math111_logo Appendix: Field

3. Linear algebra over a field

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 R.

Let F be a field. A vector space over F is a set V, its elements u called vectors, with two operations

addition
u + v: V×VV
scalar multiplication
cu: F×VV

satisfying the following properties

commutative
u + v = v + u
associative
(u + v) + w = u + (v + w)
zero
There is a special vector 0V such that u + 0 = u = 0 + u
negative
For any uV there is a vector - u such that u + (- u) = 0
distributive
c(u + v) = cu + cv
distributive
(c + d)u = cu + du
associative
(cd)u = c(du)
one
1u = 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 2a = a + a = 0 for any aF. 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)x1 + (1 - i)x2 = 2
(1 - 2i)x1 + (-3 - 2i)x2 = -5 - 2i

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

(2 + i)x1 + (1 - i)x2 = 2
  (-2 - i)x2 = -5

Then

x2= -5 = -5(-2 + i) = -5(-2 + i) = 2 - i,
-2 - i (-2 - i)(-2 + i) 5

and

x1= 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 C4. Note that in the linear dependence relation

(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 dimCcolA = 3, dimCnulA = 1, where dimC is the complex dimension. The dimensions we used before were real dimensions.

The discussion in the complex euclidean space C4 can also be translated to other complex vector spaces. For example, we know

p1(t) = 1 + t + t2 + t3,
p2(t) = (-1 + 2i)t + (1 + 3i)t2 - it3,
p3(t) = (1 + i) - t + (-2 + 2i)t2 + (2 + i)t3,
p4(t) = 1 + it - t2 - it3,

are linearly dependent and span a (complex) dimension 3 subspace of the vector space P3(C)of complex polynomials of degree ≤ 3.


[part 1] [part 2] [part 3]