Why represent a complex number $a+ib$ as $[\begin{smallmatrix}a & -b\\ b & \hphantom{-}a\end{smallmatrix}]$?

Question

I am reading through John Stillwell's Naive Lie Algebra and it is claimed that all complex numbers can be represented by a $2\times 2$ matrix $\begin{bmatrix}a & -b\\ b & \hphantom{-}a\end{bmatrix}$.

But obviously $a+ib$ is quite different from $\begin{bmatrix}a & -b\\ b & \hphantom{-}a\end{bmatrix}$, as the latter being quite clumsy to use and seldom seen in any applications I am aware of. Furthermore, it complicates simple operations such as matrix multiplication whereby you have to go one extra step and extract the complex number after doing the multiplication.

Can someone explain what exactly is the difference (if there is any) between the two different representations? In what instances is a matrix representation advantageous?

Answer 1

This representation of $\,\Bbb C\,$ arises from viewing $\,\Bbb C\cong \Bbb R^2$ as two-dimensional vector space over $\,\Bbb R,\,$ where $\rm\:\alpha = a+b\,{\it i}\:$ is an $\:\Bbb R$-linear map $\rm\:x\to \alpha\, x.\,$ Computing the coefficients of the matrix of $\,\alpha\,$ wrt to the basis $\,[1,\,{\it i}\,]^T\:$ we obtain

$$\rm (a+b\,{\it i}\,) \left[ \begin{array}{c} 1 \\ {\it i} \end{array} \right] \,=\, \left[\begin{array}{r}\rm a+b\,{\it i}\\\rm -b+a\,{\it i} \end{array} \right] \,=\, \left[\begin{array}{rr}\rm a &\rm b\\\rm -b &\rm a \end{array} \right] \left[\begin{array}{c} 1 \\ {\it i} \end{array} \right]$$

What is the point of such linear representations? By making explicit the innate linear structure we can apply the powerful techniques of linear algebra.

For example, let's look at some analogous linear algebra of Fibonacci numbers. Recall Binet's formula that $\, f_n = (\varphi^n + \bar \varphi^n)/\sqrt{5}\,$ where $\,\varphi,\bar \varphi = (1\pm\sqrt{5})/2\,$ are the roots of $\,x^2-x-1.\,$ Here it is natural to work in $\,\Bbb Q(\varphi) = \Bbb Q(\sqrt{5}) = \{a + b\sqrt{5}: a,b\in\Bbb Q\},\,$ a two-dimensional vector space over $\,\Bbb Q\,$ with basis $\,[\varphi,1].\,$ Here multiplication by $\,\varphi\,$ has the matrix $M$ displayed below

$$\rm {\it \varphi}\, \left[ \begin{array}{c} {\it \varphi} \\ 1 \end{array} \right] \,=\, \left[\begin{array}{r}\rm \varphi + 1\\\rm \varphi + 0 \end{array} \right] \,=\, \left[\begin{array}{rr}\rm 1 &\rm 1\\\rm 1 &\rm 0 \end{array} \right] \left[\begin{array}{c} \varphi \\ 1 \end{array} \right]$$

This leads to the following matrix representation of Fibonacci numbers.

$$\qquad M^n\ =\ \left[\begin{array}{ccc} \,1 & 1 \\\ 1 & 0 \end{array}\right]^n =\ \left[\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right] $$

The above allows us to quickly compute the Fibonacci's numbers by computing the powers of $\,M\,$ by repeated squaring. Further, it yields an easy proof of the Fibonacci addition law

$$\begin{eqnarray} M^{n+m} = M^n M^m &=&\, \left[\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right]\ \left[\begin{array}{ccc} F_{m+1} & F_m \\\ F_m & F_{m-1} \end{array}\right] \\ \\ \Rightarrow\ \ \left[\begin{array}{ccc} F_{n+m+1} & F_{n+m} \\\ \color{#c00}{F_{n+m}} & F_{n+m-1} \end{array}\right]\! &=&\,\left[\begin{array}{ccc} F_{n+1}F_{m+1} + F_nF_m & F_{n+1}F_m + F_nF_{m-1} \\\ \color{#C00}{F_nF_{m+1} + F_{n-1}F_m} & F_{n}F_{m} + F_{n-1}F_{m-1} \end{array}\right]\end{eqnarray}$$

which contains the sought addition law.

$$\color{#c00}{F_{n+m} = F_nF_{m+1} + F_{n-1}F_m} $$

That is but a small glimpse of the power afforded by linear representations.

Answer 2

Simply notice that $$ \begin{bmatrix}a & -b\\ b & a\end{bmatrix}=aI+bJ $$

where $ I=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} $ and $ J=\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}\;. $ Now observe that $I$ and $J$ behave like $1$ and $i$ respectevely (in fact $I$ is the identity for $2\times2$ matrices and $J^2=-I$), which are an $\Bbb R$-basis for $\Bbb C$.

Thus writing $\begin{bmatrix}a & -b\\ b & a\end{bmatrix}$ is equivalent to writing $a+ib$.

Answer 3

In analytic geometry the matrices $$ \left[ \begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array} \right] $$ Are rotations and in complex numbers the numbers $e^{i\theta}$ do the same work on vectors, so this is a good reason to take matrices of this form equal to it's complex form.

Answer 4

There is no essential difference, of course. The text describes two ways to define the complex numbers, one which is more familiar to you and one which is less so. The two structures differ only in the names of the elements. The technical term is that the two are isomorphic. The two presentations are already very close to each other that there is little technical difference. The difficulties you describe are totally inessential. This is just an example of the ability to describe what is essentially the same thing using different objects. There though quite a lot of elegance in the definition of the complex numbers as certain matrices. If nothing else, it's cool that it can be done. On a slightly deeper note, the definition via matrices makes the connection between complex numbers multiplication and rotations almost self-evident.

Answer 5

Stilwell's treatment is supposed to be elementary and therefore avoids the definitions of Lie Groups in terms of Manifolds, instead he only defines what a Matrix Lie Group is. This means that he cannot logically say that the complex plane is a Lie group unless he shows that it is a Matrix Lie Group. That is why he is writing the complex numbers and then the quaternions as matrices.

Answer 6

This definition points out that, in essence, multiplication by a complex number is a way of representing plane similarities.

Another way of defining complex numbers, pointing out its algebraic aspects, namely solving quadratic equations with a negative discriminant, if by the way of qotient ring of a polynomial ring: $$\mathbf C=\mathbf R[x]/(x^2+1),$$ i. e. a complex number is the congruence class of polynomials under the equivalence: $$f(x)\equiv g(x)\mod x^2+1\iff x^2+1 \mid f(x)-g(x). $$ By construction, if we call $\mathrm i$ the class of $x\bmod x^2+ 1$, we have: $\,\mathrm i^2+1=\overline{x^2+1}=\overline 0$.

Answer 7

I don't have enough points to post comments, so I will include this link as an answer:

Why is the complex number $z=a+bi$ equivalent to the matrix form $\left(\begin{smallmatrix}a &-b\\b&a\end{smallmatrix}\right)$

Plenty of different explanations.

Answer 8

Matrix representation of complex numbers is useful and advantageous because we can discover and explore new concepts, like this:

$\begin{bmatrix}\hphantom{-}a & b\\ -b & a\end{bmatrix}$ ---> complex numbers

$\begin{bmatrix}\hphantom{-}a & b\\ \hphantom{-}0 & a\end{bmatrix}$ ---> dual numbers

$\begin{bmatrix}\hphantom{-}a & b\\ \hphantom{-}b & a\end{bmatrix}$ ---> split complex numbers

Answer 9

Quirky little thing, there is a real version, little 2 by 2 blocks, of the Jordan Canonical Form that goes with this. For a real matrix, eigenvalues come in conjugate pairs; $$ \left( \begin{array}{rrrr} i & 1 & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & -i & 1 \\ 0 & 0 & 0 & -i \end{array} \right) $$ is similar to $$ \left( \begin{array}{rr|rr} 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ \hline 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{array} \right) $$

Then $$ \left( \begin{array}{rrrrrr} i & 1 & 0 & 0 & 0 & 0 \\ 0 & i & 1 & 0 & 0 & 0 \\ 0 & 0 & i & 0 & 0 & 0 \\ 0 & 0 & 0 & -i & 1 & 0 \\ 0 & 0 & 0 & 0 & -i & 1 \\ 0 & 0 & 0 & 0 & 0 & -i \end{array} \right) $$ is similar to $$ \left( \begin{array}{rr|rr|rr} 0 & 1 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & -1 & 0 & 0 & 1 \\ \hline 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ \end{array} \right) $$

Answer 10

Bill Dubuques's answer shows how powerful matrix representations can be. For example using the same justification an element $a+b\sqrt2$ of the quadratic number field $\mathbf{Q}[\sqrt2]$ can be written as $1\quad 2b\choose 1\quad a$. In general elements of algebraic number field of degree $n$ can be represented as $n\times n$ matrices with rational numbers as entries. That is we have an injective ring homomorphism from the number field $K$ to a (commutative) subring of non-singuler matrices over Q. The matrix of $\alpha$ being the matrix of the linear transformation corresponding to multiplication by $\alpha$. Norm and trace of the algebraic number are respectively the det and trace of the matrix.

Answer 11

If you represent every complex number this way, then doing normal matrix multiplication and addition with these matrices will yield the same results as multiplying and adding the complex numbers. Doing addition and subtraction with this notation should be trivial that it gives the correct results, namely the real and imaginary parts are separated and only the real parts are added or subtracted from each other, just as the imaginary parts.

When multiplying complex numbers it is harder to see that this yields the correct results. Lets consider the following two complex numbers $a+ib$ and $x+iy$. Multiplying these two numbers should yield,

$$ (a+ib)(x+iy) = ax-by+i(ay+bx) $$

Multiplying these two numbers using the matrix representation looks as follows,

$$ \begin{bmatrix} a & -b\\ b & a \end{bmatrix} \begin{bmatrix} x & -y\\ y & x \end{bmatrix} = \begin{bmatrix} ax-by & -ay-bx\\ ay+bx & ax-by \end{bmatrix} $$

The resulting matrix has the same structure as the first two. The indices which match the assigned location of the real part of a complex number do indeed contain the correct value for the real part of the product of the two complex numbers, just as for the imaginary parts.