Proving Covariance of a linear transformation of the Multivariate Normal Distribution

Question

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.

Answer 1

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 $B$.