Eigenvalues of a symmetric tensor? A linear map is a (1, 1)-tensor and symmetry has no meaning. A symmetric tensor is not an automorphism and thus has no eigenvectors or -values. If this can be resolved, this seems to only make sense for tensors on 3-space. This comment concerns both Rhombicity and Axiality--MarSch 12:03, 19 October 2005 (UTC)
- The tensor here is what I would call a 3-by-3 matrix. It is indeed a (1,1)-tensor, but it is acting on Euclidean space so it can be identified with a (2,0)-tensor, and for those tensors "symmetric" has a meaning. As far as I know, this is standard physics language, see for instance the discussion of the stress tensor in stress (physics) for an example within Wikipedia.
- I contacted the original author, and it seems that the term "rhombicity" is indeed only used for threedimensional space. I tried to clarify the article. Is it any better now? -- Jitse Niesen (talk) 18:17, 5 November 2005 (UTC)
- Thanks, it is better. Perhaps it would be better to call it a matrix. Also since not every matrix has a basis of eigenvectors we need to specify or allow for complex eigenvalues. --MarSch 15:39, 6 November 2005 (UTC)
- The matrix is assumed to be symmetric, hence it can be diagonalized and its eigenvalues are real. I decided that it is better to combine the two articles. Since you are a self-confessed mergist, I went straight ahead under the assumption that you wouldn't mind. I also removed the {{attention}} tag. As always, feel free to revert. -- Jitse Niesen (talk) 20:24, 6 November 2005 (UTC)