Nilpotent Operator: Can e be Less Than or Equal to dim V?

[Treadstone 71]

Suppose V is finite dimensional and N:V->V is a linear transformation such that N^e=0. Is it possible to show that e\leqdim V? Is it even true?

 

[matt grime]

Yes, it is very easy to show this, or rather the correct statement when e is the minimal positive integer such that N^e=0, so try it.

 

[Treadstone 71]

I got it. It's part of a much larger proof. Thanks.

 

[wangchong]

Yes it's true. If V were represented via its "basis" as R^n, (n=dim V) then, N can be represented as a nilpotent matrix. We knew that with appropriate representation, N is a strictly upper triangular (diagonal and lower triangular elements are all zeroes) matrix. And now we just have to show that a strictly upper triangular matrix the property that N^n is always zero. (Sorry, I use N (the operator, as the matrix).