# If A is nxn nilpotent matrix, this char(A) = x^n

1. May 10, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
If $A$ is an $n \times n$ nilpotent matrix, then the characteristic polynomial of $A$ is $x^n$

2. Relevant equations

3. The attempt at a solution
Suppose that $A$ has an eigenvalue with corresponding eigenvector such that $A v = \lambda v$. Then $A^k v = \lambda^k v = 0$, and since $v \ne \vec{0}$, $\lambda^k = 0 \implies \lambda = 0$. Since $0$ is the only eigenvalue, and since the characteristic polynomial must be of degree n, the characteristic polynomial must be $x^n$.

2. May 10, 2017

### Staff: Mentor

You should introduce k. Apart from that: right.

3. May 10, 2017

Yes.