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 δ 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 $\lambda _{i}\neq \lambda _{j}$. — 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 δ is the Kronecker delta)

then

$f(A)=\sum _{i=1}^{n}f(\lambda _{i})c_{i}r_{i}$

Part 1: (How to get to the hypothesis)

(Option A) If A is diagonalizable, $A=UDU^{-1}$, then, taking:

r_i to be the rows of U,

c_i to be the columns of U^-1,

λ_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 λ_i.

Let r_i be the corresponding row-eigenvectors

Let d_i be the corresponding column-eigenvectors.

Let $c_{i}={\frac {d_{i}}{r_{i}d_{i}}}$

(It can easily be seen that r_i c_j = 0 for i <> j.)

I'm still not convinced that Sylvester did it. — 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)