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According to the following paper of Taylor: J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.

we have

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Let $A= \begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $I= \begin{pmatrix}1&0\\0&1\end{pmatrix}$. By this answer, the Harte spectrum of $(I,A)$ is equal to $$\sigma_H(I,A)=\{(1,1);(1,-1)\}.$$ I want to understand why the taylor spectrum of $(I,A)$ which is denoted $\sigma_T(I,A)$ is equal also to $\{(1,1), (1,-1)\}$.

Proof: Let $R_X(\lambda) = (X-\lambda)^{-1}$ be the resolvent of $X$. You have the identities \begin{gather*} \begin{bmatrix} I-\lambda & A-\mu \end{bmatrix} \begin{bmatrix} R_I(\lambda) \\ 0 \end{bmatrix} = I, \\ \begin{bmatrix} R_I(\lambda) \\ 0 \end{bmatrix} \begin{bmatrix} I-\lambda & A-\mu \end{bmatrix} + \begin{bmatrix} -(A-\mu) \\ I-\lambda \end{bmatrix} \begin{bmatrix} 0 & R_I(\lambda) \end{bmatrix} = \begin{bmatrix} I & 0 \\ 0 & I \end{bmatrix} , \\ \begin{bmatrix} 0 & R_I(\lambda) \end{bmatrix} \begin{bmatrix} -(A-\mu) \\ I-\lambda \end{bmatrix} = I , \end{gather*} whenever the resolvent $R_I(\lambda)$ exists. These identities (in homological algebra, they are known as a contracting homotopy for this complex) imply that, whenever $\lambda$ is not in the spectrum of $I$ (namely, when $R_I(\lambda)$ exists), Taylor's Koszul complex is exact and hence the corresponding value of $(\lambda,\mu)$ does not belong to $\sigma_T(I,A)$. We can write similar formulas, but using $R_A(\mu)$ instead. Hence, we have reduced the calculation to $\sigma_T(I,A) \subseteq \{ (1,\mathbb{C}) \} \cap \{ (\mathbb{C},1), (\mathbb{C},-1) \} = \{ (1,1), (1,-1) \}$. Now it's just a matter of checking that for these values of $(\lambda,\mu)$ the Koszul complex really does fail to be exact, which is easy to see from the known common eigenvectors of $I$ and $A$.


For more details about the taylor spectrum in a more general context we have:

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Chose a basis so that $A\equiv\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Now note that $$(I-1)\oplus (A-1)\equiv0\oplus\begin{pmatrix}0&0\\0&-2\end{pmatrix},\qquad (I-1)\oplus (A+1)\equiv0\oplus\begin{pmatrix}2&0\\0&0\end{pmatrix}$$ both fail to be injective maps $\Bbb C^2\to\Bbb C^2\oplus \Bbb C^2$. For that reason you achieve $$\{(1,1),(1,-1)\}=\sigma(I)\times\sigma(A)\subset\sigma_T(I,A).$$ Your question contains the proof that $\sigma(a_1)\times\sigma(a_2)\supset \sigma_T(a_1,a_2)$ in general. You can generalise this specific example to see $\sigma_T(I,a)=\{1\}\times\sigma(a)$, provided that $a$ is bounded. (Here you need to use the theorem that if $a$ is a bijective linear map between Banach spaces it is an isomorphism.)

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  • $\begingroup$ Thank you very much for your answer. However, I don't see why $\sigma(a_1)\times\sigma(a_2)\supset \sigma_T(a_1,a_2)$ in general? $\endgroup$ – Student Dec 4 '18 at 5:54
  • $\begingroup$ I don't really have time to think about the case of a $d$-tuple, but the answer is probably yes. As for the case with $2$ operators it is enough to show that if $a_1,a_2$ are invertible then the sequence is non-singular. At this point injectivity at the beginning and surjectivity at the end are clear, remains to prove exactness in the middle. Let $(x_1,x_2)$ be so that $a_1x_1+a_2x_2=0$, since $a_1$ and $a_2$ are invertible you've got $y_1,y_2\in H$ with $-a_2y_1 = x_1$ and $a_1y_2=x_2$. Plug that in and you find $a_1a_2(-y_1+y_2)=0$, which is only possible if $y_1=y_2$ ... $\endgroup$ – s.harp Dec 4 '18 at 21:08
  • $\begingroup$ ... since $a_1a_2$ is invertible. With that you've found that if $(x_1,x_2)$ are in the kernel of the differential, then $(x_1,x_2)=(-a_2 y_1, a_1 y_1)$, giving exactness in the middle also. $\endgroup$ – s.harp Dec 4 '18 at 21:09
  • $\begingroup$ Thank you. For the benifits of the reader, it will be very good if you write your comments here: math.stackexchange.com/questions/3025414/… $\endgroup$ – Student Dec 5 '18 at 13:07

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