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
$$\sigma_H(I,A)=\{(1,1);(1,-1)\}.$$
Best Answer
First a remark:
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.