Krdytkyu

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

Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. 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:

Chose a basis so that $Aequivbegin{pmatrix}1&0\0&-1end{pmatrix}$. Now note that
$$(I-1)oplus (A-1)equiv0oplusbegin{pmatrix}0&0\0&-2end{pmatrix},qquad (I-1)oplus (A+1)equiv0oplusbegin{pmatrix}2&0\0&0end{pmatrix}$$
both fail to be injective maps $Bbb C^2toBbb C^2oplus Bbb C^2$. For that reason you achieve
$${(1,1),(1,-1)}=sigma(I)timessigma(A)subsetsigma_T(I,A).$$
Your question contains the proof that $sigma(a_1)timessigma(a_2)supset sigma_T(a_1,a_2)$ in general. You can generalise this specific example to see $sigma_T(I,a)={1}timessigma(a)$, 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.)