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Let $J\in\mathbb{C}^{n\times n}$ be a Jordan normal form and assume that ${\rm tr~}J<2n$. Prove or disprove that there exist $a_1,\ldots, a_{2n-1}\in\mathbb{R}$ such that \begin{equation} a_{2n-1}J^{2n-1}+a_{2n-2}J^{2n-2}+\cdots+a_1 J=I_n. \end{equation}

Any help is appreciated

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The statement is false. For example, any Jordan matrix with zeros on the diagonal satisfies the condition, but no such coefficients exist.

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  • $\begingroup$ Thanks. What if $J$ has no zero eigenvalues? $\endgroup$ – James Vestal May 23 at 22:00
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    $\begingroup$ Then this is possible, even without the condition on the trace $\endgroup$ – Ben Grossmann May 23 at 22:08
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    $\begingroup$ @JamesVestal See this post, for instance $\endgroup$ – Ben Grossmann May 23 at 22:09
  • $\begingroup$ Did you notice that J is $n$ by $n$ but we have $J^{2n-1}$ present? $\endgroup$ – James Vestal May 23 at 22:26
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    $\begingroup$ @JamesVestal Yes, and I don't see how that makes a difference. We can simply take $a_{n+1} = a_{n+2} = \cdots = a_{2n - 1} = 0$, and apply the process described in the answers to the post to find the remaining coefficients. $\endgroup$ – Ben Grossmann May 23 at 22:30

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