Problem Statement
Given $\mu, \lambda \in\mathbb{R}$ the identity matrix $I$ and a vector $x$ whose unnormalized version is $\hat{x} = x/\|x\|$ we are told that the square root of $$ A = \frac{1}{\|x\|^2}\left[\mu^2 I + (\lambda^2-\mu^2)\hat{x}\hat{x}\top\right] $$ is simply $$ A^{1/2} = \frac{1}{\|x\|}\left[\mu I + (\lambda -\mu)\hat{x}\hat{x}^\top\right]. $$ Given a matrix $S=L^\top L$ we are now asked to find (if it exists in closed form) the matrix square root of $$ B= \frac{1}{x^\top Sx}\left[\mu^2 S + (\lambda^2-\mu^2)\frac{S x x^\top S^\top}{x^\top S x}\right]. $$
Questions
- What is $B^{1/2}$?
- Is there a general technique to matrix square roots when they are available in closed form?
Attempt
First, I write $B$ explicitly using $L$ $$ B = \frac{1}{x^\top L^\top Lx}\left[\mu^2 L^\top L + (\lambda^2-\mu^2)\frac{L^\top L x x^\top L^\top L}{x^\top L^\top L x}\right] $$ Then I introduce the following notation to simplify thing $y = Lx$ so that we have $$ B = \frac{1}{\|y\|^2}\left[\mu^2 L^\top L + (\lambda^2-\mu^2)L^\top \hat{y} \hat{y}^\top L\right] $$ But then I am not sure what to do.
AHHH perhaps $$ B^{1/2} = \frac{1}{\|y\|}\left[\mu L + (\lambda-\mu)\hat{y} \hat{y}^\top L\right] $$
Let's check: $$ \begin{align} (B^{1/2})^\top B^{1/2} &= \frac{1}{\|y\|^2}\left[\mu L + (\lambda-\mu)\hat{y} \hat{y}^\top L\right]^\top \left[\mu L + (\lambda-\mu)\hat{y} \hat{y}^\top L\right] \\ &= \frac{1}{\|y\|^2}\left[\mu^2 L^\top L + 2\mu (\lambda-\mu) L^\top\hat{y}\hat{y}^\top L + (\lambda-\mu)^2 L^\top \hat{y}\hat{y}^\top L\right] \\ &= \frac{1}{\|y\|^2}\left[\mu^2 L^\top L + (\lambda^2-\mu^2)L^\top \hat{y}\hat{y}^\top L\right] \\ &= \frac{1}{x^\top S x}\left[\mu^2 S + (\lambda^2-\mu^2)\frac{Sxx^\top S^\top}{x^\top S x}\right] \\ &= B \end{align} $$ This took me a while. But is there a method to do this? For me it was just lots of trial and error.