Linear Algebra Problem #3: Proving S^n=0 for Strictly Upper Triangular Matrix

[Saladsamurai]

Here we go again.

Homework Statement

Let S be any nxn strictly upper triangular matrix; prove that S^n=0

The Attempt at a Solution

Alright so I know that if c_ij is an entry in a strictly upper triangular matrix, then c_{ij}= 0, \ i>j

I'll add more in a minute

 

[Dick]

Is this another one of those questions where you understand exactly why it's true, but can't phrase it in the form of a proof? Because it is pretty obvious why it's true. Just write down a strictly upper triangular matrix and start taking powers of it. See what happens??

 

[Defennder]

I think the problem is with formalising it into matrix entry subscript notation. I can see why it holds: Over time throughout repeated multiplication the non-zero entries of the matrix retreats into the top right hand corner and then finally becomes zero.

I've written out the matrix entries for an arbitrary c_{ij} after multiplying matrices by 3 times and I'm beginning to see nested summation series which I can't quite simplify.

PS. I know this is Saladsamurai's problem and not mine, but I'm curious as to how to prove it formally.

 

[Dick]

(c^n)_{ij}=c_{i,k_1}*c_{k_1,k_2}*c_{k_2,k_3}*...c_{k_{n-2},k_{n-1}}*c_{k_{n-1},j}. All of the k indices summed over. What condition must be true for that to be nonzero? Hey, what ever happened to your electric flux problem? It's not THAT hard.

 

[durt]

Alternatively, note that the only eigenvalue is 0. Thinking of this matrix as an operator on an n-dimensional vector space V, we must have V = dim null S^(dim V) = dim null S^n, the set of all generalized eigenvectors. So S^n = 0.