# Nilpotency of Matrix with one eigenvalue

1. Mar 11, 2009

### geor

Hello everybody,

I have a question for which I cannot find the answer around,
any help would be really appreciated.

Suppose we have a matrix A of a linear transformation of a vector space,
with only one eigenvalue, say 's'.

My question is: Is the operator (A-sI) nilpotent? ('I' is the identity matrix).

I am trying to understand the proof of the Jordan Canonical Form Theorem and there is
a fuzzy point in my notes about that...

EDIT: I am becoming pretty sure about that, but some confirmation would be great..

Last edited: Mar 11, 2009
2. Mar 11, 2009

### HallsofIvy

I assume you are talking about finite dimensional vector spaces. A linear transformation from a vector space of dimension n to itself can be represented by an n by n matrix. If an n by n matrix, over the complex numbers, has only the single eigenvalue, $\lambda$, then its characteristic equation must be $(x- \lambda)^n= 0$. Since every matrix satisfies its own characteristic equation, $(A- \lambda I)^n= 0$.

3. Mar 11, 2009

### geor

Oh yes, I see it now! Thanks so much!