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?
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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. – Marc van Leeuwen Dec 04 '11 at 11:48
2 Answers
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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 $x^n$.
David Mitra
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6This answer assumes that you are working over an algebraic closed field. – Listing Dec 04 '11 at 12:56
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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 $\chi_A(x)=x^n$.
Mariano Suárez-Álvarez
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So does that mean for $n \times n$ size matrices, there is only one similarity class of nilpotent matrices? – PythonSage Oct 17 '22 at 10:49
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1@PythonSage, nope, not at all. The classification is given by the the Jordan canonical form. – Mariano Suárez-Álvarez Oct 17 '22 at 16:34