Question: Suppose $X$ is a random vector in $\mathbb{R}^n$ with mean vector $\mu$ and covariance matrix $\Sigma$, where $\Sigma$ is invertible. Show that the random vector $Z:= \Sigma ^{-1/2} \left( X - \mu \right)$ is a mean zero, istropic random vector.
Some defintion:
A random vector $Y$ is called isotropic if $\Sigma \left( Y\right)=\mathbb{E}YY^T=I_n$, where $I_n$ denotes the indentity matrix in $\mathbb{R}^n.$
$\text{cov}(Y)=\Sigma \left( Y\right)=\Sigma$ is symmetric, positive-semidefinite.
Mean zero: $\mathbb{E}Z = \mathbb{E}\Sigma ^{-1/2} \left( X - \mu \right) = \Sigma ^{-1/2}\mathbb{E}X -\Sigma ^{-1/2} \mu = \Sigma ^{-1/2}\mu-\Sigma ^{-1/2}\mu=0$
I am stuck at showing that $\mathbb{E}ZZ^T=I_n$
My attempt:
$\mathbb{E}ZZ^T$ $=\mathbb{E} \Sigma ^{-1/2} \left( X-\mu \right) \left( X-\mu \right)^T ( \Sigma^{-1/2})^T \\ =\mathbb{E} \Sigma ^{-1/2} ( XX^T-\mu X^T- X\mu^T -\mu \mu^T) (\Sigma^{-1/2} )^T \\ =\Sigma ^{-1/2} ( \mathbb{E}XX^T ) ( \Sigma^{-1/2} )^T -\Sigma^{-1/2} \mu ( \mathbb{E}X )^T ( \Sigma^{-1/2} )^T-\Sigma^{-1/2} ( \mathbb{E}X )\mu ^T ( \Sigma^{-1/2} )^T+\Sigma^{-1/2} \mu \mu^T ( \Sigma^{-1/2} )^T\\ =\Sigma ^{-1/2} \Sigma ( \Sigma^{-1/2} )^T-2\Sigma^{-1/2} \mu \mu^T ( \Sigma^{-1/2} )^T+\Sigma^{-1/2} \mu \mu^T ( \Sigma^{-1/2} )^T \\ =\Sigma ^{-1/2} \Sigma ( \Sigma^{-1/2} )^T-\Sigma^{-1/2} \mu \mu^T ( \Sigma^{-1/2} )^T$
Then I don't know how to further simplify the expression to $I_n$. Do I have to use sepctral decomposition of $\Sigma$? Or is there any easier way to do it? Any hint or help is appreciated.
Note that $\Sigma$ is a deterministic matrix. Therefore,
$$ \mathbb{E} \sqrt{\Sigma^{-1}} (X-\mu)(X-\mu)^T\sqrt{\Sigma^{-1}}^T=\sqrt{\Sigma^{-1}} \mathbb{E} \big[(X-\mu)(X-\mu)^T\big]\sqrt{\Sigma^{-1}}^T=\sqrt{\Sigma^{-1}} \Sigma \sqrt{\Sigma^{-1}}^T $$ since $\operatorname{Cov}(X-\mu)=\operatorname{Cov}(X)$. Now, $\Sigma$ is symmetric, so $$ \sqrt{\Sigma^{-1}} \Sigma \sqrt{\Sigma^{-1}}^T=\sqrt{\Sigma^{-1}} \Sigma \sqrt{\Sigma^{-1}}=I_n $$