A formula of $A$-spectral radius for $A^{\frac{1}{2}}$-adjoint operators on semi-Hilbertian spaces

In this paper, we prove the relation $\frac{r_A(T)+r_A(T^{\diamond })+\mathrm{\mid }{r}_A(T^{\diamond })-r_A(T)\mathrm{\mid }}{2}=\sup \{\mathrm{\mid }\lambda \mathrm{\mid }:\lambda \in {\sigma }_A(T)\}$, where $A$ is a positive semidefinite operator (not necessarily to have a closed range) and $r_A(T)$ is the $A$-spectral radius of $T$ in $B_{A^{\frac{1}{2}}}(H)$. Also we prove that $\sup \{\mathrm{\mid }\lambda \mathrm{\mid }:\lambda \in {\sigma }_A(T)\}=r_A(T),\text{ when }T\in B_{A^{\frac{1}{2}}}(H)\text{ commutes with }A$. By introducing $A$-Harte spectrum ${\sigma }_{A_h}(\mathbf{T})$ of a $d$-tuple operator $\mathbf{T}=(T_1,\dots ,T_d)\in (B_{A^{\frac{1}{2}}}(H){)}^d$, we prove that $r_{A_h}(\mathbf{T})\le \sup \{\mathrm{\parallel }\lambda {\mathrm{\parallel }}_2:\lambda \in {\sigma }_{A_h}(\mathbf{T})\}$, where $r_{A_h}(\mathbf{T})$ is the $A$-Harte spectral radius of $\mathbf{T}$.