 Solution

A random vector $X$ has multivariate normal distribution $N_d(\mu ,\Sigma )$ if every real linear combination $t^{T}X$ is normally distributed, with mean $t^T\mu$ and variance $t^T\Sigma t$. The covariance matrix is symmetric positive semidefinite; zero variances permit degenerate normals. Its moment-generating function is $M_X(t)=\exp (t^T\mu +t^T\Sigma t/2)$, including the degenerate cases.

For the given $X$, let $t_j\in {\mathbb{R}}^{d_j}$. The joint moment-generating function of the $Y_j=A_{j}X$ isThe last equality uses the assumed orthogonality of the row spaces. Its factors are exactly the individual moment-generating functions, so the hint implies the random vectors $Y_j$ are independent. Also every linear combination of the stacked vector $Y$ is a linear combination of $X$, hence normally distributed. ConsequentlyThis establishes both claims without excluding singular block covariances.

For the sample mean and sample variance result, assume $n\ge 2$ and set $W=(Z_1-\mu ,\dots ,Z_n-\mu {)}^T$. Independence of the observations gives $W\sim N_n(0,{\sigma }^{2}I)$. Let $u=(1,\dots ,1{)}^T$, $A_1=u^T/n$, and $A_2=I-u{u}^T/n$. Then $A_1{A}_2^T=0$. The first transformed vector is the scalar $\overline{Z}-\mu$, and the second is the residual vector $(Z_i-\overline{Z}{)}_i$. They are independent by the just-proved result. Applying the measurable function $v\mapsto \Vert v{\Vert }^2/(n-1)$ to the residual vector preserves independence from the sample mean. Thus