# Characteristic polynomial of a nilpotent matrix

Let $A$ be $n\times n$ nilpotent matrix.

How to calculate its characteristic polynomial?

I know it should be $X^n$, but I don't know why?

• Do you know about minimal polynomials and the Cayley-Hamilton theorem? Or (if working over the complex numbers) you could argue which possible eigenvalues a nilpotent matrix can have. Dec 4 '11 at 11:48

The minimal polynomial is of the form $x^k$ for some $k$, because the matrix is nilpotent. Since the minimal polynomial is divisible by all the irreducible polynomials which divide the characteristic polynomial, we see that in fact the only irreducible polynomial which divides the latter is $x$. Thus $\chi_A(x)=x^n$.
If $A^k=0$ and $\lambda$ is an eigenvalue of $A$ with eigenvector $\bf x$:
\eqalign{ A {\bf x}=\lambda {\bf x} &\Rightarrow A^2 {\bf x}= \lambda^2 {\bf x} \cr &\Rightarrow A^3 {\bf x}=\lambda^3 {\bf x}\cr &\ {\vdots} \cr &\Rightarrow 0=\lambda^{k } {\bf x} \cr & \Rightarrow \lambda=0} \ \ \ \raise6pt{\left. {\vphantom{\matrix{1\cr1\cr1\cr1\cr1\cr}}}\right\}} \raise6pt{\scriptstyle{(k-1)-\text{times}}}
So 0 is the only eigenvalue of $A$. The characteristic polynomial of $A$ is then $x^n$.