I'm new to probability and studying multivariate normal distribution. The one thing I don't understand is the linear transformation of multivariate normal distribution.If the $X$ follows Normal distribution where $\mu$ is $0$
$$X \sim \mathcal{N}(0,\Sigma)\propto \exp(-\frac{1}{2}X^T\Sigma^{-1}X)$$ And linear transform it by matrix $A$ and get $Y=AX$.
How do you proof that $Y$ follows a Normal distribution? Thank you in advance.
More precisely, $X$ has pdf $f_X(x):=\frac{1}{\sqrt{\det(2\pi\Sigma)}}\exp-\frac{x^T\Sigma^{-1}x}{2}$. If $Y$ has pdf $f_Y$, the infinitesimal probability $f_X(x)dx=f_Y(y)dy=f_Y(y)|\det A|dx$. (This modulus of a determinant is the coordinate transformation's Jacobian.) Hence $$f_Y(y)=\frac{1}{|\det A|\sqrt{\det(2\pi\Sigma)}}\exp-\frac{y^TA^{-1T}\Sigma^{-1}A^{-1}y}{2}\\=\frac{1}{\sqrt{\det(2\pi A\Sigma A^T)}}\exp-\frac{y^T (A\Sigma A^T)^{-1}y}{2},$$i.e. $Y$ has a multivariate normal distribution of covariance matrix $A\Sigma A^T$.