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A four by four transformation matrix? Are you rotating in four dimensional space or is this a projective space? First find the eigenvalues. A rotation matrix, in four dimensions may have two real and two complex-conjugate eigenvalues or two pairs of complex eigenvalues. If there are two real eigenvalues they must be either 1 or negative one. The eigenvectors corresponding to those eigenvalues give the axes of rotation. The complex eigenvalues...

I would say for MATLAB yes! And I would like to base this on the help comments of expm and logm commands. Turns out to be coded by Higham himself. Funny coincidence. just type Code: edit expm.m

Suppose are such that . Try taking the dot product of this equation with each of the s and see what it tells you about the s.

You are correct, the range of B has the same dimension, however, B is not a linear operator, because a linear operator is a linear transformation from a space to itself, i.e. the same space, it must take P_4 -> P_4, but tA(t) takes P_4->P_5, although to a 2 dimensional subspace in P_5 it is no longer the same space. (E.g. now we have t^5 in the range of B, but by definition that does not exist in P_4, so the spaces are different).

is never constant...

I get it. (x,y,z)=(a,0,c) + (0,b,0) where (a,0,c) is in U and (0,b,0) is in Uperp. I didn't know what direct sum is, now I do! Thanks a lot rochfor1, you're awesome :)

It was pretty cool to stumble upon Euler's formula as the eigenvalues of the rotation matrix. det(Rot - kI) = (cos t - k) 2 + sin 2 t =k 2 -2(cos t)k + cos 2 t + sin 2 t =k 2 -2(cos t)k + 1 k = {2cos t +/- }/2 k = cos t +/- k = cos t +/- k = cos t +/- k = cos t +/- i sin t = e (+/-)it I was wondering what the eigenvalues are for the rotation matrix in 3D, and if there's a 3D equivalent to Euler's formula.

>> type expm Yes, I saw the name. Couldn't be coincidence. He's the expert! Who else could possibly wrote the program. I gone through my work again. Matrix A 5/2 most probably means Thank you again for your help.

Originally Posted by pamparana I am not sure I understand why we can only represent bounded functions by spherical harmonics. Is it because otherwise we would need an infinite number of the spherical basis functions? No, I think you'd potentially need an infinite number of them for a bounded function as well, so that doesn't sound like a good explanation. Originally Posted by pamparana It also says about Spherical harmonics that the...

Okay that makes sense. Thanks guys! The basis is the standard basis of {(1,0),(0,1)} And the range can be any number in R2. and yes the dim Ker(T) is 1, dim Range(T) is 2, dim Domain(T) is 3.