3d Rotations

jk22 · 2010-07-25 03:47:06

Hello,

If I consider 2 rotations in space : Rx(90°) and afterwards Ry(90°) of the vector (1,1,0)

it gives : (1,0,-1) (as by matrix multiplication Ry*Rx)

but considering that the first rotation, Rx, turns the y-axis, leads to (0,1,1)

Is there an easy way to get the latter with matrix operations, can it be shown generally, that it will be Rx*Ry, hence :

Is it that the resulting rotation of two rotations rotating the axis too, were the inverse order matrix product, is it the same as Euler-angles ?

b) do you know how many elements the group of all composition of rotations of 90° around x,y,z axis contains ?

Thx.

Last edited by jk22 (2010-07-27 19:07:52)

Bob · 2010-07-31 03:34:31

Have a look at Wiki at http://en.wikipedia.org/wiki/Rotational_matrix#Three_dimensions.

Is this what you're after?

Bob

jk22 · 2010-07-31 10:28:47

Hello bob,

yes it is, however i couldn't find the cardinality of the group of rotation of 90 around xyz which i suppose to be 48.

Bob · 2010-07-31 21:45:32

Hi jk

I'll think about it.

B

jk22 · 2010-08-03 07:25:50

Hi, thx.

I had another question : Suppose we find 2 linearly independent i=sqrt(-1), with 2x2 matrices, let say I1, I2. I don't know If their product is spanneable by {1,I1,I2}, but if it were the case, were this a sufficient condition to build an algebra ?

----------------------------------

(there is more time than patience, but the world is imaginary small)

Last edited by jk22 (2010-08-03 08:16:00)

Bob · 2010-08-03 22:09:08

hi jk22,

'the cardinality of the group of rotation of 90 around xyz which i suppose to be 48'

Unit vectors along the axes would be:
1 0 0
0 1 0
0 0 1

The origin is invariant so any transformation of these axes will have the form

x1 y1 z1
x2 y2 z2
x3 y3 z3
where

one of the xs is either 1 or -1 and the rest are zeros and similarly for the ys and the zs.

eg one transformation would be

1 0 0
0 0 -1
0 1 0

This is a rotation around the x axis of +90.

So how many such transformations are there?

Choose one from x1, x2, x3 (3 choices) and then whether it's +1 or -1 (2 choices)

Then choose from only two out of y1, y2, y3. (You cannot choose from the row that you chose your x from) (2 choices) and the whether it's +1 or -1. (2 choices).

Finally you have no choice about which of z1, z2, z3 since you must pick from the row that hasn't been chosen yet but you may chose from +1 and -1 (2 choices).

Altogether this gives 3 x 2 x 2 x 2 x 1 x 2 choices = 48.

The above rotation is one example.

But, another would be

-1 0 0
0 1 0
0 0 1

and this isn't a rotation; it's a reflection in the YZ plane.

I suspect that half the 48 are rotations and half are reflections.

So the cardinality for rotations alone would be 24.

You can prove it by making a 24 x 24 'multiplication' table of all the elements.

Here's a start:

I X X^2 Y
______________________________

I I X X^2 Y
X X X^2 X^3 P
X^2 X^2 X^3 I
Y Y

where P is

0 0 1
1 0 0
0 1 0

You're going to need a large piece of paper and be prepared for lots of matrix multiplication.

Warning. It isn't generally commutative ( XY not = YX )

If you find 24 elements in a closed set (ie. no more elements generated by those 24 in any combination) you've got your result.

Good luck.

ps. Not sure what your subsequent question means. Can you provide one example of such a 2 x 2 matrix?

Bob

Last edited by Bob (2010-08-03 22:12:17)

jk22 · 2010-08-04 07:09:47

Hi, thanks for the explanation.

Now that I have the number of elements, I tried to find the concept of dimension, but I don't know how to find it, it should of course be <=9.

For the second question I found :
[1 -1]
[2 -1] has determinant 1 and square to -1, but it's not a good example, since this 3 axes were broken at origin (x->(1,2), y->(-1,-1))

Bob · 2010-08-04 08:09:42

Hi,

"the concept of dimension, but I don't know how to find it, it should of course be <=9."

Not sure what you mean by 'dimension'. It's used such a lot in maths in different contexts.

I'm guessing you mean the minimum number of elements that will generate the whole group.

I think this would be 3. Less than 3 and you won't get into 3D.

but X, rotate 90 around the x axis, Y rotate 90 around the y axis and Z, rotate 90 around the z axis will generate the whole group, I'm sure. Again the 24 element table would prove it by inspecting what comes from {X, Y, Z}.

I'll try come back to question 2, after a longer think. If a few days go by, don't worry, I haven't forgotten you.

Bob

Last edited by Bob (2010-08-04 08:10:34)

jk22 · 2010-08-06 01:23:48

Hi, ok.

By dimension I meant : the minimal number D of elements, so that a linear combination of these independent elements (now with a sum) give all the elements.

Anyhow in this case the field of coefficient cannot be continuous, but maybe

which were not so interesting.

Last edited by jk22 (2010-08-06 01:36:40)

Bob · 2010-08-06 08:58:28

hi

For the rotations question, combinations are made by matrix multiplication. I find myself confused by your use of the term 'linear combination'.

I think, if a and b are vectors then pa + qb is a linear combination (p and q scalars).

What do you think a linear combination of two matrices would be like?

Meanwhile, for your second question:

I am assuming you are looking at 2x2 matrices that square to give
-1 0
0 -1

Let such a matrix be
a b
c d

Then squaring and setting equal to the above gives

a^2 + bc = -1 and d^2 + bc = -1 => a = + or - d

and replacing one a with -d gives

-ad +bc = -1 => ad - bc = 1 so the determinant is always 1.

A generator for all such matrices is

a b

(-1 -a^2)/b -a -a

for all a and b (would you allow complex values here?)

Thus the following examples:

(i) 3 5

-2 -3

(ii) i 0

0 i

(iii) 1 + i 2i

(-1-2i)/2i -1 -i

Since you can choose my real valued matrix and also yours as I1 and I2 you cannot generate any complex values from these.

Is that what you are after?

Bob

jk22 · 2010-08-07 07:34:21

Hi bob,

f you find 24 elements in a closed set (ie. no more elements generated by those 24 in any combination) you've got your result.

I finally got the 24, but with a computer program, after 100 multiplication, only 24 were different. Anyway, Thanks for your explanation of why it was not 48.

I haven't found yet the dimension, taken this 9 element function of 24 variable, how many are independent : it should be 4 or 8, but i need to write a code again (could it be that there were 9 linearly independent matrices in this set ?)

almost, I read we put complex numbers in the form of real matrices, but this "isomorphism" is maybe already too descriptive of the symbol i ?

Last edited by jk22 (2010-08-27 07:29:05)

Math Is Fun Forum

#1 2010-07-25 03:47:06

3d Rotations

#2 2010-07-31 03:34:31

Re: 3d Rotations

#3 2010-07-31 10:28:47

Re: 3d Rotations

#4 2010-07-31 21:45:32

Re: 3d Rotations

#5 2010-08-03 07:25:50

Re: 3d Rotations

#6 2010-08-03 22:09:08

Re: 3d Rotations

#7 2010-08-04 07:09:47

Re: 3d Rotations

#8 2010-08-04 08:09:42

Re: 3d Rotations

#9 2010-08-06 01:23:48

Re: 3d Rotations

#10 2010-08-06 08:58:28

Re: 3d Rotations

#11 2010-08-07 07:34:21

Re: 3d Rotations

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