Thread: Strictly Upper Triangular Matrix Proof Help Needed

Hey, I'm trying to prove that for a strictly upper triangular matrix A, A^n = 0. I'm pretty much stuck at writing down the definition for matrix multiplication for A*A. Any help/hints would be much appreciated. Thanks, Ultros

Originally Posted by Ultros88

Hey, I'm trying to prove that for a strictly upper triangular matrix A, A^n = 0. I'm pretty much stuck at writing down the definition for matrix multiplication for A*A. Any help/hints would be much appreciated. Thanks, Ultros

A is nilpotent. Show either that its eigenvalues are all zero or that its determinant is zero.

Is there a way to prove that A^n = 0 without using the fact that its determinant or eigenvalues are all zero? This exercise comes before those topics in the book that I am using. Thanks

Originally Posted by Ultros88

Is there a way to prove that A^n = 0 without using the fact that its determinant or eigenvalues are all zero? This exercise comes before those topics in the book that I am using. Thanks

Proof by induction will work (you have to know how to do matrix multiplication). See here: Matrix Algebra - Google Book Search