We just need to show .Let be an strictly upper triangular matrix. Show that
Open LHS to get, .
We want for which is true if and only if where is zero matrix.
Thus, it remains to prove .
(Since is strictly upper triangular it is nilpotent).
I'm trying to think of a way to start this "proof":
Source: Matrices and Linear Algebra, 2nd Ed. by Schneider and BarkerLet be an strictly upper triangular matrix. Show that
I don't want a complete proof. All I would appreciate would be an explanation on how I can start this. I will be able to take it from there.
Thank you in advance!!
I've been trying to think of a way to prove this for hours now.
This is brought up in the "Matrix Multiplication" section of my textbook. They ask me to prove this as well...but I can't use eigenvalues and determinants.
I think that mathematical induction can be used...but I'm not sure what I'm supposed to let my be.
I was thinking that , but , but that's not necessarily true.
I would appreciate any suggestions!!!
One way round that is to start the induction at n=2, where the result is easily verified.
Another way is to think more carefully about the n=1 case. A 1×1 matrix has just a single element. This lies on the main diagonal (which also contains just the one element!). So if the matrix is strictly upper triangular then that element must be 0. Therefore the matrix is the zero matrix.
Observe that A = where the bold faced 0 stands for the column vector 0 and v stands for a "n-1" length row vector and A' is a (n-1) x (n-1) strictly upper triangular matrix.
finally for induction to work, you just need to see that:
Actually the above can be proved by another induction