### Complex Diagonalization

##### 1. Complex eigenvalue and eigenvector

By the Fundamental Theorem of Algebra, any (real or complex) polynomial have roots, as long as complex numbers are allowed. In fact, if we count the repeated roots, a polynomial of degree `n` has `n` roots.

Since the eigenvalues were computed as the roots of the characteristic polynomial, we need to consider the possibility of complex eigenvalues, even for real matrices.

Example Consider the matrix

The characteristic polynomial is `det`(**A** - `λ`**I**) = (1 - `λ`)(3 - `λ`) - (-2)5 = 13 - 4`λ` + `λ`^{2}, and **A** has two eigenvalues `λ`_{1} = 2 + 3`i`, `λ`_{2} = 2 - 3`i`.

For `λ`_{1} = 2 + 3`i`, we have

**A** - (2 + 3`i`)**I** = [ |
-1 - 3`i` |
5 |
]. |

-2 |
1 - 3`i` |

The two equations (-1 - 3`i`)`x`_{1} + 5`x`_{2} = 0 and -2`x`_{1} + (1 - 3`i`)`x`_{2} = 0 are equivalent (you may verify that (1 - 3`i`)[equation 1] = 2[equation 2]). The solutions of (**A** - (2 + 3`i`)**I**)**x** = **0** are of the form **x** = `c`(1 - 3`i`, 2), `c` arbitrary (complex number). We get an eigenvector **v**_{1} = (1 - 3`i`, 2), which is really a basis of the *complex* eigenspace `nul`(**A** - (2 + 3`i`)**I**) (a subspace of **C**^{2}).

For `λ`_{1} = 2 - 3`i`, we have

**A** - (2 - 3`i`)**I** = [ |
-1 + 3`i` |
5 |
]. |

-2 |
1 + 3`i` |

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

We conclude that **A** is diagonalizable, with diagonalization

**A** = [ |
1 + 3`i` |
1 - 3`i` |
] [ |
2 + 3`i` |
0 |
] [ |
1 + 3`i` |
1 - 3`i` |
]^{-1}. |

2 |
2 |
0 |
2 - 3`i` |
2 |
2 |

We can also try to diagonalize complex matrices.

Example The matrix

**A** = [ |
`i` |
-`i` |
1 |
] |

-1 |
1 |
`i` |

0 |
0 |
1 + `i` |

has the characteristic polynomial is `det`(**A** - `λ`**I**) = [(`i` - `λ`)(1 - `λ`) - `i`](1 + `i` - `λ`) = - `λ`(1 + `i` - `λ`)^{2}. We get two eigenvalues `λ`_{1} = 0, `λ`_{2} = 1 + `i`.

For `λ`_{1} = 0, we solve `Ax` = **0** and find solutions **x** = `x`_{2}(1, 1, 0), `x`_{2} arbitrary. Thus **v**_{1} = (1, 1, 0) form a basis of the eigenspace `nul`**A**.

For `λ`_{2} = 1 + `i`, we have

**A** - (1 + `i`)**I** = [ |
-1 |
-`i` |
1 |
]. |

-1 |
-`i` |
`i` |

0 |
0 |
0 |

The solutions of (**A** - (1 + `i`)**I**)**x** = **0** are of the form **x** = `x`_{2}(-`i`, 1, 0), `x`_{2} arbitrary. Thus **v**_{2} = (-`i`, 1, 0) form a basis of the eigenspace `nul`(**A** - (1 + `i`)**I**).

The matrix is not diagonalizable.