If I have $X\sim$ MVNn$(\mu,\Sigma)$ for positive definitie $\Sigma$ and $A$ is a full rank $n\times n$ matrix, how do I prove Cov$(AX)=A\Sigma A^T$?
I have tried constructing $A\Sigma A^T$ and working backwards to Cov$(AX)$, resulting in the matrix
$A\Sigma A^T=\begin{bmatrix} \sum_{j=1}^n a_{1j}\sum_{i=1}^n a_{1i}\Sigma_{ij} & \dots & \sum_{j=1}^n a_{nj}\sum_{i=1}^n a_{1i}\Sigma_{ij}\\ \vdots & \ddots & \vdots\\ \sum_{j=1}^n a_{1j}\sum_{i=1}^n a_{ni}\Sigma_{ij} & \dots & \sum_{j=1}^n a_{nj}\sum_{i=1}^n a_{ni}\Sigma_{ij} \end{bmatrix}$
Where $\Sigma_{ij}=\mathbb{E}[(X_i-\mu_i)(X_j-\mu_j)]$. However, I have no idea how to mainpulate this into the form Cov$(AX)$. I feel like I'm missing some trick but I cannot find what it is through googling.