Question about Multivariate normal distribution

If $X_1$, … , $X_n$ are iid standard normals, does it follow that $(Xbar, X_1 - Xbar, … , X_n - Xbar)$ are MVN, where Xbar is the mean of the $X_i$? I don’t think so, since taking a linear combination of 0 times the first component and 1 times the other components gives 0. However, I read the following in Blitzstein and Hwang:

Write a vector $X=[X_1,...,X_n]'$ and a second vector $\bar{X}=[Xbar,X_1-Xbar,...,X_n-Xbar]'$

Then note that there is a matrix

$M=\begin{pmatrix} 1/n & 1/n & \cdots & 1/n \\ 1-1/n & -1/n & \cdots & -1/n \\ -1/n & 1-1/n & \cdots & -1/n \\ \vdots & \vdots & \ddots & \vdots \\ -1/n & -1/n & \cdots & 1-1/n \end{pmatrix}$

such that $\bar{X}=MX$. Then if $X \sim N(\mu,\Sigma)$, $\bar{X} \sim N(M\mu,M'\Sigma M)$

This is a standard result that linear combinations of normals remain normal.

It is true that $M'\Sigma M$ is not of full rank -- that's because $\bar{X}$ has more elements than $X$. If you dropped the last element of $\bar{X}$, then it would be of full rank again. But none of this changes the multivariate normality.