I would like to prove such a matrix as a positive definite one,


$$
(\omega^T\Sigma\omega) \Sigma - \Sigma\omega \omega^T\Sigma
$$
where $\Sigma$ is a positive definite symetric covariance matrix while $\omega$ is weight column vector (without constraints of positive elements)


I would apply an arbitrary $x$ belonging to $R^n$ to the following formula, 
$$
x^T((\omega^T\Sigma\omega) \Sigma - \Sigma\omega \omega^T\Sigma)x > 0
$$


But how could I go further to prove such a inequality above?


Thanks,



Proof of A Positive Definite Covariance Matrix
To elaborate a bit on Mahdi's comment, recall that a positive definite matrix $\Sigma$ can be used to define a scalar product, i.e. $\langle a,b \rangle := a^\top\Sigma\, b$, and $\langle a,a\rangle = ||a||^2$. 


You can multiply out the left side of the inequality you end up with scalar products and norms. That's when you apply [Cauchy-Schwarz](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality). 


