Does matrix multiplication of a random vector preserve independence?

Question

Let $\boldsymbol{X},\boldsymbol{Y} \in \mathbb{R}^n$ be random vectors and let $\boldsymbol{A} \in \mathbb{R}^{m \times n}$ be a non-square matrix of constants i.e. m < n. Suppose that each element of $\boldsymbol{X}$ is known to be independent of each element of the vector $\boldsymbol{Y}$. Are the elements of the vector $\boldsymbol{X}^* = \boldsymbol{A}\boldsymbol{X}$ also independent of the elements of $\boldsymbol{Y}$?

Answer 1

Since $\mathbf{A}$ is constant, then $\mathbf{X}^*=\mathbf{A X}$ is a function of $\mathbf{X}$. So this follows from the general result that functions of independent random variables are independent. So, the answer is yes, without further conditions on the matrix $\mathbf{A}$ (such as rank restrictions).

As to your further information in comment: The only important fact about the matrix $\mathbf{A}$ is that it is non-random. Square or rectangular plays no role at all.