Let $A$ be any complex $n\times n$ matrix. Can I then find a Jordan basis $$ B_1\cup\ldots\cup B_r $$ for $A$ with $B_k = \{b_1^k,\ldots,b_{n_k}^k\}$ ($b_{n_k}^k$ being an eigenvector of $A$ w.r.t. to the eigenvalue $\lambda_k$) such that $$ (A^*- \overline{\lambda_k})b_1^k = 0 $$ for each $k=1,\ldots,r$? I figured out that the answer is yes if $k=1$, but I doubt it when there is more than one Jordan chain. Can anyone help?
2026-03-30 09:44:56.1774863896
Specifically designed Jordan basis
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