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Relation of this antisymmetric matrix $r = \!\left(\begin{smallmatrix}0 &1\\-1 & 0\end{smallmatrix}\right)$ to $i$

On Wikipedia, it says that:

Matrix representation of complex numbers
Complex numbers $z=a+ib$ can also be represented by $2\times2$ matrices that have the following form: $$\pmatrix{a&-b\\b&a}$$

I don't understand why they can be represented by these matrices or where these matrices come from.

Christian Chapman
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NFDream
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8 Answers8

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No one seems to have mentioned it explicitly, so I will. The matrix $J = \left( \begin{smallmatrix} 0 & -1\\1 & 0 \end{smallmatrix} \right)$ satisfies $J^{2} = -I,$ where $I$ is the $2 \times 2$ identity matrix (in fact, this is because $J$ has eigenvalues $i$ and $-i$, but let us put that aside for one moment). Hence there really is no difference between the matrix $aI + bJ$ and the complex number $a +bi.$

Christian Chapman
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Geoff Robinson
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    Yes, indeed, so, if there had been any doubt whether there could be a legitimate square root of $-1$, here is one! :) – paul garrett Aug 10 '12 at 00:04
  • Very clear and easy understandable,thank you – NFDream Aug 10 '12 at 00:16
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    We should not forget that in the nineteenth century this sort of thing was considered an important legitimization of complex numbers. Also, Kronecker's $\mathbb C=\mathbb R[x]/\langle x^2+1\rangle$. – paul garrett Aug 10 '12 at 00:29
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    Jumping in on a month old question here, but is there a way to show that $J$ is unique in this regard? – crf Sep 05 '12 at 05:48
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    Well, no, it isn't unique. You can replace it with a conjugate within ${\rm GL}(2,\mathbb{R}),$ but it is unique up to conjugacy within the group of invertible real $2 \times 2$ matrices. – Geoff Robinson Sep 05 '12 at 07:43
  • @paulgarrett What's wrong with just identifying $a+bi$ with the ordered pair $(a,b)$? – Akiva Weinberger Aug 19 '15 at 03:34
  • @AkivaWeinberger That's the modern approach. Matrices were a more familiar terrain back then. – Dustan Levenstein Jan 13 '16 at 16:06
  • @AkivaWeinberger ordered pairs don't have arithmetic operations such as addition and (especially important) multiplication defined automatically. It's natural to $\textit{additionally}$ impose the definition $(a,b)+(c,d) = (a+c,b+d)$, but imposing the multiplication rule $(a,b)(c,d) = (ac-bd, ad+bc)$ isn't super intuitive in pair notation. The more natural multiplication rule of $(a,b)(c,d)=(ac,bd)$ is not useful because the result depends on your coordinates. Personally, I explicitly think of complex numbers as "two dimensional numbers that come with consistent multiplication built in". – Zxv Feb 03 '20 at 01:28
  • @Zxv What does "the result depends on your coordinates" mean? Isn't that also true for the ordinary complex addition and multiplication rules? – IssaRice Jun 23 '22 at 21:16
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    @riceissa "Isn't that also true for the ordinary complex addition and multiplication rules?" No, because the naïve multiplication rule I listed of $(a,b)*(c,d)=(ac,bd)$ isn't how complex multiplication is defined. Complex numbers have the rule of $i^2=-1$ which naturally generates the correct rule. The naïve rule is actually known as the Hadamard product. It has properties and uses (e.g. in image compression & neural networks), but it is not as widespread as the standard matrix product or complex numbers in maths and physics because it famously isn't preserved under transformations. (1/3) – Zxv Jun 24 '22 at 02:57
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    As for why it's correct and what I mean by "depends on coordinates": consider a point in $(a,b) \in \mathbb R^2$. Remember that points and coordinate representations of a point are strictly two different things. You can refer to this point using two different coordinate systems, denoted $(x-y)$ and $(u-v)$ where the latter is a plane rotated by 45°. Explicitly, $\vec e_u=\frac{\vec e_x + \vec e_y}{\sqrt2}$ and $\vec e_v=\frac{-\vec e_x + \vec e_y}{\sqrt2}$, so $(a,b) = a\vec e_x + b\vec e_y$ and $(a,b) = \frac{a+b}{\sqrt2}\vec e_u + \frac{-a+b}{\sqrt2}\vec e_v$. (2/3) – Zxv Jun 24 '22 at 02:58
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    Now compute $(a,b)(c,d)$ in both coordinate frames. You will get $(a,b)(c,d) = (ac,bd)$ in the $(x-y)$ plane, but $(a,b)*(c,d) \neq (ac,bd)$ in the $(u-v)$ plane, verified by converting the first result from $(x-y)\rightarrow (u-v)$ by inverting the rules above (ultimately, it will be missing the cross-terms $ad$ and $bc$). This is a problem, because a point and operations on a point should not depend on the coordinates used to represent that point. (3/3) – Zxv Jun 24 '22 at 02:59
  • @Zxv Thank you! I think I understand now. Just to phrase it in my own words: Given two vectors and a basis, we can define a multiplication on the coordinate representation of those vectors. The resulting coordinate representation can be converted back into a vector. If we define the multiplication rule using ordinary complex multiplication, the final vector is the same regardless of the chosen basis. But this is not true if we define multiplication using the Hadamard product. – IssaRice Jun 25 '22 at 05:22
  • And this happens because complex multiplication can be described in a coordinate-free way, by saying "multiply the lengths and add the angles". Similarly, complex addition can be described in a coordinate-free way by saying "put the tail of one vector at the tip of the other to create a new vector". But the Hadamard product has no such description. – IssaRice Jun 25 '22 at 05:23
  • @riceissa Exactly. Very well said. The Hadamard product intrinsically is defined with regards to the coordinates of the vector, so it is only really useful in situations where your basis will not change and the transformation issues won't be a factor. – Zxv Jun 27 '22 at 09:03
  • @GeoffRobinson Does this mean $e^{ \left( \begin{smallmatrix} 0 & -1\1 & 0 \end{smallmatrix} \right)}= \left( \begin{smallmatrix} 1 & \frac1e\e & 1 \end{smallmatrix} \right) =e^i$? – Тyma Gaidash Nov 13 '22 at 14:18
  • No, the exponential of that matrix is $\cos 1 I+ \sin 1 J$, which is what corresponds to $e^{i}$ in sense of my answer. – Geoff Robinson Nov 14 '22 at 18:11
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Look at the arithmetic operations and their actions. With + and *, these matrices form a field. And we have the isomorphism $$a + ib \mapsto \left[\matrix{a&-b\cr b &a}\right].$$

ncmathsadist
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What I think of a complex number is a scaling and 2D rotation operation, where the absolute value $r$ is scaling factor, and the phase $\theta$ is the rotation angle.

The same operation can be described by scalar multiplication of a rotation matrix as $$r\begin{pmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$

Since $r e^{i\theta}=r\cos \theta + ir \sin \theta = a +ib$, we have $$a +ib = \begin{pmatrix}a & -b \\ b & a \end{pmatrix}$$

chaohuang
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As to the "where did it come from?", rather than verifying that it does work: this is a special case of a "rational representation" of a bigger collection of "numbers" as matrices with entries in a smaller collection. ("Fields" or "rings", properly, but it's not clear what our context here is.)

That is, the collection of complex numbers is a two-dimensional real vector space, and multiplication by $a+bi$ is a real-linear map of $\mathbb C$ to itself, so, with respect to any $\mathbb R$-basis of $\mathbb C$, there'll be a corresponding matrix. For example, with $\mathbb R$-basis $e_1=1,\,e_2=i$, $$ (a+bi)\cdot e_1 = a+bi = ae_1+be_2 \hskip40pt (a+bi)\cdot e_2 = (a+bi)i = -b+ai = -be_1+ae_2 $$ So $$ \pmatrix{e_1 \cr e_2}\cdot (a+bi) \;=\; \pmatrix{a & b \cr -b & a}\pmatrix{e_1\cr e_2} $$ Oop, I guessed wrong, and got the $b$ and $-b$ interchanged. Maybe using $e_2=-i$ instead will work... :)

But this is the way one finds such representations.

paul garrett
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  • Nice one, basically using Eigenvalues. That also explains the wrong sign, you "guessed" the "wrong" Eigenvector and reverse-constructed a transposed (but equally valid) Matrix representation – Tobias Kienzler Aug 10 '12 at 07:44
  • Nice answer but I don't understand why you write you guessed wrong: I don't understand why we cannot represent a complex number $a + bi$ as the matrix in your answer. It seems to me that we can interchange $b$ and $-b$ as we like. What am I missing? – Rudy the Reindeer Aug 19 '15 at 04:17
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    @RudytheReindeer, you are right that the $\pm b$ can be interchanged without harm, which amounts to switching $\pm i$. My "guessed wrong" was only that I was aiming to "hit" one of the two choices, but got the other one by my choices of convention. Doesn't really matter. – paul garrett Aug 19 '15 at 12:49
  • It seems that it is just the way of writing the matrix representation, for instance, $a$ and $b$ must come in one column and $-b$ and $a$ in the other column. – Himanshu Feb 12 '22 at 06:16
  • Question, so we obtain the 'final' matrix representation of complex numbers by abstracting away the matrix of the basis elements? – Gerald Sep 08 '22 at 06:22
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I had something written up on this lying around. The $-$ sign is off, but it's more or less the same, I hope it helps.

Let $M$ denote the set of such matrices. Define a function $\phi\colon M\to\mathbb{C}$ by $$ \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha\end{pmatrix}\mapsto \alpha+i\beta. $$ Note that this function has inverse $\phi^{-1}$ defined by $\alpha+i\beta\mapsto\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha\end{pmatrix}$. This function is well defined, since $\alpha+i\beta=\gamma+i\delta$ if and only if $\alpha=\gamma$ and $\beta=\delta$, and thus it is never the case that a complex number can be written in two distinct ways with different real part and different imaginary part. So $\phi$ is invertible.

Now let $$ A=\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha\end{pmatrix},\qquad B=\begin{pmatrix} \gamma & \delta \\ -\delta & \gamma\end{pmatrix}. $$ Then $$ \phi(A+B)=\phi\begin{pmatrix} \alpha+\gamma & \beta+\delta \\ -\beta-\delta & \alpha+\delta\end{pmatrix}=(\alpha+\gamma)+i(\beta+\delta)=(\alpha+i\beta)+(\gamma+i\delta)=\phi(A)+\phi(B). $$ Also, $$ \phi(AB)=\phi\begin{pmatrix} \alpha\gamma-\beta\delta & \alpha\delta+\beta\gamma \\ -\beta\gamma-\alpha-\delta & -\beta\delta+\alpha\gamma\end{pmatrix}=(\alpha\gamma-\beta\delta)+i(\alpha\delta+\beta\gamma)=(\alpha+i\beta)(\gamma+i\delta)=\phi(A)\phi(B). $$ So $\phi$ respects addition and multiplication. Lastly, $\phi(I_2)=1$, so $\phi$ also respects the multiplicative identity. Hence $\phi$ is a field isomorphism, so $M$ and $\mathbb{C}$ are isomorphic as fields.

yunone
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The matrix rep of $\rm\:\alpha = a+b\,{\it i}\:$ is simply the matrix representation of the $\:\Bbb R$-linear map $\rm\:x\to \alpha\, x\:$ viewing $\,\Bbb C\cong \Bbb R^2$ as vector space over $\,\Bbb R.\,$ Computing the coefficients of $\,\alpha\,$ wrt to the basis $\,[1,\,{\it i}\,]^T\:$

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

As above, any ring may be viewed as a ring of linear maps on its additive group (the so-called left-regular representation). Informally, simply view each element of the ring as a $1\!\times\! 1$ matrix, with the usual matrix operations. This is a ring-theoretic analog of the Cayley representation of a group via permutations on its underlying set, by viewing each $\,\alpha\,$ as a permutation $\rm\,x\to\alpha\,x.$

When, as above, the ring has the further structure of an $\rm\,n$-dimensional vector space over a field, then, wrt a basis of the vector space, the linear maps $\rm\:x\to \alpha\, x\:$ are representable as $\rm\,n\!\times\!n\,$ matrices; e.g. any algebraic field extension of degree $\rm\,n.\,$ Above is the special case $\rm n=2.$

Bill Dubuque
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The matrices $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $J=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ commute (everything commutes with $I$), and $J^2=-I$. Everything else follows from the standard properties (associativity, commutativity, distributivity, etc.) that matrix operations have.

Thus, $aI+bJ=\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ behaves exactly like $a+bi$ under addition, multiplication, etc.

robjohn
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Since you put the tag quaternions, let me say a bit more about performing identifications like that:

Recall the quaternions $\mathcal{Q}$ is the group consisting of elements $\{\pm1, \pm \hat{i}, \pm \hat{j}, \pm \hat{k}\}$ equipped with multiplication that satisfies the rules according to the diagram

$$\hat{i} \rightarrow \hat{j} \rightarrow \hat{k}.$$

Now what is more interesting is that you can let $\mathcal{Q}$ become a four dimensional real vector space with basis $\{1,\hat{i},\hat{j},\hat{k}\}$ equipped with an $\Bbb{R}$ - bilinear multiplication map that satisfies the rules above. You can also define the norm of a quaternion $a + b\hat{i} + c\hat{j} + d\hat{k}$ as

$$||a + b\hat{i} + c\hat{j} + d\hat{k}|| = a^2 + b^2 + c^2 + d^2.$$

Now if you consider $\mathcal{Q}^{\times}$, the set of all unit quaternions you can identify $\mathcal{Q}^{\times}$ with $\textrm{SU}(2)$ as a group and as a topological space. How do we do this identification? Well it's not very hard. Recall that

$$\textrm{SU}(2) = \left\{ \left(\begin{array}{cc} a + bi & -c + di \\ c + di & a-bi \end{array}\right) |\hspace{3mm} a,b,c,d \in \Bbb{R}, \hspace{3mm} a^2 + b^2 + c^2 + d^2 = 1 \right\}.$$

So you now make an ansatz (german for educated guess) that the identification we are going to make is via the map $f$ that sends a quaternion $a + b\hat{i} + c\hat{j} + d\hat{k}$ to the matrix $$\left(\begin{array}{cc} a + bi & -c + di \\ c + di & a-bi \end{array}\right).$$

It is easy to see that $f$ is a well-defined group isomorphism by an algebra bash and it is also clear that $f$ is a homeomorphism. In summary, the point I wish to make is that these identifications give us a useful way to interpret things. For example, instead of interpreting $\textrm{SU}(2)$ as boring old matrices that you say "meh" to you now have a geometric understanding of what $\textrm{SU}(2)$. You can think about each matrix as being a point on the sphere $S^3$ in 4-space! How rad is that?

On the other hand when you say $\Bbb{R}^4$ has now basis elements consisting of $\{1,\hat{i},\hat{j},\hat{k}\}$, you have given $\Bbb{R}^4$ a multiplication structure and it becomes not just an $\Bbb{R}$ - module but a module over itself.

  • Yes ,the question is the basic of the real-number matrix form of quaternions which is I really want to know next step. – NFDream Aug 10 '12 at 04:46
  • @NFDream What do you mean by next step? Can you elaborate a bit more? Would you like me to put more details in my answer above? –  Aug 10 '12 at 07:24
  • Actually I ask this question because I want to know how to get the real-number matrix form of quaternions, thanks. – NFDream Aug 12 '12 at 08:38