Suppose that we have two different discreet signal vectors of $N^\text{th}$ dimension, namely $\mathbf{x}[i]$ and $\mathbf{y}[i]$, each one having a total of $M$ set of samples/vectors.
$\mathbf{x}[m] = [x_{m,1} \,\,\,\,\, x_{m,2} \,\,\,\,\, x_{m,3} \,\,\,\,\, ... \,\,\,\,\, x_{m,N}]^\text{T}; \,\,\,\,\,\,\, 1 \leq m \leq M$
$\mathbf{y}[m] = [y_{m,1} \,\,\,\,\, y_{m,2} \,\,\,\,\, y_{m,3} \,\,\,\,\, ... \,\,\,\,\, y_{m,N}]^\text{T}; \,\,\,\,\,\,\,\,\, 1 \leq m \leq M$
And, I build up a covariance matrix in-between these signals.
$\{C\}_{ij} = E\left\{(\mathbf{x}[i] - \bar{\mathbf{x}}[i])^\text{T}(\mathbf{y}[j] - \bar{\mathbf{y}}[j])\right\}; \,\,\,\,\,\,\,\,\,\,\,\, 1 \leq i,j \leq M $
Where, $E\{\}$ is the "expected value" operator.
What is the proof that, for all arbitrary values of $\mathbf{x}$ and $\mathbf{y}$ vector sets, the covariance matrix $C$ is always semi-definite ($C \succeq0$) (i.e.; not negative definte; all of its eigenvalues are non-negative)?