If \(N\) is a nilpotent linear operator on an \(n\)-dimensional vector space \(V\), then the characteristic polynomial for \(N\) is \(x^n\).
Solution
Choose \(m\) with \(N^m=0\). Consider the increasing chain \[ \{0\}=\ker N^0\subseteq\ker N\subseteq\ker N^2\subseteq\cdots \subseteq\ker N^m=V. \] Choose a basis of \(\ker N\), extend it to a basis of \(\ker N^2\), continue in this way, and finally obtain a basis of \(V\) adapted to this flag [@HK71].
If \(v\in\ker N^j\), then \[ Nv\in\ker N^{j-1}. \] Therefore, in the adapted basis, the matrix of \(N\) is strictly upper triangular: every basis vector is sent into the span of vectors occurring in earlier stages of the flag, and all diagonal entries are zero.
Hence \(xI-[N]\) is upper triangular with every diagonal entry equal to \(x\). Its determinant is therefore \[ \chi_N(x)=\det(xI-[N])=x^n. \] Thus \[ \boxed{\chi_N(x)=x^n}. \]