Talk:Frobenius covariant

Normalization of eigenvectors
[This section was taken from Talk:Sylvester's formula --Jorge Stolfi (talk) 22:40, 30 December 2009 (UTC)]

I think the article should explain how to normalize the eigenvectors. -- Jitse Niesen (talk) 14:27, 5 January 2007 (UTC)


 * In context, the normalization is "obviously" such that $$r_i c_j = \delta_{ij}$$, (where &delta; is the Kronecker delta), but a reference is still required. I think I have one, but it's not called "Sylvester's" in that reference.  If the eigenvalues are distinct, then it's easy to construct such normalized eigenvectors, as $$r_i c_j = 0\, $$ if $$\lambda_i \neq \lambda_j$$. &mdash; Arthur Rubin |  (talk) 15:00, 5 January 2007 (UTC)
 * The reference I thought was there didn't have it. However, the actual results seems to be the following:


 * Part 2: If
 * $$A = \sum_{i=1}^n \lambda_i c_i r_i, $$ and
 * $$r_i c_j = \delta_{ij}$$ (where &delta; is the Kronecker delta)
 * then
 * $$ f(A)=\sum_{i=1}^n f(\lambda_i)c_ir_i$$


 * Part 1: (How to get to the hypothesis)
 * (Option A) If A is diagonalizable, $$ A = U D U^{-1} $$, then, taking:
 * ri to be the rows of U,
 * ci to be the columns of U-1,
 * &lambda;i to be the diagonal entries of D
 * the hypothesis of Part 2 can easily be seen to be met.
 * (Option B, which may be where Sylvester got into it). If A has n distinct (left-)eigenvalues, denote them &lambda;i.
 * Let ri be the corresponding row-eigenvectors
 * Let di be the corresponding column-eigenvectors.
 * Let $$c_i = \frac {d_i} {r_i d_i}$$
 * (It can easily be seen that ri cj = 0 for i <> j.)


 * I'm still not convinced that Sylvester did it. &mdash; Arthur Rubin |  (talk) 21:43, 5 January 2007 (UTC)

Projection of what?
The text says that the Frobenius covariants are projections onto the eigenspaces etc.. Pojections of what, exactly? All the best, --Jorge Stolfi (talk) 22:41, 30 December 2009 (UTC)
 * Projections. — Arthur Rubin  (talk) 07:05, 19 April 2015 (UTC)