Yes. That affects things. The key is that inverse(diag(d_1 ... d_n)) = diag(1/d_1 ... 1/d_n)
that means that inverse(D V') = V inverse(D). If you have X' = DV' you need to compute inverse(X') = X D^-2 On Fri, Feb 25, 2011 at 1:25 PM, Chris Schilling <[email protected]> wrote: > One more linear algebra question. So, does this still hold when the > diag(d) matrix is multiplied through the right hand side? Is that an affect > I should worry about when trying to compute u?
