Linear Algebra: Field
When studying linear algebra, we are continuously faced with the word field. In most cases, we use the set of real numbers, which is an example of a field.
But what is a field? A field is a set that satisfies certain conditions. Let’s dive deeper into it. Just a brief read is fine. For now, it’s enough to keep in mind that the set of real numbers and the set of complex numbers are examples of fields.
The Official Rules: Defining a Field
For a set, let’s call it \(F\), to officially be a field, it must have two operations: addition \((+)\) and multiplication \((\cdot)\), that follow a specific set of rules, for all elements \(a, b, c \in F\).
Let’s break down these rules, which are laid out in the definition below:
Note: It’s a key point that these operations are defined within the context of the set F. The symbols \((+)\) and \((\cdot)\) are just placeholders. For example, the rule for adding two real numbers is different from the rule for adding two complex numbers. A field is always a package deal: a set plus its two specific operations.
1. Commutativity
The order doesn’t matter for addition or multiplication:
$$
a + b = b + a
$$
$$
a \cdot b = b \cdot a
$$
2. Associativity
When you’re just adding or just multiplying, the grouping of numbers doesn’t affect the result:
$$
(a + b) + c = a + (b + c)
$$
$$
(a \cdot b) \cdot c = a \cdot (b \cdot c)
$$
3. Identity Elements
The set must contain unique elements for addition and multiplication that don’t change other elements.
- Additive Identity: There exists an element \(0 \in F\) such that $$
a + 0 = a \quad \forall a \in F
$$- Here, the symbol \(\forall\) means “for all” or “for every”. And \(a \in F\) means ‘a is an element of the set F’. So, together \(\forall a \in F\) read as: For every element \(a\) in the set \(F\).
- Multiplicative Identity: There exists an element \(1 \in F\) such that $$
a \cdot 1 = a \quad \forall a \in F
$$
4. Inverse Elements
Every element must have an “opposite” for both addition and multiplication.
- Additive Inverse: For any \(a \in F\), there exists an element \(c\) (we usually denote \(c\) as \(-a\)) such that $$
a + c = 0
$$ - Multiplicative Inverse: For any nonzero \(b \in F\), there exists an element \(d\) (we usually denote \(d\) as \(b^{-1}\) or \(1/b\)) such that $$
b \cdot d = 1
$$
5. Distributivity
Multiplication distributes over addition, linking the two operations:
$$
a \cdot (b + c) = (a \cdot b) + (a \cdot c)
$$
Common Examples of Fields
The Set of Real Numbers \((\mathbb{R})\)
This is the default field for most introductory linear algebra. Let’s see how it fits the rules:
• Commutativity:
$$
5 + 7 = 7 + 5, \quad 5 \cdot 7 = 7 \cdot 5
$$
• Associativity:
$$
(2 + 3) + 4 = 2 + (3 + 4)
$$
• Identities:
- The additive identity is \(0\):
$$
12 + 0 = 12
$$
- The multiplicative identity is \(1\):
$$
12 \cdot 1 = 12
$$
• Inverses:
- The additive inverse of \(9\) is \(-9\):
$$
9 + (-9) = 0
$$
- The multiplicative inverse of \(9\) is \(1/9\):
$$
9 \cdot {1}/{9} = 1
$$
• Distributivity:
$$
3 \cdot (4 + 5) = (3 \cdot 4) + (3 \cdot 5)
$$
Great! Let’s rewrite the Complex Numbers example with LaTeX formulas for clarity:
The Set of Complex Numbers \((\mathbb{C})\)
Another crucial field, especially in physics and engineering. Let
$$
z_{1} = a + bi, \quad z_{2} = c + di
$$
• Commutativity:
$$
z_{1} + z_{2} = (a + c) + (b + d)i
$$
$$
= (c + a) + (d + b)i = z_{2} + z_{1}
$$
• Associativity (See Appendix):
$$
(z_{1} + z_{2}) + z_{3} = z_{1} + (z_{2} + z_{3})
$$
• Identities:
- Additive identity is
$$
0 + 0i
$$
Because
$$
z + (0+0i) = (a+bi) + (0+0i)
$$
$$= (a+0) + (b+0)i$$
$$= a + bi = z.$$
- Multiplicative identity is
$$
1 + 0i
$$
Because
$$
z \cdot (1+0i) = (a+bi)(1+0i)
$$
$$= a\cdot 1 – b\cdot 0 + (a\cdot 0 + b\cdot 1)i$$
$$ = a + bi = z$$
• Inverses:
- Additive inverse of \(a + bi\):
$$
-(a + bi) = -a – bi
$$
- Multiplicative inverse of a nonzero \(a + bi\):
$$
\frac{1}{a + bi} = \frac{a}{a^{2} + b^{2}} – \frac{b}{a^{2} + b^{2}} i
$$
• Distributivity (See Appendix):
$$
z_{1}(z_{2} + z_{3}) = z_{1}z_{2} + z_{1}z_{3}
$$
A Non-Example: The Integers \((\mathbb{Z})\)
The integers are not a field because they fail the multiplicative inverse rule.
For example, for the integer \(2\), there is no integer \(x\) such that
$$
2 \cdot x = 1
$$
We’d need \(x = 1/2\), but \(1/2 \notin \mathbb{Z}\).
This shows clearly why \(\mathbb{R}\) and \(\mathbb{C}\) are fields, but \(\mathbb{Z}\) is not.
Appendix: Proofs for Complex Number Properties
Let
$$
z_{1} = a + bi, \quad z_{2} = c + di, \quad z_{3} = e + fi
$$
Proof of Associativity of Addition
We want to show:
$$
(z_{1} + z_{2}) + z_{3} = z_{1} + (z_{2} + z_{3})
$$
1. Left-Hand Side (LHS):
$$
(z_{1} + z_{2}) + z_{3}
$$
$$= \big((a+bi) + (c+di)\big) + (e+fi)$$
$$
= \big((a+c) + (b+d)i\big) + (e+fi)
$$
$$
= \big((a+c) + e\big) + \big((b+d) + f\big)i
$$
2. Right-Hand Side (RHS):
$$
z_{1} + (z_{2} + z_{3})
$$
$$ = (a+bi) + \big((c+di) + (e+fi)\big)$$
$$
= (a+bi) + \big((c+e) + (d+f)i\big)
$$
$$
= \big(a + (c+e)\big) + \big(b + (d+f)\big)i
$$
Conclusion:
Since both LHS and RHS simplify to the same expression, we have
$$
(z_{1} + z_{2}) + z_{3} = z_{1} + (z_{2} + z_{3})
$$
Thus, addition of complex numbers is associative. ✅
Proof of Distributivity of Multiplication over Addition
We want to show:
$$
z_{1}(z_{2} + z_{3}) = z_{1}z_{2} + z_{1}z_{3}
$$
Let
$$
z_{1} = a + bi, \quad z_{2} = c + di, \quad z_{3} = e + fi
$$
Left-Hand Side (LHS):
$$
z_{1}(z_{2} + z_{3}) = (a+bi)\big((c+di) + (e+fi)\big)
$$
$$
= (a+bi)\big((c+e) + (d+f)i\big)
$$
$$
= a(c+e) + a(d+f)i
$$
$$~~~~~~~~~~ + bi(c+e) + bi(d+f)i$$
$$
= \big(ac + ae\big) + \big(ad + af\big)i
$$
$$~~~~~~~~~~+ \big(bc + be\big)i + \big(bd + bf\big)i^{2}$$
Since \(i^{2} = -1\):
$$
= (ac + ae – (bd + bf))
$$
$$~~~~~~~~~~+ (ad + af + bc + be)i$$
Right-Hand Side (RHS):
$$
z_{1}z_{2} + z_{1}z_{3} = (a+bi)(c+di)
$$
$$~~~~~~~~~~+ (a+bi)(e+fi)$$
Expand the first term:
$$
(a+bi)(c+di) = ac + adi + bci + bdi^{2}
$$
$$= (ac – bd) + (ad + bc)i$$
Expand the second term:
$$
(a+bi)(e+fi) = ae + afi + bei + bfi^{2}
$$
$$= (ae – bf) + (af + be)i$$
So, adding them:
$$
z_{1}z_{2} + z_{1}z_{3} = (ac – bd + ae – bf)
$$
$$~~~~~~~~~~ + (ad + bc + af + be)i$$
Comparison:
LHS:
$$
(ac + ae – (bd + bf)) + (ad + af + bc + be)i
$$
RHS:
$$
(ac – bd + ae – bf) + (ad + bc + af + be)i
$$
They are identical. ✅
Conclusion:
$$
z_{1}(z_{2} + z_{3}) = z_{1}z_{2} + z_{1}z_{3}
$$
Thus, multiplication distributes over addition in the complex numbers.