I have a covariance matrix C. I have then formulated an quadratic optimization problem that involves the following matrix in the quadratic form:



> 
> [ C C ]  
> 
>  [ C C ]
> 
> 
> 


However, the quadratic solver complains that this matrix is not positive definite. I can also reformulate the optimization problem so that it uses the following matrix in quadratic form:



> 
> [ C -C ]  
> 
>  [-C  C ]
> 
> 
> 


This matrix is also not positive definite. Now, I know that the problem I am trying to solve might not be possible to set up for quadratic optimization. However, I was wondering if maybe someone encountered a similar setup before, and can give me any hints? Maybe reformulate the problem? Or do some approximation? Make the matrix positive definite somehow? I know, now very clear, but I don't know what else I could add. Thank you!



Block Covariance Matrix - Positive Definite? (Quadratic Optimization)
If $C$ is positive semidefinite, then so is $\begin{bmatrix} C & C\\ C & C\end{bmatrix}$ for the simple reason that it is nothing but the Kronecker product of $\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$ with $C$. 


*EDIT*


For those who don't like Kronecker products, here is an alternative proof using block-matrices:


Since $C>0$, we can write $C^{1/2}$. Then, we can write
\begin{equation\*}
\begin{bmatrix} C & C\\ C & C\end{bmatrix} = \begin{bmatrix} C^{1/2} & 0\\ C^{1/2} & 0\end{bmatrix} \begin{bmatrix} C^{1/2} & C^{1/2}\\ 0 & 0\end{bmatrix},
\end{equation\*}
which is a Gram matrix, hence semidefinite.


