# Spectrum of a pair of commuting operators

Definition: Let $${\bf A}=(A_1,A_2,\cdots,A_d)\in \mathcal{B}(\mathcal{H})^d$$ be a $$d$$-tuple of commuting operators on a complex Hilbert space $$\mathcal{H}.$$ Let $$\sigma_H({\bf A})$$ denotes the Harte spectrum of $${\bf A}$$. $$(\lambda_1,\lambda_2,\cdots,\lambda_d)\notin \sigma({\bf A})$$ if there exist operators $$U_1,\cdots,U_d,V_1,\cdots,V_d \in \mathcal{B}(\mathcal{H}))$$ such that $$\sum_{1\leq k \leq d}U_k(A_k-\lambda_k I)=I\;\hbox{and}\;\;\sum_{1\leq k \leq d}(A_k-\lambda_k I)V_k =I.$$

Let $$A= \begin{pmatrix}0&1\\1&0\end{pmatrix}$$ and $$I= \begin{pmatrix}1&0\\0&1\end{pmatrix}$$. I don't understant how to prove that $$\sigma_H(I,A)=\{(1,1);(1,-1)\}.$$

If $$(\lambda_1,...,\lambda_n)\in \sigma_H(A_1,..,A_n)$$ then $$\lambda_i\in\sigma(A_i)$$ for all $$i$$. If this is not the case, ie there is a $$\lambda_i$$ with $$A_i-\lambda_i I$$ invertible, then $$U_i=V_i=(A_i-\lambda_i I)^{-1}$$ and the other $$V_j=U_j=0$$ shows that $$(\lambda_1,..,\lambda_n)\notin\sigma_H(A_1,...,A_n)$$.
This means $$\sigma_H(A_1,..,.A_n)\subset \sigma(A_1)\times...\times \sigma(A_n)$$. In our case we've got then that $$\sigma_H(I,A)\subset \{1\}\times \{1,-1\}$$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $$I-1\cdot I$$ is always $$0$$ and the question reduces to showing that $$A\pm I$$ is not invertible.