Given two real zero mean random variables $\mathbf{x}$ and $\mathbf{y}$, the linear LMS of $\mathbf{x}$ given $\mathbf{y}$ such that $$\hat{\mathbf{x}}=\mathbf{K_o}\mathbf{y}$$
is given by
$$\mathbf{K_o}=\mathbf{R_{xy}}\mathbf{R_y^{-1}}$$
I'm trying to prove this but I get stuck in the final step. We want to find the matrix $\mathbf{K_o}$ such that the mean-square error is minimized.
$$\begin{align}\mathbb{E}[(\mathbf{x}-\hat{\mathbf{x}})(\mathbf{x}-\hat{\mathbf{x}})^T] &= \mathbb{E}[(\mathbf{x}-\mathbf{K_o}\mathbf{y})(\mathbf{x}-\mathbf{K_o}\mathbf{y})^T] \\&= \mathbb{E}[\mathbf{x}\mathbf{x}^T]-\mathbb{E}[\mathbf{K_o}\mathbf{y}\mathbf{x}^T]-\mathbb{E}[\mathbf{x}(\mathbf{K_o}\mathbf{y})^T]+\mathbb{E}[(\mathbf{K_oy})(\mathbf{K_oy})^T] \\&= \mathbf{R_{x}}-\mathbf{K_o}\mathbf{R_{yx}} - \mathbf{R_{xy}}\mathbf{K_o}^T + \mathbf{K_o}\mathbf{R_{y}}\mathbf{K_o}^T \end{align} $$
If these values were all scalars, then it is easy to see that the minimum would be achieved (i.e. the derivative would equal zero) in $k_o = r_{xy}/r_y$, as stated at the beginning.
But how can one continue the sketch of proof above when all variables are matrices?
I think what you want to minimize is $\mathbb{E}\left[\text{Tr}\left((\mathbf{x}-\mathbf{\hat x)(\mathbf{x}-\mathbf{\hat x})}^T\right) \right]$, otherwise what does minimize mean for a matrix?
If you agree, you can do the following $$ f(\boldsymbol{K_0}) = \text{Tr}\left(\boldsymbol{R_{x}}-\boldsymbol{K_0}\boldsymbol{R_{yx}}-\boldsymbol{R_{xy}}\boldsymbol{K_0}^T+\boldsymbol{K_0}\boldsymbol{R_{y}} \boldsymbol{K_0}^T\right) $$ Find the derivative with respect to $\boldsymbol{K_0}$ and set it to zero, yielding $$ \boldsymbol{0}-\boldsymbol{R_{yx}}^T-\boldsymbol{R_{xy}}+2\boldsymbol{K_0}\boldsymbol{R_y}=0 $$ yielding, since $\boldsymbol{R_{yx}^T}=\boldsymbol{R_{xy}}$ $$ \boldsymbol{K_0} = \boldsymbol{R_{xy}}\boldsymbol{R_y}^{-1} $$which is the minimum since the second derivative is $\boldsymbol{R_y}\geq0$.