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Establishing equality of matrix expressions

Posted 10 months ago

Looking at a basis and their transformed.
I have two independently evaluated matrix expressions, one matrix is formed out of pairwise dot products of a transformed basis:

nmatrix=Outer[Simplify[
   TensorExpand[#2 . #1]] &, basisnTransformed, basisnTransformed]  

and the matrix product of the actual transformation matrix R and its transpose:

R . Transpose[R]

both turn out to be equal and Mathematica gives me the correct answer:

nmatrix  == R.Transpose[R]  -> True

which is correct.

Now I want to get a step further and say

orthonormalTransformedConditions = {nmatrix == IdentityMatrix[3]}

So nmatrix or the pairwise dot products of a transformed basis is assumed to be the IdentityMatrix[3} in other words I assume the transformed basis is orthnormal and then show that R is also orthonormal by proving

R.Transpose[R] == IdentityMatrix[3]

but the test:

Simplify[
 Equal[R . Transpose[R], IdentityMatrix[3], 
  Assumptions -> orthonormalTransformedConditions]]-> Flase

Why ?

POSTED BY: Andreas Walter
3 Replies

What is the formula for nmatrix?

POSTED BY: Gianluca Gorni
Posted 10 months ago
nmatrix == R . Transpose[R]

gives True

orthonormalTransformedConditions = {nmatrix == IdentityMatrix[3]}; Simplify[
 Equal[R . Transpose[R], IdentityMatrix[3], 
  Assumptions -> orthonormalTransformedConditions]]

gives False I other words I have already established identity between A, B. Now I assume B==C but the logical conclusion A==C is not made. All done usine sybolic matrix calculations. nmatriix and likewis R.Transpose.R look like this (SubSuperscript indexes):

enter image description here

POSTED BY: Andreas Walter

Are you saying that

nmatrix := r . Transpose[r];
orthonormalTransformedConditions :=
  {nmatrix == IdentityMatrix[3]};
Simplify[r . Transpose[r] == IdentityMatrix[3],
 Assumptions -> orthonormalTransformedConditions]

gives False?

POSTED BY: Gianluca Gorni
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