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Re: Square of a lower triangular matrix
Posted:
Apr 14, 2013 6:12 PM


HardySpicer <gyansorova@gmail.com> wrote in news:41941a8abdaf4676bd6e f1decee0b343@ka6g2000pbb.googlegroups.com:
> If I have a lower (say n square) triangular matrix T and I compute > T^2 is there a proof which says that it too must be a lower triangular > matrix with diagonal elements the square of the original?
Each element above the diagonal in the product is the dot product of a row ending with m zeros and a column beginning with k zeros where m+k >= n. The diagonal entries are dot products with only one nonzero term.



