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Let $(T_1,T_2)$ be a pair of commuting operators acting on $\mathcal{H}$. $(\lambda_1,\lambda_2)\in \sigma_T(T_1,T_2)$ if and only if the Koszul-Complex of $(T_1-\lambda_1,T_2-\lambda_2)$ is not exact. This is equivalent that at least one of the following properties fails:

  • $(a)$ The equation system $$(T_1-\lambda_1)\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=0\;\; \wedge \;\; (T_2-\lambda_2)\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=0$$ has the unique solution $(x_1,x_2)=(0,0)$.

  • $(b)$ The only solutions for the equation $$(T_1-\lambda_1)\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}+(T_2-\lambda_2)\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}=0$$ are of the form $$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=-(T_2-\lambda_2)\begin{pmatrix} s_1 \\ s_2 \end{pmatrix}\;\;\text{and}\;\;\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}=(T_1-\lambda_1)\begin{pmatrix} s_1 \\ s_2 \end{pmatrix},$$ for some $(s_1,s_2)\in \mathbb{C}^2$.

  • $(c)$ The equation $$(T_1-\lambda_1)\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}+(T_2-\lambda_2)\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}=\begin{pmatrix} b_1 \\ b_2 \end{pmatrix},$$ has a solution for every $(b_1,b_2)\in \mathbb{C}^2$.

I want to prove that $$\sigma_T(T_1,T_2)\subset\sigma(T_1)\times\sigma(T_2)$$

For more informations about the Taylor spectrum see:

J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.

we have

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You seem to be assuming that $\mathcal H=\mathbb C^2$. There is no need for that.

Take $(\lambda,\mu)$ such that at least $\lambda\not\in\sigma(T_1)$ or $\mu\not\in\sigma(T_2)$. Suppose first that $\lambda\not\in\sigma(T_1)$ (the other case is similar). So $T_1-\lambda I$ is invertible. For typing simplicity, from now on I will write $T_1$ instead of $T_1-\lambda I$, and $T_2$ instead of $T_2-\mu I$.

As $T_1$ is invertible, $\ker T_1=\{0\}$, so $$\tag1 \ker T_1\cap \ker T_2=\{0\}. $$ For the second condition, if $T_1x+T_2y=0$, then $x=-T_1^{-1}T_2y=T_2T_1^{-1}y$. Let $z=T_1^{-1}y$. Then $$\tag2 x=-T_2z,\ \ \ y=T_1z, $$ so the second condition is satisfied.

Finally, since $T_1$ is onto, the sum of the images of $T_1$ and $T_2$ is $\mathcal H$. Together with $(1)$ and $(2)$, this shows that if $(\lambda,\mu)\not\in\sigma(T_1)\times\sigma(T_2)$, then $(\lambda,\mu)\not\in\sigma_T(T_1,T_2)$. Thus $$ \sigma_T(T_1,T_2)\subset\sigma(T_1)\times \sigma(T_2). $$

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  • $\begingroup$ If $\mathcal H$ is not $\mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $\mathcal H =\mathbb{C}^2$? Thank you for your help. $\endgroup$ – Student Dec 4 '18 at 18:49
  • $\begingroup$ In the paper of Taylor $X$ is a Banach space. $\endgroup$ – Student Dec 4 '18 at 18:53
  • $\begingroup$ Did you read my answer? In your question, you used $\mathcal H=\mathbb C^2$; in my answer, I'm saying that such a thing is not necessary. $\endgroup$ – Martin Argerami Dec 4 '18 at 18:53
  • $\begingroup$ Now I understand. Yes in my question I use $\mathbb{C}^2$ in order to understand with an example. $\endgroup$ – Student Dec 4 '18 at 18:58
  • $\begingroup$ I don't really know what $\sigma_T(\mathbf T)$ is. I would still expect the same argument to work, though. $\endgroup$ – Martin Argerami Dec 4 '18 at 20:44

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