RE: [Algorithms] transforming vectors...

[Jamie F. <ja...@qu...>]

No.

	(M^{-1})^t = (S^{-1} A^t)^t
		     = (S^{-1})^t (A^t)^t
		     = S^{-1} A

is not true. t(AB) = t(B)t(A), so 

t(inv(M)) = t(inv(S) t(A))
	= t(t(A)) t(inv(S))
	= A inv(S)

(_not_ inv(S) A)


Regardless, my previous post showed that if

	a = Sx
	b = S^{-1}x

then a and b are not in general colinear.

Jamie


-----Original Message-----
From: gda...@li...
[mailto:gda...@li...]On Behalf Of
Willem H. de Boer
Sent: 31 March 2003 17:16
To: gda...@li...
Subject: RE: [Algorithms] transforming vectors...


>And it wouldn't work for any scalings (apart from the trivial
>one: all 1's) either. Because a matrix that contains a scaling
>in whatever direction is not an orthogonal matrix. Before anyone
>starts objecting: With orthogonal I mean M^T M = I, it's what
>many people would call orthonormal.

Actually, it _would_ work for scalings, as I've just found out.

[People not interested in maths should click away, NOW]

Suppose M is a transformation (an NxN matrix), composed of two 
other transformations M = SA, where both S and A are NxN matrices.
S is a scaling matrix, and is thus diagional. And A is an orthogonal
matrix (or orthonormal, if you will). Now we want to "prove"
that for a given vector v:

	Mv . Nv = |Mv| |Nv|				(1)

Or, the vector a=Mv, is colinear but not necessarily of same magnitude
to the vector b=Nv, where N is the inverse transpose of M. Let's
calculate the inverse transpose of M, firstly, its inverse:

	M^{-1} = (S A)^{-1} = S^{-1} A^{-1} = S^{-1} A^t

The last bit comes from the fact that the inverse of an orthogonal
matrix is just its transpose. Now let's take the transpose of M^{-1}

	(M^{-1})^t = (S^{-1} A^t)^t
		     = (S^{-1})^t (A^t)^t
		     = S^{-1} A

The last bit coming from the fact that the transpose of a diagonal
matrix (S^{-1}, in this case, the inverse of S) is the same
matrix. And the transpose of the transpose of a matrix A, is
just A.

So, now back to the original statement (1). We can rewrite the vector
a as:
	a = Mv = (SA)v
and b:
	b = M^-tv = (S^{-1} A)v

Now we now that both a and b transform similarily under A:
so, let's call Av, x, both equations reduce to:	

	a = Sx
	b = S^{-1}x

So, yes, you're assumption was right. a, and b will be colinear
but of different magnitudes, if and only if M is a transformation
that consists only of rotations, and scalings (even non-uniform ones).

Willem


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