Possible Duplicate:
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.

  • 16
    $\begingroup$ Hint: $\mathbb{C}$ is a 2 dimensional vector space of $\mathbb{R}$. Take $\{1,i\}$ as a basis for $\mathbb{C}$ over $\mathbb{R}$. Multiplication by a complex number $a+bi$ is a linear operator. How does the matrix representing it look like? $\endgroup$
    – user3533
    Aug 9, 2012 at 22:23
  • $\begingroup$ Here a very nice article by Rod Carvalho: Representing complex numbers as 2×2 matrices $\endgroup$ Aug 9, 2012 at 22:44
  • 4
    $\begingroup$ I can't see the article linked. Is it just me? $\endgroup$ Aug 4, 2014 at 4:54
  • 4
    $\begingroup$ Me neither. How can we do? $\endgroup$
    – Joe
    May 3, 2015 at 22:45
  • 4
    $\begingroup$ @SalechAlhasov Your link leads merely to a login page. Maybe you can post a new link? $\endgroup$ Aug 19, 2015 at 3:30

8 Answers 8


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 $a +bi.$

  • 3
    $\begingroup$ Yes, indeed, so, if there had been any doubt whether there could be a legitimate square root of $-1$, here is one! :) $\endgroup$ Aug 10, 2012 at 0:04
  • 9
    $\begingroup$ We should not forget that in the nineteenth century this sort of thing was considered an important legitimization of complex numbers. Also, Kronecker's $\mathbb C=\mathbb R[x]/\langle x^2+1\rangle$. $\endgroup$ Aug 10, 2012 at 0:29
  • 1
    $\begingroup$ Jumping in on a month old question here, but is there a way to show that $J$ is unique in this regard? $\endgroup$
    – crf
    Sep 5, 2012 at 5:48
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 5, 2012 at 7:43
  • 1
    $\begingroup$ @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) $\endgroup$
    – Zxv
    Jun 24, 2022 at 2:57

Look at the arithmetic operations and their actions. With + and *, these matrices form a field. And we have the isomorphism $$a + ib \mapsto \left[\matrix{a&-b\cr b &a}\right].$$

  • 1
    $\begingroup$ I think this is a better answer because it points out the isomorphism. $\endgroup$ Dec 11, 2013 at 3:23

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 $$a +ib = \begin{pmatrix}a & -b \\ b & a \end{pmatrix}$$


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.

  • $\begingroup$ 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 $\endgroup$ Aug 10, 2012 at 7:44
  • $\begingroup$ 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? $\endgroup$ Aug 19, 2015 at 4:17
  • 1
    $\begingroup$ @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. $\endgroup$ Aug 19, 2015 at 12:49
  • $\begingroup$ 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. $\endgroup$
    – Himanshu
    Feb 12, 2022 at 6:16
  • $\begingroup$ Question, so we obtain the 'final' matrix representation of complex numbers by abstracting away the matrix of the basis elements? $\endgroup$
    – Gerald
    Sep 8, 2022 at 6:22

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.


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 $\rm n=2.$

  • $\begingroup$ See also this answer for a finite field analogue in $\,\Bbb F_9 \cong \Bbb F_3[i].\ \ $ $\endgroup$
    – Math Gems
    Feb 5, 2013 at 15:42
  • $\begingroup$ What is the domain and codomain of your $\mathbb R$-linear map? $\endgroup$
    – user5826
    Apr 1, 2021 at 22:01
  • $\begingroup$ Would you suggest me to use this way for undergraduate students who are passing Calculus in University? $\endgroup$
    – Mikasa
    Sep 19, 2023 at 7:14

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.

  • 1
    $\begingroup$ ... and matrix transposition behaves like complex conjugation. $\endgroup$
    – Mark Viola
    Jul 8, 2016 at 2:48
  • $\begingroup$ @Dr.MV: Indeed! Swapping $J$ and $J^T$ gives the same isomorphism that swapping $i$ and $-i$ does. $\endgroup$
    – robjohn
    Jul 8, 2016 at 2:53

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.

  • $\begingroup$ Yes ,the question is the basic of the real-number matrix form of quaternions which is I really want to know next step. $\endgroup$
    – NFDream
    Aug 10, 2012 at 4:46
  • $\begingroup$ @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? $\endgroup$
    – user38268
    Aug 10, 2012 at 7:24
  • $\begingroup$ Actually I ask this question because I want to know how to get the real-number matrix form of quaternions, thanks. $\endgroup$
    – NFDream
    Aug 12, 2012 at 8:38

Not the answer you're looking for? Browse other questions tagged .