Days late, but worth the response for accuracy's sake since I just popped in on HN now:
> 1. A linear mapping is not a "kind of equivalence" by any > reasonable definition.
By definition, ANY mapping is an equivalence relation--even if that relation results in a 0 or NAN value.
> 2. The eigenvectors are not the rows of the diagonalized > matrix...
Correct. It depends on one's orientation (columns vs. rows), but the matrix which does the coordinate transforms contains the eigenvectors. Bad wording on my part. The key behind diagonalization is it removes any orientation issues from the relationship by establishing eigenweights across a given vector space.
> 3. I think the paragraph beginning "For the common 3D isometric ..."
Your interpretation is correct and, indeed, my choice of verbiage was poor. I really should have used the words "ortho-normal to a given vector space" which collapses to the common XYZ unit vectors in a linearly partitioned vector space of 6 degrees of freedom (3 translations & 3 rotations)--e.g. classic Cartesian space.
> Tensors are just as linear as matrices.
Most tensor fields are modeled using linear approximations (i.e. matricies of n-dimensions), but the very fact that a tensor itself is being used in the characteristic equations is typically indicative of non-linear behavior in the overall system. For example, in fluid dynamics used to model airflow across a wing or boundary values issues when the Cauchy stress tensor is used for structures undergoing plastic deformation.
I believe that you are conflating Manifolds with Tensors. The latter is a refinement and/or characteristic relation defined upon the former. A non-linear Tensor is defined upon a manifold with one or more non-linear relations. Perhaps you are used to dealing exclusively with metric tensors?
> The probabilities in a Markov chain's stationary state are not eigenvalues.
Here you are spot-on. Indeed it's the eigenvalues of the transition matrix (or convergence for ergodic ones) to which I was referring. Thanks (again) for the correction and further clarification.
> 1. A linear mapping is not a "kind of equivalence" by any > reasonable definition.
By definition, ANY mapping is an equivalence relation--even if that relation results in a 0 or NAN value.
> 2. The eigenvectors are not the rows of the diagonalized > matrix...
Correct. It depends on one's orientation (columns vs. rows), but the matrix which does the coordinate transforms contains the eigenvectors. Bad wording on my part. The key behind diagonalization is it removes any orientation issues from the relationship by establishing eigenweights across a given vector space.
> 3. I think the paragraph beginning "For the common 3D isometric ..."
Your interpretation is correct and, indeed, my choice of verbiage was poor. I really should have used the words "ortho-normal to a given vector space" which collapses to the common XYZ unit vectors in a linearly partitioned vector space of 6 degrees of freedom (3 translations & 3 rotations)--e.g. classic Cartesian space.
> Tensors are just as linear as matrices.
Most tensor fields are modeled using linear approximations (i.e. matricies of n-dimensions), but the very fact that a tensor itself is being used in the characteristic equations is typically indicative of non-linear behavior in the overall system. For example, in fluid dynamics used to model airflow across a wing or boundary values issues when the Cauchy stress tensor is used for structures undergoing plastic deformation.
I believe that you are conflating Manifolds with Tensors. The latter is a refinement and/or characteristic relation defined upon the former. A non-linear Tensor is defined upon a manifold with one or more non-linear relations. Perhaps you are used to dealing exclusively with metric tensors?
> The probabilities in a Markov chain's stationary state are not eigenvalues.
Here you are spot-on. Indeed it's the eigenvalues of the transition matrix (or convergence for ergodic ones) to which I was referring. Thanks (again) for the correction and further clarification.