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Let $A$ be a compact linear operator on a Hilbert space $\mathcal H.$ Let $p(A,A^*) = \sum\limits_{i,j = 1}^{k} a_{ij} A^i {A^*}^j$ be a polynomial in $A$ and $A^*.$ Here $a_{ij} \in \mathbb C$ for all $i,j = 1,2, \cdots, k.$ Prove or disprove

$$\sigma (p(A,A^*)) = \left \{p(\lambda, \overline {\lambda})\ \big |\ \lambda \in \sigma (A) \right \}.$$

I can able to show that $$\left \{p(\lambda, \overline {\lambda})\ \big |\ \lambda \in \sigma (A) \right \} \subseteq \sigma (p(A,A^*)).$$ But I don't think that the other part of the inclusion holds true. But I couldn't able to find a suitable counter-example. Could anyone please help me in this regard?

Thanks a bunch.

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1 Answer 1

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Let $p(X,Y) = -X+Y$, $\mathcal H = \mathbb C^2$ and $A= \begin{pmatrix}0&1\\0&0\end{pmatrix}$. Then, we have: $$\sigma(A) = \sigma(A^*)= \{0\} \qquad \text{and} \qquad \{p(\lambda,\bar\lambda):\lambda\in\sigma(A)\} = \{0\}$$ while : $$p(A,A^*) = \begin{pmatrix} 0&-1\\ 1&0\end{pmatrix}$$ and : $$\sigma(p(A,A^*)) = \{i,-i\}$$

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  • $\begingroup$ Wow. How easy it was! Thanks for your help. $\endgroup$
    – RKC
    Sep 2, 2021 at 17:30
  • $\begingroup$ So actually none of the two inclusions is true, contrary to what the OP claimed. $\endgroup$
    – PhoemueX
    Sep 2, 2021 at 18:31

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