is actually presented as a basis of `nul`(**A** - `λ`_{i}**I**), called the eigenspace of **A** with eigenvalue `λ`_{i}.

Example Consider the matrix

of the linear transformation **T**(`x`_{1}, `x`_{2}) = (13`x`_{1} - 4`x`_{2}, - 4`x`_{1} + 7`x`_{2}) on **R**^{2}. The characteristic polynomial is

`det`(**A** - `λ`**I**) = [ |
13 - `λ` |
-4 |
] = (13 - `λ`)(7 - `λ`) - (-4)(-4) = 75 - 20`λ` + `λ`^{2}. |

-4 |
7 - `λ` |

Since 75 - 20`λ` + `λ`^{2} = (5 - `λ`)(15 - `λ`), we get two eigenvalues `λ`_{1} = 5, `λ`_{2} = 15.

For `λ`_{1} = 5, we have

**A** - 5**I** = [ |
13 - 5 |
-4 |
] = [ |
8 |
-4 |
]. |

-4 |
7 - 5 |
-4 |
2 |

The solutions of (**A** - 5**I**)**x** = **0** are of the form **x** = `c`(1, 2), `c` arbitrary. We get an eigenvector **v**_{1} = (1, 2), which is really a basis of the eigenspace `nul`(**A** - 5**I**).

For `λ`_{2} = 15, we have

**A** - 15**I** = [ |
13 - 15 |
-4 |
] = [ |
-2 |
-4 |
]. |

-4 |
7 - 15 |
-4 |
-8 |

The solutions of (**A** - 15**I**)**x** = **0** are of the form **x** = `c`(-2, 1), `c` arbitrary. We get an eigenvector **v**_{2} = (-2, 1), which is really a basis of the eigenspace `nul`(**A** - 15**I**).

Example Consider the matrix

**A** = [ |
1 |
3 |
-3 |
]. |

-3 |
7 |
-3 |

-6 |
6 |
-2 |

The characteristic polynomial has been computed in an earlier example: `det`(**A** - `λ`**I**) = (4 - `λ`)^{2}(- 2 - `λ`). We have two eigenvalues `λ`_{1} = 4, `λ`_{2} = -2.

For `λ`_{1} = 4, we have

**A** - 4**I** = [ |
1 - 4 |
3 |
-3 |
] = [ |
-3 |
3 |
-3 |
]. |

-3 |
7 - 4 |
-3 |
-3 |
3 |
-3 |

-6 |
6 |
-2 - 4 |
-6 |
6 |
-6 |

The solutions of (**A** - 4**I**)**x** = **0** are of the form **x** = `x`_{2}(1, 1, 0) + `x`_{3}(1, 0, -1), `x`_{2} and `x`_{3} arbitrary. We get two eigenvectors **v**_{1} = (1, 1, 0), **v**_{2} = (1, 0, -1), which form a basis of the eigenspace `nul`(**A** - 4**I**).

For `λ`_{2} = -2, we have

**A** + 2**I** = [ |
1 + 2 |
3 |
-3 |
] = [ |
3 |
3 |
-3 |
]. |

-3 |
7 + 2 |
-3 |
-3 |
9 |
-3 |

-6 |
6 |
-2 + 2 |
-6 |
6 |
0 |

The solutions of (**A** + 2**I**)**x** = **0** are of the form **x** = `c`(1, 1, 2), `c` arbitrary. We get an eigenvector **v**_{3} = (1, 1, 2), which is really a basis of the eigenspace `nul`(**A** + 2**I**).

We may verify that **v**_{1}, **v**_{2}, **v**_{3} actually form a basis of **R**^{3}. Geometrically, we have complete undestanding of the linear transformation given by **A**.

From the examples, we saw that for an `n` by `n` matrix **A**, `det`(**A** - `λ`**I**) is a polynomial of degree `n`. The general reason can be found here.

If **A** is the matrix of a linear transformation **T**: **V** → **V** on a finite dimensional vector space with respect to one basis, then according to the discussion here, the matrix of **T** with respect to another basis is **B** = `PAP`^{-1}. By this property, we have

`det`(**B** - `λ`**I**) = `det`(`PAP`^{-1} - `λ`**I**) = `det`(**P**(**A** - `λ`**I**)**P**^{-1}) = `det`(**P**^{-1}**P**(**A** - `λ`**I**)) = `det`(**A** - `λ`**I**).

Thus the characteristic polynomial of a linear transformation does not depend on the choice of the basis. We also call `det`(**A** - `λ`**I**) the characteristic polynomial of **T**.