Linear algebra can be introduced over any field because we can define the concept of vector space over any field. All we need to do is to change real number **R** to a field ** F** in the definition of vector spaces over

Let ** F** be a field. A vector space over

- addition
+`u`:`v`×`V`→`V``V`- scalar multiplication
`c`:**u**×`F`→`V``V`

satisfying the following properties

- commutative
+`u`=`v`+`v``u`- associative
- (
+`u`) +`v`=`w`+ (`u`+`v`)`w` - zero
- There is a special vector
**0**∈such that`V`+`u`**0**==`u`**0**+`u` - negative
- For any
∈`u`there is a vector -`V`such that`u`+ (-`u`) =`u`**0** - distributive
`c`(+`u`) =`v``c`+`u``c``v`- distributive
- (
`c`+`d`)=`u``c`+`u``d``u` - associative
- (
`cd`)=`u``c`(`d`)**u** - one
- 1
=`u``u`

The claim is that, with one exception, everything we did so far is still valid over field. The one exception has to do with fields of characteristic 2. We say an field has characteristic 2 if 2`a` = `a` + `a` = 0 for any `a` ∈ ** F**. There are certain arguments (see this exercise, for example) in which we need to divide the number 2 (= 1 + 1, where 1 is the unit element of the field). Such argument is only valid in a field of characteristic ≠ 2.

We will do some examples to practice linear algebra over **C**.

Example We try to solve the system

(2 + i)x_{1} |
+ (1 - i)x_{2} |
= | 2 |

(1 - 2i)x_{1} |
+ (-3 - 2i)x_{2} |
= | -5 - 2i |

Noting that 1 - 2`i` = -`i`(2 + `i`), we do `i`[equation 1] + [equation 2] and get

(2 + i)x_{1} |
+ (1 - i)x_{2} |
= | 2 |

(-2 - i)x_{2} |
= | -5 |

Then

x_{2}= |
-5 | = | -5(-2 + i) |
= | -5(-2 + i) |
= 2 - i, |

-2 - i |
(-2 - i)(-2 + i) |
5 |

and

x_{1}= |
2 - (1 - i)(2 - i) |
= | (1 + 3i)(2 - i) |
= | 5 + 5i |
= 1 + i. |

2 + i |
(2 + i)(2 - i) |
5 |

Alternatively, we may also use the inverse formula and get

[ | 2 + i |
1 - i |
]^{-1} = |
1 | [ | -3 - 2i |
-1 + i |
] |

1 - 2i |
-3 - 2i |
(2 + i)(-3 - 2i) - (1 - i)(1 - 2i) |
-1 + 2i |
2 + i |

= | -3 + 4i |
[ | -3 - 2i |
-1 + i |
] = | 1 | [ | 17 - 6i |
-1 - 7i |
]. |

(-3 - 4i)(-3 + 4i) |
-1 + 2i |
2 + i |
25 | -5 - 10i |
-10 + 5i |

Then the solution is

=x |
1 |
[ | -3 - 2i |
-1 + i |
] [ | 2 |
] = | -3 + 4i |
1 - 7i |
] = [ | 1 + i |
]. |

-3 - 4i |
-1 + 2i |
2 + i |
-5 - 2i |
(-3 - 4i)(-3 + 4i) |
-10 - 5i |
2 - i |

Example Consider a matrix

= [A |
1 | 0 | 1 + i |
1 | ]. |

1 | -1 + 2i |
-1 | i |
||

1 | 1 + 3i |
-2 + 2i |
-1 | ||

1 | -i |
2 + i |
-i |

By (-1)[row 1] + [row 2], (-1)[row 1] + [row 3], (-1)[row 1] + [row 3], we have

[ | 1 | 0 | 1 + i |
1 | ] |

0 | -1 + 2i |
-2 - i |
-1 + i |
||

0 | 1 + 3i |
-3 + i |
-2 | ||

0 | -i |
1 | -1 - i |

By [row 2] ↔ [row 4], (3 - `i`)[row 2] + [row 3], (2 + `i`)[row 2] + [row 4], we have

[ | 1 | 0 | 1 + i |
1 | ] |

0 | -i |
1 | -1 - i |
||

0 | 0 | 0 | -6 -2i |
||

0 | 0 | 0 | -2 - 2i |

By [row 3] ↔ [row 4], (-2 + `i`)[row 3] + [row 4], we have

[ | 1 | 0 | 1 + i |
1 | ] |

0 | -i |
1 | -1 - i |
||

0 | 0 | 0 | -2 -2i |
||

0 | 0 | 0 | 0 |

Thus the four columns of ** A** are linearly dependent and do not span

(1 + `i`)[col 1] + `i`[col 2] + (-1)[col 3] + 0[col 4] = 0,

it is often necessary to take complex coefficients. We also know `dim`_{C}`col A` = 3,

The discussion in the complex euclidean space **C**^{4} can also be translated to other complex vector spaces. For example, we know

`p`_{1}(`t`) = 1 + `t` + `t`^{2} + `t`^{3},

`p`_{2}(`t`) = (-1 + 2`i`)`t` + (1 + 3`i`)`t`^{2} - `it`^{3},

`p`_{3}(`t`) = (1 + `i`) - `t` + (-2 + 2`i`)`t`^{2} + (2 + `i`)`t`^{3},

`p`_{4}(`t`) = 1 + `it` - `t`^{2} - `it`^{3},

are linearly dependent and span a (complex) dimension 3 subspace of the vector space `P`_{3}(**C**)of complex polynomials of degree ≤ 3.