Re: Rigid transform

In fact, if M is a real symmetric matrix it does follow.
That can be proved using eigenvalues and eigenvectors. But for your
particular problem there's a more elementary proof: use the fact that
4 x^T y = (x+y)^T (x+y) - (x-y)^T (x-y) (and similarly with Ax and Ay)
to show that (Ax)^T Ay = x^T y for all x and y.

Thanks for the reply.

Checking:

(x + y)^T (x + y) - (x - y)^T (x - y)
= x^T x + 2 x^T y + y^T y - (x^T x - 2 x^T y + y^T y)
= 4 x^T y

It works, let's try it:

(Ax)^T Ay
= [(Ax + Ay)^T (Ax + Ay) - (Ax - Ay)^T (Ax - Ay)] * (1/4)
= [(x + y)^T A^T A (x + y) - (x - y)^T A^T A (x - y)] * (1/4)
(assuming A^T A = I)
= [(x + y)^T (x + y) - (x - y)^T (x - y)] * (1/4)
= x^T y

Thus if I assume A^T A = I, the "<=" part follows. But if I assume (Ax)
^T Ay = x^T y, what forces to conclude A^T A = I (the "=>" part)?

I am also interested in the eigenvalue/eigenvector technique for the
other problem:
Let M be real and symmetric. Then I recall it can be diagonalized as:
M = Q^T D Q
Q orthogonal (eigenvectors), D diagonal (eigenvalues)

for all x:
x^T M x =
x^T Q^T D Q x =
(Qx)^T D (Qx) = 0

The column vectors of Q form a basis and thus the Qx span R^n.

<=>
for all x:
x^T D x = 0

Using the standard basis vectors e_i, it is easy to show d_ii = 0 and
thus D = 0. Thus M = 0. Correct?

Hmm.. (A^T A - I) is symmetric and real. This would prove the original
claim!

--
http://kaba.hilvi.org
.

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