Hello!

The course is Linear Algebra, the subject is vector-spaces, basis, dimensions.

The question is:

A (from the field) F^nxn is a nilpotent matrix.

(Nilpotent matrix - Wikipedia, the free encyclopedia)

A square matrix is nilpotent from the order k if are all different from zero, but

Prove: A^n=0

Now, there's a clue :

let k be the smallest number for which . There is a 'v' for which isn't equal to zero.

prove that is a linear independent.

Okay, now I didn't want to look for any other way to solve this (meaning without using the clue), because I also wanted to use what I've learned and practiced.

So, what I did was:

Let , =/0 (is not zero)

therefore, let there be v=/0, for which =/0, and for any other number that is lower than k. (since A is a nilpotent matrix)

I have to prove that is linear independent. So, let{α1,α2, . . . ,αk} be parameters. Let's see when (for which alphas):

α1v+α2Av+. . . +α(k)A^(k-1)v=0(in order to prove that it's linear independent I need to show that the ONLY case that it's equal to zero is when α1=α2=α3=...=αk=0)

So, I multiply it by the matrix A:α1Av+α2vA^2+ . . . +α(k-1)A^(k-1)v+α(k)A^k=0

^according to what we assumed, A^k=0, so it's equal to:

α1Av+α2A^2*v+ . . . +α(k-1)A^(k-1)v=0

we'll multiply again by A:

α1A^2*v+α2A^3*v+ . . . +α(k-2)A^(k-1)v=0

.

.

.

and so on. In total, we multiply it by A (k-1) times, and in the end we get:

α1A^(k-1)*v=0

In such case, α1canbe equal to zero, while the other alphas can be everything. which means this is not necessarily linear independent! Argh...Can anyone please help me with this ?

Thank you very much !!!