# Proving Covariance of a linear transformation of the Multivariate Normal Distribution

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.

The covariance of a randvom variable $$Y$$ is defined as $$\mathrm{Cov}(Y) = \mathbb E[(Y - \mathbb E[Y])(Y - \mathbb E[Y])']$$. Now consider $$Y = AX$$: \begin{align*}\mathrm{Cov}(AX) &= \mathbb E[(AX - \mathbb E[AX])(AX - \mathbb E[AX])'] \\ &= \mathbb E\left[\big(A(X - \mathbb E[AX])\big)\big(A(X - \mathbb E[X])\big)'\right] \\ &= \mathbb E\left[A(X - \mathbb E[X])(X - \mathbb E[X])'A'\right] \\ &= A\mathbb E\left[(X - \mathbb E[X])(X - \mathbb E[X])'\right]A' \\ &= A\mathrm{Cov}A',\end{align*} where it was used that $$\mathbb E[AX] = A\mathbb E[X]$$ for non-random $$A$$ and the shoe-socks rule $$(AB)' = B'A'$$ for matrices $$A$$ and $$B$$.