Hermitian rank-one resolvent update and trace-norm bound

By [F1], write $B=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)U^{\ast }$ with ${\lambda }_j\ge 0$. If $c=U^{\ast }x$, then $q:=x^{\ast }Rx={\sum }_j\frac{\mathrm{\mid }{c}_j{\mathrm{\mid }}^2}{{\lambda }_j-z},\mathrm{Im}q=\eta {\sum }_j\frac{\mathrm{\mid }{c}_j{\mathrm{\mid }}^2}{\mathrm{\mid }{\lambda }_j-z{\mathrm{\mid }}^2}=\eta \mathrm{\parallel }Rx{\mathrm{\parallel }}_2^2\mathrm{.}$ For $x\mathrm{\ne }0$ this is positive, so $1+q\mathrm{\ne }0$; for $x=0$ the denominator is one.

Direct multiplication, using step 1.1, gives $(B+x{x}^{\ast }-zI)(R-\frac{Rx{x}^{\ast }R}{1+x^{\ast }Rx})=I\mathrm{.}$ The Hermitian matrix $B+x{x}^{\ast }$ has a resolvent at nonreal $z$, so the displayed inverse is $R_x$.

The difference in step 2.1 is a rank-one matrix. Its trace norm equals the product of the two vector norms divided by $\mathrm{\mid }1+q\mathrm{\mid }$; normality of $R$ makes both vector norms $\mathrm{\parallel }Rx{\mathrm{\parallel }}_2$. Since $\mathrm{\mid }1+q\mathrm{\mid }\ge \mathrm{Im}q=\eta \mathrm{\parallel }Rx{\mathrm{\parallel }}_2^2$, step 1.1 gives $\mathrm{\parallel }{R}_x-R{\mathrm{\parallel }}_1\le 1\mathrm{/}\eta$. The inequality $\mathrm{\mid }\mathrm{tr}M\mathrm{\mid }\le \mathrm{\parallel }M{\mathrm{\parallel }}_1$ then gives the trace bound. The case $x=0$ has both differences zero.