Let $A$ be a nilpotent operator. Show how to obtain, from aJordan basis for $A$, aJordan basis of $\wedge^2A$.

Solution. Since $A$ is nilpotent, each eigenvalue of $A$ is zero, and thus there exists an basis $e_1,\cdot,e_n$ of $\scrH$ such that $$\bex A(e_1,\cdots,e_n)=(e_1,\cdots,e_n) \sex{\ba{cccc} 0_s&&&\\ &J_1&&\\ &&\ddots&\\ &&&J_t \ea},\quad J_{i}=\sex{\ba{cccc} 0&1&&\\ &\ddots&\ddots&\\ &&\ddots&1\\ &&&0 \ea}_{n_i\times n_i} \eex$$ with $$\bex s+\sum_{i=1}^t n_i=n. \eex$$ Hence $Ae_i=0$ for $$\bex i\in S=\sed{1\leq i\leq s+1, s+\sum_{i=1}^jn_i+1,\ j=1,\cdots,t-1}, \eex$$ and $Ae_k=0$ for $$\bex k\in T=\cup_{j=1}^t T_j,\quad T_j=\sed{s+\sum_{i=1}^{j-1}n_i+2\leq k\leq s+\sum_{i=1}^j n_i+2}. \eex$$ Thus $$\bex k\neq j,\ k,j\in T\lra 0\neq \wedge^2A(e_k\wedge e_l)=e_{k-1}\wedge e_{l-1}. \eex$$ Hence $\wedge^2 A$ has a Jordan basis $$\bex e_i\wedge e_j;(i\in S,i<j\leq n) \eex$$ $$\bex e_k\wedge e_{k+1};\quad\sex{k\in T}; \eex$$ $$\bex e_k\wedge e_{k+2};\quad\sex{k\in T}; \eex$$ $$\bex \cdots,\quad e_{s+2}\wedge e_n. \eex$$

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