Rotate Basis Vectors Programmatically

Question

Mohammad MSBR 2022-9-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1814840-rotate-basis-vectors-programmatically

评论： Mohammad MSBR 2022-10-2

I have six 6-dimensional basis vectors, i.e., that are orthogonal. I wonder how I can rotate these 6 vectors programatically in the 6D space to build new basis vectors. In other words, is there a way to parameterize these basis vectors so that I can change them without losing orthogonality?

A = [a1,a2,...,a6];

B = [b1,b2,...,b6];

C = [c1,c2,...,c6];

D = [d1,d2,...,d6];

E = [e1,e2,...,e6];

F = [f1,f2,...,f6];

Thank you.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Jan 2022-9-30

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1814840-rotate-basis-vectors-programmatically#answer_1063845

See: FEX: Rotation Matrix This creates N-dimensional rotation matrices.

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Mohammad MSBR 2022-10-1

Thank you, Jan.

The rotation matrix constructed in your suggested code works well. many thanks.

请先登录，再进行评论。

Answer 2

Chunru 2022-9-30

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1814840-rotate-basis-vectors-programmatically#answer_1063780

编辑：Chunru 2022-9-30

V1 = orth(randn(6))         % your original orthonormal basis
V1 = 6×6
   -0.4706    0.7349   -0.0145    0.2908   -0.2682    0.2859
   -0.3058   -0.0469    0.1456    0.2725   -0.2311   -0.8692
   -0.0524   -0.3243    0.7312    0.0009   -0.5190    0.2967
   -0.2590    0.3046    0.5051   -0.5918    0.4612   -0.1489
   -0.7004   -0.4887   -0.0798    0.2369    0.3948    0.2287
   -0.3532   -0.1447   -0.4272   -0.6594   -0.4865   -0.0169
% Then you can apply any other orthonormal basis to it
% For example,
V2 = orth(randn(6));        % get another orthonormal basis
Vnew = V2*V1;               % this is the transform of the original orthonormal basis
Vnew*Vnew'                  % to demonstrate the oorthonormal property
ans = 6×6
    1.0000    0.0000    0.0000   -0.0000   -0.0000   -0.0000
    0.0000    1.0000   -0.0000    0.0000    0.0000    0.0000
    0.0000   -0.0000    1.0000    0.0000    0.0000    0.0000
   -0.0000    0.0000    0.0000    1.0000   -0.0000   -0.0000
   -0.0000    0.0000    0.0000   -0.0000    1.0000   -0.0000
   -0.0000    0.0000    0.0000   -0.0000   -0.0000    1.0000
% If you want to control the rotation with angle in N-D space
% Rotate on hyperplane i-j by theta
i=2; j=4;           % for example
theta = 5;          % deg
R = eye(6);         % 6D
R([i j], [i j]) = [cosd(theta) -sind(theta); sind(theta) cosd(theta)]
R = 6×6
    1.0000         0         0         0         0         0
         0    0.9962         0   -0.0872         0         0
         0         0    1.0000         0         0         0
         0    0.0872         0    0.9962         0         0
         0         0         0         0    1.0000         0
         0         0         0         0         0    1.0000
% Then you can have a series of rotation matrices and you can put them
% together as one rotation matrices

5 个评论
显示 3更早的评论隐藏 3更早的评论

Chunru 2022-10-2

Rotation on any hyper plane will not change orthogonity.

Mohammad MSBR 2022-10-2

Correct.

请先登录，再进行评论。

Answer 3

Mohammad MSBR 2022-9-30

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1814840-rotate-basis-vectors-programmatically#answer_1064180

Thank you both for very helpful information.

The only missing point is the order of rotations. For a 6D space, I have 6 rotattions. Is there a way to pick a sequence without losing generality?

In other words, can I lead to the same beses by starting with rotations in different hyperplanes, or does starting with a specific hyperplane always lead to a unique configuration?

Thank you so much.

4 个评论
显示 2更早的评论隐藏 2更早的评论

Jan 2022-10-1

Sorry, there are only 15 hyperplanes in 6D: 6 choices for the first vector, 5 possible choices for the 2nd one, but the order does not matter, so divide by 2.

nchoosek(1:6, 2)
ans = 15×2
   2
   3
   4
   5
   6
   3
   4
   5
   6
   4

Mohammad MSBR 2022-10-2

Thank you.

请先登录，再进行评论。