# Prove that $\sigma_T(T_1,T_2)\subset\sigma(T_1)\times\sigma(T_2)$

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 $$\sigma_T(T_1,T_2)\subset\sigma(T_1)\times\sigma(T_2)$$

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

we have

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 $$\sigma_T(T_1,T_2)\subset\sigma(T_1)\times \sigma(T_2).$$

• 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. – Student Dec 4 '18 at 18:49
• In the paper of Taylor $X$ is a Banach space. – Student Dec 4 '18 at 18:53
• 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. – Martin Argerami Dec 4 '18 at 18:53
• Now I understand. Yes in my question I use $\mathbb{C}^2$ in order to understand with an example. – Student Dec 4 '18 at 18:58
• I don't really know what $\sigma_T(\mathbf T)$ is. I would still expect the same argument to work, though. – Martin Argerami Dec 4 '18 at 20:44