Transformation Matrix

Reuel · 2013-12-17 02:36:00

Hello.

How does one go about transforming an arbitrary matrix, say, one of the form

to the form

where the only change are the Greek components?

I tried just working out a matrix P on the spot to multiply M by to get N but I didn't get what I wanted at all. I have had a little experience with matrix transformations but I don't know how to derive the transformation itself. If someone could give me the matrix and explain how you got it so I understand, I would appreciate it.

Thank you for your help.

Last edited by Reuel (2013-12-17 02:36:24)

bobbym · 2013-12-17 02:49:44

This is just an observation but it does appear that

are the only values they can take.

Reuel · 2013-12-17 02:59:44

Why isn't it enough to just make the transformation matrix

which is what I originally tried? Wouldn't you just multiply through and get N? Or is multiplying 3x3 matrices that different from 2x2?

bobbym · 2013-12-17 03:10:18

M X P yields:

anonimnystefy · 2013-12-17 03:11:52

Multiplying 3x3 matrices is pretty much the same as multiplying 2x2 matrices.

Where does this come from?

Reuel · 2013-12-17 03:12:12

Yeah, I got that too.

So the answer is there is no transformation that will work?

bobbym · 2013-12-17 03:16:16

There are an infinite number of P's that will work.

Reuel · 2013-12-17 03:18:20

Such as?

bobbym · 2013-12-17 03:29:47

That is where I am having a problem. Each solution insists that alpha and beta are 1 and only 1. Therefore I can not come up with an answer that involves them.

anonimnystefy · 2013-12-17 03:37:37

anonimnystefy wrote:

Where does this come from?

Reuel · 2013-12-17 03:51:48

The problem I introduced stemmed from experimentation with representing that hyperbolic function I keep bringing up in matrix form. As bob bundy implied, the function

may be rewritten as

which, according to wikipedia, is of the form

where 2B=1, 2D=-a/c, and 2E=b/c, and everything else is zero. Therefore we can state the original f(x)=y in matrix form as

where

and

In coming here I just used Greek letters to make the problem more simple rather than using fractions for every term. I left the 1/2's in case they held significance.

So then the problem became finding a matrix P that would transform a more basic equation y=x into the equation f(x) given at the start. The function y=x has the matrix

but that didn't work at all which is why I added in the two extra 1/2's in to M because if those terms were left as 0s then no matter what I multiplied by it, I would still get 0. And so, essentially, my original question was to ask for a transformation matrix P that would transform R in to M; that is, to go from

to

using matrices and a matrix transformation.

Last edited by Reuel (2013-12-17 03:54:53)

bobbym · 2013-12-17 03:53:49

You may have distilled the problem down too much. The system of linear equations your distilled version produces tend to obliterate alpha and beta.

Reuel · 2013-12-17 04:01:38

Quite possible. I was just trying to simplify the model for the sake of those reading it but I may have messed it up in doing so.

If simplification complicated the issue, the function f(x) might even be able to expand with some manipulation. For instance letting c=a-b or something to eliminate c while adding in additional a and b terms. Or not. Actually, that might make it worse.

Last edited by Reuel (2013-12-17 04:04:19)

bobbym · 2013-12-17 04:18:01

I am pretty sure that the problem as posted in post #1 will always yield a value of alpha and beta equaling 1.

This may not be bad just set those fractions equal to 1.

Reuel · 2013-12-17 04:32:05

You mean rewrite M as

and then N as

?

That makes it look cleaner. I still don't know how to find P to go from M to N, but I am experimenting with it on paper.

Last edited by Reuel (2013-12-17 04:41:15)

bobbym · 2013-12-17 04:36:50

Why change M, it does not have alpha or beta in it?

Reuel · 2013-12-17 04:39:38

Consistency. And it still works in that it still leads to y=x.

bobbym · 2013-12-17 04:41:13

Okay, I am big fan of doing what works. Play with it yourself and if you can not get a solution I will try.

Reuel · 2013-12-17 04:43:11

What works and whatever is most simple.

I'll keep working on it and come back either with a solution or more whining.

bobbym · 2013-12-17 04:47:45

Math makes everybody whine. Good luck.

Reuel · 2013-12-17 09:49:55

I did it but it's more typing than I have time for right now. You have to multiply M by a matrix with 9 unknowns so that you end up with a system of nine equations with 11 unknowns and then solve it for those unknowns entirely in terms of a and b only. It was quite a bit of substitution and algebraic manipulation and the resulting transformation matrix is enormous.

Thanks for your input!

bobbym · 2013-12-17 10:27:51

Hi;

Glad you did it. Come back if anything goes wrong. Good luck.

Math Is Fun Forum

#1 2013-12-17 02:36:00

Transformation Matrix

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