# Nilpotent Matrices

Im just new to linear algebra, and am having trouble proving that a nilpotent matrix must be singular. Can anyone give me a hand in the right direction slash provide the proof?
Cheers

### Re: Nilpotent Matrices

Let A be a nilpotent matrix. Then by definition for some natural n, A^n = 0.
Let d = |A| (the determinant). |0| = 0 = |A^n| = |A|^n = d^n. Hence, d = 0 and A is singular.

### Re: Nilpotent Matrices

Another way to think about this is to suppose instead that we had an invertible matrix A such that A^n=0. A product of invertible matrices is invertible, so A^n=0 is invertible, but 0 is not invertible.

### Re: Nilpotent Matrices

Yet another approach. Consider N nilpotent (but non-zero) with n the smallest integer (>1) such that N^n = 0. Assume N is non-singular with inverse M, then N^(n-1) = M N^n = M 0 = 0, but n was suppose to be the smallest integer...proof by contradiction.