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Matrix representation of complex numbers [math-001F]

  • 2025-12-03

To see why

Rθ=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]R_\theta = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

behave same as the complex numbers zθ=cos⁡θ+isin⁡θz_\theta = \cos\theta + i\sin\theta is to write RθR_\theta as a linear combination:

Rθ=cos⁡θ[1001]+sin⁡θ[0−110]R_\theta = \cos\theta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \sin\theta \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}

and hence we are wondering, what if we define

1=[1001]andi=[0−110]\bold{1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \quad \text{and} \quad \bold{i} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}

and multiplication as matrix multiplication, addition as matrix addition? Are these behave same as complex numbers? Since 1\bold{1} is the identity matrix, we simply get followings

  1. 12=1\bold{1}^2 = \bold{1}
  2. 1i=i1=i\bold{1}\bold{i} = \bold{i}\bold{1} = \bold{i}

we would like to know if i2=−1\bold{i}^2 = -\bold{1} :

i2=[0−110][0−110]=[−100−1]=−[1001]=−1\bold{i}^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -\bold{1}

as desired. Then we also know linear combinations based on 1\bold{1} and i\bold{i} maps to complex numbers bijectively:

[a−bba]=a1+bi≃a+bi\begin{bmatrix} a & -b \\ b & a \end{bmatrix} = a\bold{1} + b\bold{i} \simeq a + bi

so now we can see this indeed is a representation of complex numbers.