Fields

Before we talk about vector spaces, we first need to talk about scalars. Up to this point, we have been treating our scalars as real numbers.

But, it turns out that real numbers are a special case of more general notion. Other examples of this more general notion can be quite interesting.

Definition

A field $k$ is a set equipped with two operations $\begin{aligned} + : k \times k & \to k \\ \times : k \times k & \to k \end{aligned}$ satisfying the following the conditions:

$+$ and $\times$ are associative. This means that for any $a,b,c \in k$ we have $(a + b) + c = a + (b + c)$ and $(a \times b) \times c = a \times (b \times c)$ In other words, order of application does not affect the result of applying either $+$ or $\times$ to a list of elements.
$+$ and $\times$ are commutative. This means that for any $a,b \in $k $a + b = b + a$ and $a \times b = b \times a$ In other words, we can swap the inputs for both $+$ and $\times$ and the outputs remain the same.
We can distribute $\times$ over $+$. For any $a,b,c \in k$, we have $a \times (b + c) = a \times b + a \times c$
There is a special element denoted $0 \in k$ with the property that for any $a \in k$ we have $a + 0 = 0 + a = a$ This element is called the identity for $+$.
There is special element often denoted $1 \in k$ with the property that $a \times 1 = 1 \times a = a$ for any $a \in k$. This element is called the identity for $\times$.
For any element $a \in k$, there is another element $-a \in k$ with so that $a + (-a) = (-a) + a = 0$
Finally for any $a \in k$ with $a \neq 0$, there is another element $a^{-1} \in k$ satisfying $a \times a^{-1} = a^{-1} \times a = 1.$

Examples

The most familiar example is the real numbers $\mathbb{R}$ where $+$ is addition and $\times$ is multiplication. $0$ is zero and $1$ is one. The additive inverse to $r$ is just $-r$ or negative $r$ while the multiplicative inverse $r^{-1}$ is $1/r$.
In Sage, we refer to $\mathbb{R}$ with RR.
In $\mathbb{R}$, we have the field of rational numbers of $\mathbb{Q}$. These are all real numbers that can be represented as a ratio $r = m/n$ of integers $m,n$ with $n \neq 0$.
Given a rational number $q$, $-q$ is rational as is $1/q$ if $q \neq 0$. Both $0$ and $1$ also rational numbers.
It is useful to note that not all real numbers are rational. For example, $\pi$ is known to be irrational.
In Sage, we refer to $\mathbb{Q}$ with QQ.
A class of examples that are more novel are finite fields. We can consider the set $\lbrace 0,1 \rbrace$ alone. Declaring that $0$ is the additive identity determines $+$ most pairs $\begin{aligned} 0 + 0 & = 0 \\ 1 + 0 & = 1 \\ 0 + 1 & = 1 \end{aligned}$ We also need an additive inverse to $1$. It cannot be $0$ since $0 + 1 = 1$. So it must be $1$: $1 + 1 = 0$ Similarly, having $1$ be the multiplicative identity forces $\begin{aligned} 1 \times 1 & = 1 \\ 1 \times 0 & = 0 \\ 0 \times 1 & = 0 \end{aligned}$ What could $0 \times 0$ be? If $0 \times 0 = 1$, then we have a problem with associativity: $(0 \times 0) \times = 1 \times 1 = 1$ while $0 \times (0 \times 1) = 0 \times 1 = 0$ which are not equal. So $0 \times 0 = 0$.
The set $\lbrace 0,1 \rbrace$ with this choice of $+$ and $\times$ is also a field, often written $\mathbb{F}_2$ since it has two elements.
An interesting incarnation of $\mathbb{F}_2$ is obtained by replacing $0$ with False or $F$ and $1$ with True or $T$. Then $+$ is the exclusive or (XOR) operation while $\times$ is the and operation on the set $\lbrace F, T \rbrace$.
For any prime number $p$ and any natural number $r > 0$, there are also fields $\mathbb{F}_{p^r}$ with $p^r$ many elements.
In Sage, we would refer to $\mathbb{F}_{p^r}$ with GF(p^r).
Finally, the complex numbers $\mathbb{C}$ are a field built by taking $\mathbb{R}$ and pretending we have a solution $\imath$ to $x^2 = -1$.
The elements of $\mathbb{C}$ are written as $a + b \imath$ for $a, b \in \mathbb{R}$. $0$ is $0 + 0 \imath$ and $1$ is $1 + 0 \imath$. We add complex numbers as $(a + b \imath ) + (c + d\imath) = (a+c) + (b+d) \imath$ and we multiply them via $(a + b \imath ) (c + d\imath) = (ac - bd) + (ad + bc) \imath$ You can quickly convince yourself that this is the same as allowing us to distribute multiplication over addition and using multiplication in $\mathbb{R}$ plus $\imath^2 = -1$.
The additive inverse to $a + b \imath$ is $-a - b\imath$ but what is the multiplicative inverse? $(a + b\imath)^{-1} = \frac{a - b \imath}{a^2 + b^2}$ which makes sense for all $a + b\imath \neq 0$.
Adding $\imath$ to $\mathbb{R}$ to make $\mathbb{C}$ is surprisingly powerful. The complex numbers make appearances in surprising places from understanding the motion of water waves to the fabric of our universe.
CDF is Sage for $\mathbb{C}$.
Something that is not field is the set of $n \times n$ matrices. We have matrix addition as $+$ and matrix multiplication for $\times$. We have the zero matrix $0$ and the identity matrix $I$.
What we are missing are multiplicative inverses. As we have seen, in general, for a nonzero matrix $A$ there is not another matrix $B$ with $AB = BA = I$
Square matrices of a given size form something more general than a field, called a ring.

Gaussian Elimination

While we focused on $k = \mathbb{R}$, Gaussian elimination can be carried out for any $m \times n$ matrix with entries in $k$.

For most of the examples we have seen, we only used rational numbers in each step.

But we can carry them out over more exotic fields. Let’s take the field ${F,T}$ and reduce

\[A = \begin{pmatrix} T & F & F & T \\ F & F & T & T \\ T & T & T & T \end{pmatrix}\]

Expand for computations.

Notice first that addition and subtract are the same thing in this field and we cannot scale by anything unless it is $F$, ie $0$. This, in fact, simplifies the game. To eliminate the $T$ in $a_{1,3}$ we add $\mathbf{R}_1$ to $\mathbf{R}_3$ gives $$ \begin{pmatrix} T & F & F & T \\ F & F & T & T \\ F & T & T & F \end{pmatrix} $$ Now, we have finished the first column and we turn to the second. We can swap $\mathbf{R}_2$ and $\mathbf{R}_3$ to get $$ \begin{pmatrix} T & F & F & T \\ F & T & T & F \\ F & F & T & T \end{pmatrix} $$ Finally, to arrive at reduced row echelon form, we want to convert $a_{3,2}$ to a $F$. We add $\mathbf{R}_3$ to $\mathbf{R}_2$. $$ \begin{pmatrix} T & F & F & T \\ F & T & F & T \\ F & F & T & T \end{pmatrix} $$

The resulting matrix in reduced row echelon form is $\begin{pmatrix} T & F & F & T \\ F & T & F & T \\ F & F & T & T \end{pmatrix}$

We can use all the tools we developed previously to understand and solve linear systems over a general field $k$.