Wikipedia:Reference desk/Archives/Mathematics/2013 April 21

= April 21 =

Matrix estimation
In Estimation_of_covariance_matrices, 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)
 * 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)

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)
 * 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)