Recent content by Azad Koshur

Yes I was thinking the norm will still be given by the same way i did in the old basis. In this new basis I would convert the vectors back to standard basis by a matrix ##P##, then the norm will be just ##x^Tx## (considering real elements only).

I think represent those into orthogonal basis coordinates and then calculate the norm using the usual way

Yes got it now.

But I don't see where the argument went wrong?

Wikipedia says : "In linear algebra, two n-by-n matrices A and B are called similar if there exists an invertible n-by-n matrix P such that ##{\displaystyle B=P^{-1}AP.}## Similar matrices represent the same linear map under two (possibly) different bases, with P being the change of basis...

##B^TB=I## is a solution but I'm not sure it's the only one.

I've a transformation ##T## represented by an orthogonal matrix ##A## , so ##A^TA=I##. This transformation leaves norm unchanged. I do a basis change using a matrix ##B## which isn't orthogonal , then the form of the transformation changes to ##B^{-1}AB## in the new basis( A similarity...