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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).

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  • $\begingroup$ Hi, complex noob here, can you point me to a link where I can read about matrix representation of complex numbers? Thanks! $\endgroup$ – Gaurang Tandon Feb 12 '18 at 15:51
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    $\begingroup$ @GaurangTandon Ok. See en.wikipedia.org/wiki/… $\endgroup$ – Widawensen Feb 12 '18 at 15:52
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    $\begingroup$ It is not really clear what you are asking. The usual matrix representation of a complex number is not arbitrary, it is invertible iff $(a,b) \neq 0$. $\endgroup$ – copper.hat Feb 12 '18 at 15:54
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    $\begingroup$ The point is not calling them numbers or not, but the properties they satisfy or fail to satisfy. Many choices are vector spaces over the field of coefficients, some are associative algebras too using the matrix multiplication, fewer are division algebras. $\endgroup$ – user530511 Feb 12 '18 at 15:55
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    $\begingroup$ What is "number"? $\endgroup$ – Dietrich Burde Feb 12 '18 at 15:56
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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.

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  • $\begingroup$ Thank you for the explanation and in-depth links, however I'm not sure when we can exclude a given form of matrix from consideration it as a number...perhaps I'm to weak in abstract algebra.. $\endgroup$ – Widawensen Feb 12 '18 at 16:12
  • $\begingroup$ @Widawensen I don't believe there's a simple answer to that other than working out the arithmetic to see which properties are preserved. If you understand matrix arithmetic then it is straightforward (but often tedious) to check. For instance moving from $\mathbb{R}$ to $\mathbb{C}$ is straightforward but then moving from $\mathbb{C}$ to $\mathbb{H}$ you lose some arithmetic properties (in particular you lose the commutative property). $\endgroup$ – law-of-fives Feb 12 '18 at 16:17
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    $\begingroup$ +1 out of curiosity, are multicomplex numbers also special cases of this? $\endgroup$ – user541686 Feb 12 '18 at 22:57
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    $\begingroup$ @Mehrdad It is straightforward. If you picture extending your base field by the quadratic element $\xi_i$ giving a 2x2 matrix, then these 2x2 matrices are your numbers, and you can extend quadratically again getting a 4x4 matrix. Something like $a \binom{\sqrt{-1}}{\rightarrow} \pmatrix{a & -b\\ b & a} \binom{\sqrt{-1}}{\rightarrow} \pmatrix{A & -B \\ B & A}$. But solving "the same" equation twice causes other problems as your link shows. $\endgroup$ – law-of-fives Feb 13 '18 at 0:19
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    $\begingroup$ If I understand you correctly, you are interested in $\mathbb R$-algebras that are subalgebras of $Mat(n \times n, \mathbb R)$ for some $n \in \mathbb N$. Perhaps surprisingly, the only $\mathbb R$-algebras that can be construct this way, which also happen to be Skew-fields (i.e, every element is invertible, but multiplication is not necessarily commutative) are isomorphic to either $\mathbb R$, $\mathbb C$ or $\mathbb H$ (This is known as Frobenius theorem). $\endgroup$ – H1ghfiv3 Feb 13 '18 at 11:33
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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.

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  • $\begingroup$ Very interesting approach.. so these pairs of numbers can be treated here separately, but in the case of complex representation I suppose not, is a good conclusion? $\endgroup$ – Widawensen Feb 12 '18 at 16:24
  • $\begingroup$ @Widawensen Yes. To say that the complex numbers can't be turned into pairs treated separately we'd formally say that $\mathbb{C}$ is not isomorphic to $\mathbb{R} \times \mathbb{R}$. This can be proven from the observation (which you made) that all complex numbers are invertible, while not all these pairs are invertible, e.g. the pair $(2,0)$ cannot multiply by anything to give $(1,1)$. $\endgroup$ – BallBoy Feb 12 '18 at 16:30
  • $\begingroup$ Such observations as you've made are clarifying the situation, that's what I'm looking for .... for essential differences in both cases... it's interesting that all explanations given here are lightening the problem as if from different points of view.. $\endgroup$ – Widawensen Feb 12 '18 at 16:39
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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.

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  • $\begingroup$ Aha, connections with abstract algebra are essential here, that's when I'm weak. But I memorize these "abelian groups" and I will rethink them.. $\endgroup$ – Widawensen Feb 12 '18 at 16:04
  • $\begingroup$ I'm also interested how the construct of field is referenced to the quaternions? $\endgroup$ – Widawensen Feb 12 '18 at 16:17
  • $\begingroup$ You have been right .. maybe the "field" is the central construct here ...according to Wikipedia necessary is condition of existence "multiplicative inverse $b^−1$ for every nonzero element $b$", so "numbers" are only numbers when are appearing on a field.. quaternions have also always (except 0) inverse elements therefore they are the numbers.. $\endgroup$ – Widawensen Feb 12 '18 at 17:05
  • $\begingroup$ Hamilton's real quaternions do not form not a field, but they are still close to it. They are sometimes called "a number system that extends the complex numbers", see here. So you are right in a way. $\endgroup$ – Dietrich Burde Feb 12 '18 at 19:50
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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.

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  • $\begingroup$ Hmm, this gives a freedom in the decision what the number is.. really we have so much freedom? $\endgroup$ – Widawensen Feb 12 '18 at 16:43
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    $\begingroup$ @Widawensen No, you don't. There was a time when people argued whether or not $0$ and $1$ were numbers. Look at the names of some numbers, they all have distainful connotations: negative, complex, imaginary, surd (as in a number that is not surd must be an absurd number), irrational. Ultimately, the decision whether a particular set is a set of numbers is a matter of (mathematical) public opinion based mostly on utility. $\endgroup$ – steven gregory Feb 12 '18 at 17:03
  • $\begingroup$ ..but what with the proposition of Dietrich that numbers are the numbers only on - so called - field ? Maybe this is what is widely accepted nowadays? $\endgroup$ – Widawensen Feb 12 '18 at 17:08
  • $\begingroup$ That's his definition. How well has it been received? $\endgroup$ – steven gregory Feb 12 '18 at 17:49
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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).

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    $\begingroup$ Thank you for the answer, I understand correspondence between complex numbers and these matrices because we have rules of translation one representation to the other. Any way I appreciate your answer... $\endgroup$ – Widawensen Feb 12 '18 at 16:08
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Just to maybe help widen the number concept.. =)


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?

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    $\begingroup$ Functions as numbers? Is it not going too far ?? $\endgroup$ – Widawensen Feb 13 '18 at 11:34
  • $\begingroup$ @Widawensen the really cool stuff does not happen until you find ways to do the same for subsets of $\mathbb R$, then you can do some real magic. ;) $\endgroup$ – mathreadler Feb 13 '18 at 22:07
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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: $$ f(x + \varepsilon) = x^2 + 2x\varepsilon + \varepsilon^2 + 5x + 5\varepsilon - 1 = (x^2 + 5x - 1) + (2x + 5)\varepsilon$$

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  • $\begingroup$ Presented connection with differentation is really cool .. it seems that everything can be "replaced" with matrices :) $\endgroup$ – Widawensen Feb 13 '18 at 11:45

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