# 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 ${\displaystyle r_{i}c_{j}=\delta _{ij}}$, (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 ${\displaystyle r_{i}c_{j}=0\,}$ if ${\displaystyle \lambda _{i}\neq \lambda _{j}}$. — 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
${\displaystyle A=\sum _{i=1}^{n}\lambda _{i}c_{i}r_{i},}$ and
${\displaystyle r_{i}c_{j}=\delta _{ij}}$ (where δ is the Kronecker delta)
then
${\displaystyle f(A)=\sum _{i=1}^{n}f(\lambda _{i})c_{i}r_{i}}$
Part 1: (How to get to the hypothesis)
(Option A) If A is diagonalizable, ${\displaystyle A=UDU^{-1}}$, then, taking:
ri to be the rows of U,
ci to be the columns of U-1,
λi 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 λi.
Let ri be the corresponding row-eigenvectors
Let di be the corresponding column-eigenvectors.
Let ${\displaystyle c_{i}={\frac {d_{i}}{r_{i}d_{i}}}}$
(It can easily be seen that ri cj = 0 for i <> j.)
I'm still not convinced that Sylvester did it. — 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)