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?

  • $\begingroup$ 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. $\endgroup$ 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 the overlying field is the complex numbers (see Listing's and Mark's comment):

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 $x^n$.

  • 4
    $\begingroup$ This answer assumes that you are working over an algebraic closed field. $\endgroup$
    – Listing
    Dec 4 '11 at 12:56
  • $\begingroup$ @Listing Thanks, I edited the post to reflect this. $\endgroup$ Dec 4 '11 at 13:00

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