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April 21

Matrix estimation

In Estimation_of_covariance_matrices#Maximum-likelihood_estimation_for_the_multivariate_normal_distribution, part "using the spectral theorem", why do you need to use the cyclic property of the trace? Why can't you simply define $B=S\Sigma ^{-1}$ and work form there? All of the working would be exactly the same until "concluding steps" when you simply solve $\Sigma =B^{-1}S$ instead. Why split the matrix up into square roots? AnalysisAlgebra (talk) 23:31, 21 April 2013 (UTC)[reply]

I just had a look and didn't see why either, so I've stuck a note on the talk page of the person I think stuck it in as they don't normally do things wrong. Dmcq (talk) 10:49, 23 April 2013 (UTC)[reply]

The matrix $S^{1/2}\Sigma ^{-1}S^{1/2}$ is symmetric and its entries are real. And it's non-negative-definite. Hence by the (finite-dimensional) spectral theorem, it can be diagonalized and the diagonal entries are non-negative. Is there a simpler way to show that's it's diagonalizable and the diagonal entries are non-negative? Michael Hardy (talk) 00:50, 25 April 2013 (UTC)[reply]

Thanks, yes the major point here is the product of two symmetric matrices need not be symmetric. Dmcq (talk) 08:33, 25 April 2013 (UTC)[reply]