Talk:Sylvester's matrix theorem

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[edit] Normalization

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 ricj = δij, (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
ricj = δij (where δ 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 = 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 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)

[edit] Added "disputed" tag

It seems only to apply (as written) if the matrix is diagonalizable with distinct eigenvalues, and can be adapted only if the matrix is diagonalizable. I'll get back to it, later, but there may very well be a problem. — Arthur Rubin | (talk) 14:37, 5 January 2007 (UTC)