Rank-one resolvent update and trace bound for a positive semidefinite matrix

By [F1], write $B=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)U^{\mathsf{T}}$ with ${\lambda }_j\ge 0$. Thus $R$ exists, is normal, and $x^{\mathsf{T}}Rx={\sum }_j\frac{(U^{\mathsf{T}}x{)}_j^2}{{\lambda }_j-z},\mathrm{Im}(x^{\mathsf{T}}Rx)=\eta {\sum }_j\frac{(U^{\mathsf{T}}x{)}_j^2}{({\lambda }_j-u{)}^2+{\eta }^2}\mathrm{.}$ If $x\mathrm{\ne }0$, this imaginary part is strictly positive, so $1+x^{\mathsf{T}}Rx\mathrm{\ne }0$; if $x=0$, the denominator is $1$.

Direct multiplication verifies $(B+x{x}^{\mathsf{T}}-zI)(R-\frac{Rx{x}^{\mathsf{T}}R}{1+x^{\mathsf{T}}Rx})=I\mathrm{.}$ Since $B+x{x}^{\mathsf{T}}$ is also real symmetric and $z\mathrm{\notin }\mathbb{R}$, its resolvent exists; hence the displayed expression is $R_x$.

Taking traces in step 2.1 gives $\mathrm{tr}(R_x-R)=-\frac{x^{\mathsf{T}}{R}^{2}x}{1+x^{\mathsf{T}}Rx}\mathrm{.}$ In the eigenbasis of [F1], $\mathrm{\mid }{x}^{\mathsf{T}}{R}^{2}x\mathrm{\mid }\le {\sum }_j\frac{(U^{\mathsf{T}}x{)}_j^2}{({\lambda }_j-u{)}^2+{\eta }^2}=\frac{\mathrm{Im}(x^{\mathsf{T}}Rx)}{\eta }\mathrm{.}$ Also $\mathrm{\mid }1+x^{\mathsf{T}}Rx\mathrm{\mid }\ge \mathrm{Im}(x^{\mathsf{T}}Rx)$ when $x\mathrm{\ne }0$; for $x=0$ the trace difference is zero. Combining these inequalities proves the stated $1\mathrm{/}\eta$ bound.