Multiple correlation formula Given $v \sim N_{p+1}(\mu, \Sigma), v = (x, y)', \mu = (\mu_x, \mu_y)', \Sigma = \begin{pmatrix}
\Sigma_{xx} & \sigma_{xy} \\
\sigma_{yx} & \sigma_{yy} 
\end{pmatrix}$, how do you prove the formula for multiple correlation (the Pearson correlation between $\mu_{y|x}$ and $y$): $$\rho(y,x) = Corr(y, E[y|x]) =  \left( \frac{\sigma_{yx}\Sigma_{xx}^{-1}\sigma_{xy}}{\sigma_{yy}} \right)^{1/2}$$
I tried to use $\mu_{y|x} = \mu_y - \sigma_{yx}\Sigma_{xx}^{-1}(x-\mu_x)$ and plug it into $Corr(y, E[y|x]) = \frac{ Cov(y, E[y|x])}{\sqrt{V(y)}\sqrt{V(E[y|x])}}$ without any success.

The initial approach can be used as well:
$$\rho(x, y) = \frac{\text{Cov}(\mu_{y|x}, y)}{\sqrt{V(\mu_{y|x}) V(y)}}.$$
Then, $$V(y) = \sigma_{yy},$$
\begin{align*}
V(\mu_{y|x}) &= V(\mu_y + \sigma_{yx}\Sigma_{xx}^{-1}(x-\mu_x)) \\
&= V(\sigma_{yx}\Sigma_{xx}^{-1}x) && \small{ | \mu_y \text{ and } \sigma_{yx}\Sigma_{xx}^{-1}\mu_x \text{ are constants} }\\
&= \sigma_{yx}\Sigma_{xx}^{-1}V(x)(\sigma_{yx}\Sigma_{xx}^{-1})'  && \small{ | \sigma_{xy} = \sigma_{yx}' \text{ and } \Sigma_{xx}^{-1} =  (\Sigma_{xx}^{-1})' } \\
&= \sigma_{yx}\Sigma_{xx}^{-1}\Sigma_{xx}\Sigma_{xx}^{-1}\sigma_{xy} \\
&= \sigma_{yx}\Sigma_{xx}^{-1}\sigma_{xy}
\end{align*} and
\begin{align*}
\text{Cov}(\mu_{y|x}, y) &= \text{Cov}(\mu_{y} + \sigma_{yx}\Sigma_{xx}^{-1}(x-\mu_x), y) \\
&= \text{Cov}(\sigma_{yx}\Sigma_{xx}^{-1}x, y) \\
&= \sigma_{yx}\Sigma_{xx}^{-1} \text{Cov}(x, y) \\
&= \sigma_{yx}\Sigma_{xx}^{-1} \sigma_{xy}
\end{align*}
Here, we used $\text{Cov}(ax + b, cy) = a\text{Cov}(x, y)c'$ and $V(x) = \text{Cov}(x,x)$.
We can plug these results back into the formula and simplify:
\begin{align}
\rho(x, y) &= \frac{\text{Cov}(\mu_{y|x}, y)}{\sqrt{V(\mu_{y|x}) V(y)}} \\
&= \frac{\sigma_{yx}\Sigma_{xx}^{-1} \sigma_{xy}}{\sqrt{\sigma_{yx}\Sigma_{xx}^{-1}\sigma_{xy} \sigma_{yy}}} \\
&= \sqrt{\frac{\sigma_{yx}\Sigma_{xx}^{-1} \sigma_{xy}}{\sigma_{yy}}}
\end{align}
