10-23-2013, 12:43 PM
The Prime has an eigenvalue solver as part of it's matrix utilities that will solve the common eigenvalue problem Det (A - lambda) = 0, where A is a square matrix and lambda is the eigenvalue. The problem I have, however, is in the form Det (A - lambda * B), where B is a diagonal square matrix, which does not work with the eigenvalue solver. I could convert it to the base form only if all the elements of B were the same. In the past under RPN and RPL, I used the root solver to do this by creating a function Det (A - lambda * B) = 0 to solve for lamda, but it doesn't seem possible to do this with the Prime solver. Perhaps someone can show me how to do this on the Prime.
Edited: 23 Oct 2013, 12:44 p.m.