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

In summary, To prove that any nxn strictly upper triangular matrix S^n=0, it is necessary to show that all the entries of the matrix become zero after repeated multiplication, which can be done by proving that the only eigenvalue of the matrix is 0. This can be shown by thinking of the matrix as an operator on an n-dimensional vector space and using the fact that the dimension of the null space of S^(dim V) is equal to the dimension of the null space of S^n.
  • #1
Saladsamurai
3,020
7
Here we go again.:redface:

Homework Statement



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

The Attempt at a Solution



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

I'll add more in a minute:smile:
 
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  • #2
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??
 
  • #3
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.
 
  • #4
[tex](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}[/tex]. 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.
 
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  • #5
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.
 

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