Let $ E$ be a complex Hilbert space. Let $ A=(A_1,\cdots,A_d)\in \mathcal{L}(E)^d$ . Consider \begin{eqnarray*} W_{max}(A) &=&\{\alpha\in \mathbb{C}^d:\;\exists\,(z_n)\subset E\;\;\hbox{such that}\;\|z_n\|=1,\displaystyle\lim_{n\rightarrow+\infty}\langle A_j z_n,z_n\rangle=\alpha_j,\ &&\phantom{++++++++++}\;\hbox{and}\;\displaystyle\lim_{n\rightarrow+\infty}\|A_jz_n\|= \|A_j\|,\;\forall j=1,\cdots,d \}. \end{eqnarray*} If $ 0\notin W_{max}(A)$ , then $ \forall\{z_n\}\in E$ such that $ \|z_n\|=1$ we have, $ $ \langle A_j z_n,z_n\rangle\nrightarrow 0 \quad \mbox{or}\quad\|A_jz_n\|\nrightarrow \|A_j\|,\,\, 1\leq j \leq d.$ $ Let us rotate $ A$ , that is we consider $ {\bf \widehat{A}}=(e^{i\alpha_1}A_1,…,e^{i\alpha_d}A_d)$ . Moreover, if $ \lambda=(\lambda_1,…,\lambda_d)\in W_{max}( \widehat{A})$ , then there exists a sequence $ \{y_n\}$ in $ E$ such that $ $ \|y_n\|=1,\,\,\langle \widehat{A}_j y_n,y_n\rangle\rightarrow \lambda_j \quad \mbox{and}\quad\|\widehat{A}_jy_n\|\rightarrow \|\widehat{A}_j\|,\,\, 1\leq j \leq d.$ $ i.e. $ $ \|y_n\|=1,\,\,e^{-i\alpha_j}\langle A_j y_n,y_n\rangle\rightarrow \lambda_j \quad \mbox{and}\quad\|A_jy_n\|_A\rightarrow \|A_j\|,\,\, 1\leq j \leq d.$ $ Since, $ \displaystyle\lim_{n\rightarrow\infty}\langle A_j y_n,y_n\rangle\neq0$ , then $ \lambda_j\neq0$ for all $ 1\leq j \leq d.$ Assume that there exists $ \tau_j$ such that $ \Re e(\lambda_j)\geq\tau_j>0$ for all $ j$ .

Why for all $ z\in E$ such that $ \|z\|=1$ we have $ \Re e\langle \widehat{A}_j z ,z\rangle\geq\tau_j>0$ ) ?

Notice that $ \pi_k(W_{max}( \widehat{A}))$ is convex, with $ \pi_k$ denotes the canonical projection from $ \mathbb{C}^d$ to $ \mathbb{C}$ .

Thank you for your help.