Numbers defined with matrices

Question

We know that a complex number, written as $c=(a,b)$, can be expressed with the help of a matrix as $$\begin{bmatrix}a & -b\\ b & a\end{bmatrix}$$ and operations on such matrices resemble operations on complex numbers.

However with $2 \times 2$ matrices we could imagine a definition of another type "number" $x=(a,b)$, for example $$(a,b) \longleftrightarrow \begin{bmatrix}a & b\\b & a\end{bmatrix}.$$

Here the operations are quite well defined - multiplication and addition are commutative - the only difference to the complex numbers it seems is that not all numbers have their inverses - for example for $(a,a)$ or $(a,-a)$ it's hard to say what is its inverse.

Why don't we use such "numbers"? Are they numbers at all? When can we say that a given matrix represents number?

The same is true for $ 4 \times 4$ matrices ... it seems only one way of defining numbers - known as quaternions - has found its way into the numbers world... (even though the number of possible ways for constructing matrices with $4$ values when every value is repeated in the matrix $4$ times is much greater).

Answer 1

This is to some extent a question of representation theory. So we do use these things.

For example, suppose you want to extend the field of rational numbers to include some weird number $\xi$ which satisfies $\xi^2 - N = 0$. You can represent multiplication by a number $a+b\xi$ with the 2x2 matrix $\pmatrix{a & Nb \\ b & a}$. Then yours is just a special case of $N = -1$, which (if the underlying elements $a$ and $b$ are in $\mathbb{R}$) is one way to represent complex numbers. If $N = 0$ then we have the dual numbers, and if $N=1$ then we have the split-complex numbers.

Note that extensions like this can cause problems and we may lose field properties, for instance there is no way to divide a real number by a pure dual number since eliminating the dual from the denominator constitutes division by zero. So always check that the basic rules of arithmetic are preserved, or if we need additional constraints. Just because we lose field properties doesn't mean the algebraic structure isn't interesting or useful.

This kind of extension can continue. If your underlying field is $\mathbb{Q}$ and you extend it to include $\xi_N$ (as above) and then want to extend it again to include another $\xi_M$ then you have a 4x4 matrix representation.

Answer 2

The point is that we want to have a field. At least, this is what is suggested here by speaking of "numbers". Now in general, subalgebras of $M_2(K)$ need not be a field. In the first case, however, we obtain a field, namey the field of complex numbers $\mathbb{C}$ . In the second example, we do not obtain a field. For a field $(K,+,\cdot)$, we need that $(K,+)$ and $(K^*,\cdot)$ are both abelian groups. This is not the case here.

Answer 3

Not sure if I fully understand your question, but I'll give it a try. The reason that we can represent a complex number $a+bi$ as $$\begin{bmatrix}a&-b\\b&a\end{bmatrix}=a\begin{bmatrix}1&0\\0&1\end{bmatrix}+b\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ is because they behave the same way. For instance $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}^2=(-1)\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ behaves just like $i$, since $i^2=-1$.

You can also do this with the real numbers by identifying $a\in\mathbb{R}$ with the matrix $$\begin{bmatrix}a&0\\0&a\end{bmatrix}=a\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ but the crucial point is that the "numbers" and the "matrices" have to behave the same way (this is called an isomorphism).

Answer 4

To put what you've done in algebraic context: you've identified a subset of the ring of 2x2 matrices (with real number entries I assume, since those are the type we use to represent the complex numbers) and showed that within this subset, multiplication is commutative, and the subset is closed under addition and multiplication, so it's a commutative subring. In quid's answer to this related question, quid shows that your subring is isomorphic to the direct product of two copies of $\mathbb{R}$. I refer you there to the algebraic proof: I'll show it for your example in a more concrete way.

Associate to each matrix $\begin{bmatrix} a & b \\ b & a \end{bmatrix}$ the pair $(a+b,a-b)$. It's clear that, conversely, given such a pair, you can recover the matrix. Let's check how matrix addition and multiplication affect the pair. $$\begin{bmatrix} a & b \\ b & a \end{bmatrix} + \begin{bmatrix} a' & b' \\ b' & a' \end{bmatrix} = \begin{bmatrix} a+a' & b+b' \\ b+b' & a+a' \end{bmatrix}$$, so $$(a+b,a-b)+(a'+b',a'-b') = (a+a'+b+b',a+a'-b-b')$$. And $$\begin{bmatrix} a & b \\ b & a \end{bmatrix} \cdot \begin{bmatrix} a' & b' \\ b' & a' \end{bmatrix} = \begin{bmatrix} aa'+bb' & ab'+ba' \\ ab'+ba' & aa'+bb'' \end{bmatrix}$$, so $$(a+b,a-b)\cdot(a'+b',a'-b') = (aa'+bb'+ab'+ba',aa'+bb'-ab'-ba')= ((a+b)(a'+b'),(a-b)(a'-b'))$$. Note that the matrix addition and multiplication on the matrices induce regular addition and multiplication on the pairs, performed independently on the first and second entries.

So it turns out these "numbers" are equivalent to pairs of numbers on which addition and multiplication are performed separately.

Answer 5

The matrix representation of the complex numbers shows that there exists an isomorphism between the complex numbers and that particular subset of matrices in $\mathbb R^{2 \times 2}$.

The point is, they started with the complex numbers and then searched for a matrix representation.

I don't want to get all Zen on you***, but what is a number?

The integers, whole number, natural numbers, real numbers, complex numbers, quaternions, and so on all have different properties. Yet we call them all numbers.

What about differential forms? What about $m \times n$ arrays of real numbers? What about the set of all permutations of the set $\{1,2,3 \dots \}$? Are they numbers?

You wonder, since there exists a particular subset of $\mathbb R^{2 \times 2}$ that behaves exactly like the field of complex numbers, can that process be turned around to create new $\text{$``$numbers$"$}$?

Sure it can. I suppose that you should expect at least for some form of closure to happen but everything else is just a matter of what you discover to be true and decide is important.

The big question is, when you show what you have discovered to the rest of the world, will they call them numbers too? That is to say, what is a number?

*** Whenever someone says "I don't want to...", they want to.

Answer 6

Just to maybe help widen the number concept.. =)

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Any function from a finite subset of $\mathbb Z$ onto itself can be represented as a binary matrix with column sum = 1. For example square modulo 4: $\cases{0\to 0\\1\to 1\\2\to 0\\3\to 1}$

$$\left[\begin{array}{cccc}1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0\end{array}\right] \text{ assuming column vectorization of variable:}\left[\begin{array}{cccc}mod=0\\mod=1\\mod=2\\mod=3\end{array}\right]$$

Now matrix multiplication represents function concatenation so we can actually treat functions as numbers. Cool huh?

Answer 7

You can use whatever "numbers" you like! At a bare minimum, you probably want to be able to add, subtract, and multiply your "numbers". Division might also be nice too. It turns out that many of these "numbers" can be thought of as matrices.

Here is an interesting example: the "dual numbers". A dual number $a + b \varepsilon$ has a real part $a$, and an infinitesimal part $b\varepsilon$. The interesting thing about the dual numbers is that $\varepsilon^2 = 0$ (think of $\varepsilon$ as being really really small, so when it gets squared it goes away). In general, the multiplication rule is $$(a + b \varepsilon)(c + d \varepsilon) = ac + bc\varepsilon + ad\varepsilon + bd\varepsilon^2 = ac + (bc + ad)\varepsilon$$

We can embed the dual numbers into the $2 \times 2$ matrices, via $$ a + b \varepsilon = \begin{bmatrix}a & b \\ 0 & a \end{bmatrix}$$ and double-check that this multiplication works: $$ (a + b \varepsilon)(c + d \varepsilon) = \begin{bmatrix}a & b \\ 0 & a \end{bmatrix}\begin{bmatrix}c & d \\ 0 & c \end{bmatrix}=\begin{bmatrix}ac & ad+bc \\ 0 & ac \end{bmatrix}$$

One super-cool thing about the dual numbers is that they somehow know how to differentiate: We know via calculus that if $f(x) = x^2 + 5x - 1$, then $f'(x) = 2x + 5$. But the dual numbers just know this anyway: $$ f(x + \varepsilon) = x^2 + 2x\varepsilon + \varepsilon^2 + 5x + 5\varepsilon - 1 = (x^2 + 5x - 1) + (2x + 5)\varepsilon$$

Answer 8

As others already mentioned, your proposed system is isomorphic to split-complex numbers.

I just wanted to add that if you allow $a$ and $b$ in your matrix to be complex numbers, you will get tessarines, a 4-dimensional algebra, that is simply addition of complex and split-complex unities to the reals simultaneously (the fourth basis vector will be $ij$).

It is commutative, associative and algebraically closed in a sense (if to consider polynomials with coefficients which are not zero divisors).