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