Complex matrices

[A23]

I'm new to that stuff : what are the relation between complex numbers and matrices ?

 

I read that complex numbers could be identified to a set of 2x2 matrices, where

 

[math] 1:\equiv\mathbf{1}_2=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)[/math]

 

and [math]i:\equiv Sqrt(-\mathbf{1}_2)=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)[/math]

 

Let [math]B=\left(\begin{array}{cc} a&b\\c&d\end{array}\right)[/math]

 

Then we get the system of equations :

 

[math] \left\{\begin{array}{l} a^2+bc=-1\\b(a+d)=0\\c(a+d)=0\\d^2+bc=-1\end{array}\right.[/math]

 

with the "norm", determinant of B equals 1.

 

a) If I can find other matrices, let say [math]B[/math], such that [math]B^2=-1[/math], e.g. () :

 

[math] B=\left(\begin{array}{cc} 1 & \sqrt{2}\\ -\sqrt{2} & -1\end{array}\right)[/math],

 

and that [math]\{1,i,B\}[/math] are linearly independent, can [math]B[/math] be seen as another root of minus 1 ?

 

(or is the 'cycle' important 1,B,-1,-B to determine if the root were 'different') ?

 

In this 2x2 matrix representation, there should exist 3 'different' roots of minus -1 (since this space is 4 dimensional, and 1 represent already one dimension, the real axis), so that

 

Taken time arrow, on a real axis, in a metric description, space dimensions could be seen as the "roots of the going back in time dimension" ?

 

(root gives often irrational, like fears..or fear to loose things)

 

in fact i found those matrices built a space that was only 2 dimensional,

 

b) can they be made up to build a group ?

 

 

c) Were then the following matrix equation right : [math]\left(\begin{array}{cc} 0 & -i \\ i&0 \end{array}\right)=i^2=-1[/math] ?

 

BTW : Why does \mathbb{1}=[math]\mathbb{1}[/math] not give the correct vertical bar 1 for the unit 2x2 matrix ?

[CraigD]

I read that complex numbers could be identified to a set of 2x2 matrices, where

 

[math] 1:\equiv\mathbf{1}_2=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)[/math]

 

and [math]i:\equiv Sqrt(-\mathbf{1}_2)=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)[/math]

 

Let [math]B=\left(\begin{array}{cc} a&b\\c&d\end{array}\right)[/math]

 

Then we get the system of equations :

 

[math] \left\{\begin{array}{l} a^2+bc=-1\\b(a+d)=0\\c(a+d)=0\\d^2+bc=-1\end{array}\right.[/math]

With you so far :thumbs_up

I think you can go a step further, asserting [math]a=-d[/math] and [math]b=-c[/math], to get simply [math]b^2-a^2=1[/math]

a) If I can find other matrices, let say [math]B[/math], such that [math]B^2=-1[/math], e.g. () :

 

[math] B=\left(\begin{array}{cc} 1 & \sqrt{2}\\ -\sqrt{2} & -1\end{array}\right)[/math],

 

and that [math]\{1,i,B\}[/math] are linearly independent, can [math]B[/math] be seen as another root of minus 1 ?

I’d stick with the naming you started in your first equation, and write [math]\{1_2,i_2,B\}[/math], then note that I don’t think their linear (AKA combinatory) independence is important, then agree, yes, any matrix [math]B[/math] as described above can considered a representation of [math]i[/math], and used like it in a 2x2 matrix representation of a complex number.

b) can they be made up to build a group ?

I don’t think so, as I can’t see how to define a binary operator [imath]\cdot[/imath] such that [imath]B_a \cdot B_b = B_c[/imath] but no [imath]B_d[/imath] exists such that [imath]B_a \cdot B_d = B_c[/imath]

c) Were then the following matrix equation right : [math]\left(\begin{array}{cc} 0 & -i \\ i&0 \end{array}\right)=i^2=-1[/math] ?

No, because [math]\left(\begin{array}{cc} 0 & -i \\ i&0 \end{array}\right) \left(\begin{array}{cc} 0 & -i \\ i&0 \end{array}\right)= \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) \not= \left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right) [/math]

 

I think you were trying for [math]\left(\begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right) or \left(\begin{array}{cc} -i & 0 \\ 0 & i \end{array}\right)[/math]

 

I find your question (a) the most interesting, A23, as I think it leads to a general form of a 2x2 matrix representation of [math]i[/math]

[math] i \equiv \left(\begin{array}{cc} a\sqrt{n} & b\sqrt{n+1}\\ -b\sqrt{n+1} & -a\sqrt{n} \end{array}\right)[/math]

where [math]a,b = \pm 1[/math]

 

... of which all the valid representations of [math]i[/math] above are special cases :)

 

BTW : Why does \mathbb{1}=[math]\mathbb{1}[/math] not give the correct vertical bar 1 for the unit 2x2 matrix ?

That’s pretty weird!

 

\mathbb is usually used to render characters “blackboard bold”, eg \mathbb{ABC} as [math]\mathbb{ABC}[/math]

 

The usual way to render a matrix with surrounding marks is by the argument to the \begin, eg:

\begin{vmatrix} a & b \\ c & d \end{vmatrix} to render [math]\begin{vmatrix} a & b \\ c & d \end{vmatrix}[/math]

 

Nonethless, I’m curious, and will look into why our LaTeX renderer is rendering a \mathbb{123} as “[math]\mathbb{123}[/math]”. (There does seem to be some pattern to this unexpected bahavior)

 

PS:

:confused: This bit of you post I found a bizarre and incomprehensible, A23:

In this 2x2 matrix representation, there should exist 3 'different' roots of minus -1 (since this space is 4 dimensional, and 1 represent already one dimension, the real axis), so that

 

Taken time arrow, on a real axis, in a metric description, space dimensions could be seen as the "roots of the going back in time dimension" ?

 

(root gives often irrational, like fears..or fear to loose things)

 

:thumbs_up This bit, I found entirely sensible:

in fact i found those matrices built a space that was only 2 dimensional,

2x2 matrixes and complex numbers are both commonly used to represent rotations and other transformations in 2 dimensional space, 2 vectors points, velocities, etc.

[A23]

:thumbs_up This bit, I found entirely sensible:

 

2x2 matrixes and complex numbers are both commonly used to represent rotations and other transformations in 2 dimensional space, 2 vectors points, velocities, etc.

 

 

This is very useful in technical calculation that become then easier.

 

Just to finish had I 3 last points :

 

a) If we find that the "imaginary" part has to linearly independent direction (in this representation), does it mean that this space made of (1,sqrt(-1)) is 3 dimensional ?

 

If we used other higher dimensional matrices, then the imaginary space made of roots of -1 would contain a lot of dimensions ?

 

 

 

b) Group structure :

 

Suppose we had 2 roots of -1 as matrices [math]B_a, B_b[/math]

 

To build a group, would it suffice to add the equation : [math]B_a\cdot B_b=\alpha 1+\beta B_a+\gamma B_b[/math]

 

with a normal matrix product, so that this operation remains intern to the space spanned by the two "imaginary units" found ?

 

 

 

c) "weird calculation"

 

No, because [math]\left(\begin{array}{cc} 0 & -i \\ i&0 \end{array}\right) \left(\begin{array}{cc} 0 & -i \\ i&0 \end{array}\right)= \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) \not= \left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right) [/math]

 

In fact I wanted to write : [math]\left(\begin{array}{cc} 0 & -i \\ i&0 \end{array}\right)=i\underbrace{\left(\begin{array}{cc} 0 & -1 \\ 1&0 \end{array}\right)}_{i\textrm{ as above}}=i*i=-1[/math]

 

but I don't know if it's allowed to compute like this, since it seems contradictory :

 

-1 is not [math]\sigma_y[/math], maybe because the representation of i as a matrix is not always fitting ?