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powerof
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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<br /> &0 \\ <br /> 0 & <br /> 1\end{bmatrix} +y\cdot \begin{bmatrix}1<br /> &0 \\ <br /> 0 & <br /> 1\end{bmatrix} \begin{bmatrix}0<br /> &-1 \\ <br /> 1 & <br /> 0\end{bmatrix}=\begin{bmatrix}x<br /> &0 \\ <br /> 0 & <br /> x\end{bmatrix} +\begin{bmatrix}0<br /> &-y \\ <br /> y & <br /> 0\end{bmatrix}=\begin{bmatrix}x<br /> &-y \\ <br /> y & <br /> 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<br /> &0 \\ <br /> 0 &1 <br /> \end{bmatrix}=\begin{bmatrix}a<br /> &b \\ <br /> c&d <br /> \end{bmatrix}\begin{bmatrix}<br /> a&b \\ <br /> c&d <br /> \end{bmatrix}=\begin{bmatrix}a^2+bc<br /> &b(a+d) \\ <br /> c(a+d)& d^2+bc<br /> \end{bmatrix}\Rightarrow \left\{\begin{matrix}a^2+bc=-1<br /> \\ d^2+bc=-1<br /> \\ b(a+d)=0 <br /> \\ c(a+d)=0 <br /> \end{matrix}\right.
There aren't just 2 solutions now. Why is the matrix \begin{bmatrix}<br /> 0& -1\\ <br /> 1&0 <br /> \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.
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<br /> &0 \\ <br /> 0 & <br /> 1\end{bmatrix} +y\cdot \begin{bmatrix}1<br /> &0 \\ <br /> 0 & <br /> 1\end{bmatrix} \begin{bmatrix}0<br /> &-1 \\ <br /> 1 & <br /> 0\end{bmatrix}=\begin{bmatrix}x<br /> &0 \\ <br /> 0 & <br /> x\end{bmatrix} +\begin{bmatrix}0<br /> &-y \\ <br /> y & <br /> 0\end{bmatrix}=\begin{bmatrix}x<br /> &-y \\ <br /> y & <br /> 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<br /> &0 \\ <br /> 0 &1 <br /> \end{bmatrix}=\begin{bmatrix}a<br /> &b \\ <br /> c&d <br /> \end{bmatrix}\begin{bmatrix}<br /> a&b \\ <br /> c&d <br /> \end{bmatrix}=\begin{bmatrix}a^2+bc<br /> &b(a+d) \\ <br /> c(a+d)& d^2+bc<br /> \end{bmatrix}\Rightarrow \left\{\begin{matrix}a^2+bc=-1<br /> \\ d^2+bc=-1<br /> \\ b(a+d)=0 <br /> \\ c(a+d)=0 <br /> \end{matrix}\right.
There aren't just 2 solutions now. Why is the matrix \begin{bmatrix}<br /> 0& -1\\ <br /> 1&0 <br /> \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.