# Help with matrix form of the imaginary unit, i

by powerof
Tags: form, imaginary, matrix, unit
 P: 25 While investigating more about complex numbers today I ran across the 2x2 matrix representation of a complex number, and I was really fascinated. You can read what I read here. As I understand it, you write z in its binomial form but instead of "1" you use the identity matrix, I, and for i you use a matrix m such that $M^2=-I$, like with the more usual definition of i. $z =x\cdot 1+y\cdot 1 \cdot i\sim z=x\cdot \begin{bmatrix}1 &0 \\ 0 & 1\end{bmatrix} +y\cdot \begin{bmatrix}1 &0 \\ 0 & 1\end{bmatrix} \begin{bmatrix}0 &-1 \\ 1 & 0\end{bmatrix}=\begin{bmatrix}x &0 \\ 0 & x\end{bmatrix} +\begin{bmatrix}0 &-y \\ y & 0\end{bmatrix}=\begin{bmatrix}x &-y \\ y & x\end{bmatrix}$ In the "normal" way (the one you learn first, at least in my case) i is defined as follows: $i^{2}=-1$, the positive solution to the equation $x^2+1=0$. This equation has two solutions, and by convention i is the positive one. If we try to solve $M^2=-I$, then we get infinite possibilites: $M^2=-I \Rightarrow -\begin{bmatrix}1 &0 \\ 0 &1 \end{bmatrix}=\begin{bmatrix}a &b \\ c&d \end{bmatrix}\begin{bmatrix} a&b \\ c&d \end{bmatrix}=\begin{bmatrix}a^2+bc &b(a+d) \\ c(a+d)& d^2+bc \end{bmatrix}\Rightarrow \left\{\begin{matrix}a^2+bc=-1 \\ d^2+bc=-1 \\ b(a+d)=0 \\ c(a+d)=0 \end{matrix}\right.$ There aren't just 2 solutions now. Why is the matrix $\begin{bmatrix} 0& -1\\ 1&0 \end{bmatrix}$ chosen over any of the rest of the matrices that satisfy m^2=-I? What is so convenient about that form over any other? Perhaps i isn't defined only as the solution to m^2=-I, but if so, what am I missing? Thank you for reading.