1. Feb 8, 2010

### Seb97

1. The problem statement, all variables and given/known data
For an n x n-matrix A we have A^N = 0 for some N. Prove that A^n = 0.

2. Relevant equations

3. The attempt at a solution
I took the inverse of A^n and multiplied across by it. and I got the Identity equal to zero. But that cant happen. Any help would be much appreciated

Last edited: Feb 8, 2010
2. Feb 8, 2010

### Staff: Mentor

What exactly is the problem statement? Are you supposed to prove that A^n = 0 for all integer values of n? (This is not true - counterexample is the 2x2 matrix with 1 in the upper right corner and 0 elsewhere. A^2 = 0, but A^1 != 0.)
Or for all integer values of n >= N?

You said you took the inverse of A^n and did something with it. What guarantee do you have that A or A^n have inverses?

3. Feb 8, 2010

### Seb97

No thats precisely what the question says. I was hoping for a different view on it. Now im really confused because I taught the inverses had something to do with it. What do you is the right approach or what would you do?

4. Feb 8, 2010

### Seb97

Hi Mark44
How would you prove that the rank of the matrix would be zero because if I can do that then it would be proved wouldnt it??

5. Feb 8, 2010

### Staff: Mentor

OK, it went right past me that the matrix is n x n and you're trying to show that A^n = 0.
I think you can assume that 2 <= N <= n. If N > n, then it might be possible that A^N = 0, but for a smaller exponent, n, A^n != 0. If it's not reasonable to make that assumption, you can at least break down the problem into two cases: 2 <= N <= n, and N > n.

The first case is pretty easy to prove. A^1 = whatever, A^2 = whatever, ..., A^N = 0, A^(N + 1) = A^N * A = 0*A = 0, and so on.

6. Feb 8, 2010

### Seb97

Hi Mark44

Thanks a million your a legend. I can now go to sleep tonight. Thanks again

7. Feb 8, 2010

### Staff: Mentor

You'll need to say it more elegantly than I did. I was just thinking out loud, so to speak.