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Help with matrix form of the imaginary unit, i

by powerof
Tags: form, imaginary, matrix, unit
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Jun12-14, 02:41 PM
powerof's Avatar
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 [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 \\
0 &
x\end{bmatrix} +\begin{bmatrix}0
&-y \\
y &
&-y \\
y &

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
&b \\
a&b \\
&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

There aren't just 2 solutions now. Why is the matrix [itex]\begin{bmatrix}
0& -1\\
\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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Jun12-14, 02:58 PM
P: 327
Any matrix of the form ##U^{-1}\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]U##, where ##U## is a unitary matrix, can be used to represent the imaginary unit.

Mathematical groups, of which the complex numbers are one example, generally have many different representations as sets of matrices.

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