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.(adsbygoogle = window.adsbygoogle || []).push({});

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 [itex]M^2=-I[/itex], like with the more usual definition of i.

[itex]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}[/itex]

In the "normal" way (the one you learn first, at least in my case) i is defined as follows: [itex]i^{2}=-1[/itex], the positive solution to the equation [itex]x^2+1=0[/itex]. This equation has two solutions, and by convention i is the positive one.

If we try to solve [itex]M^2=-I[/itex], then we get infinite possibilites:

[itex]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.[/itex]

There aren't just 2 solutions now. Why is the matrix [itex]\begin{bmatrix}

0& -1\\

1&0

\end{bmatrix}[/itex] 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.

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# Help with matrix form of the imaginary unit, i

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