Suppose $h=(h_1,\ldots,h_n)$, $\sqrt{h}=(\sqrt{h_1},\ldots,\sqrt{h_n})$, $H=\text{diag}(h)$, $a\otimes a=aa'$.
What is the easiest way to show the following for operator norm $\|.\|$ and largest eigenvalue $\lambda_\text{max}$?
$$\lambda_{max} \left(H+0.5(h\otimes h)H^{-1}\right)=\left\|H+0.5\sqrt{h}\otimes\sqrt{h}\right\|$$
Motivation: reciprocal of this gives the largest learning which guarantees convergence of LMS when observations come from zero-centered Gaussian with diagonal covariance matrix and entries $(h_1,\ldots,h_n)$
Note that $H + 0.5(h \otimes h)H^{-1}$ is similar to $$ H^{-1/2}(H + 0.5(h \otimes h)H^{-1})H^{1/2} = \\ H^{-1/2}HH^{1/2} + 0.5H^{-1/2}h h' H^{-1}H^{1/2} = \\ H + 0.5H^{-1/2}h h' H^{-1/2}=\\ H + 0.5(H^{-1/2}h)(H^{-1/2}h)' =\\ H + 0.5\sqrt{h} \otimes \sqrt{h}. $$ Thus, we have $\lambda_{\max}(H + 0.5(h \otimes h)H^{-1}) = \lambda_{\max}(H + 0.5\sqrt{h} \otimes \sqrt{h})$. However, $H + 0.5\sqrt{h} \otimes \sqrt{h}$ is symmetric, which means that $\lambda_{\max}(H + 0.5\sqrt{h} \otimes \sqrt{h}) = \|H + 0.5\sqrt{h} \otimes \sqrt{h}\|$.