Independence and uncorrelatedness between two normal random vectors. If $X$ and $Y$ are normal random vectors in $\mathbb R^n$ and in $\mathbb R^m$, and they are jointly normally distributed i.e. $(X,Y)$ is normally distributed in $\mathbb R^{n+m}$, then are the following equivalent


*

*$\operatorname{Cov}(X,Y)=0$;

*$X$ and $Y$ are independent.


Note that it is true when $n=m=1$. Thanks.

$\newcommand{\E}{\operatorname{E}}\newcommand{\var}{\operatorname{var}}\newcommand{\cov}{\operatorname{cov}}$
In comments you say you have a theorem that if two multivariate normal distributions have the same mean and the same variance, then they are the same distribution.
You have
$$
\E\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \E X \\ \E Y \end{bmatrix}
$$
and
\begin{align}
\var \begin{bmatrix} X \\ Y \end{bmatrix} & = \E\left( \left(\begin{bmatrix} X \\  Y \end{bmatrix} -\E\begin{bmatrix} X \\ Y \end{bmatrix} \right) \left(\begin{bmatrix} X \\  Y \end{bmatrix} -\E\begin{bmatrix} X \\ Y \end{bmatrix} \right)^T \right) \\[10pt]
& = \begin{bmatrix} \var X & \cov(X,Y) \\ \cov(X,Y)^T & \var(Y) \end{bmatrix}.
\end{align}
The two off-diagonal matrices are $0$, by hypothesis.
Now consider another normal distribution of $(n+m)\times 1$ column vectors: The first $n$ components are distributed exactly as $X$ is distributed and the last $m$ components as $Y$, and they are independent.  That multivariate normal distribution has the same mean (in $\mathbb R^{n+m}$) and the same variance (in $\mathbb R^{(n+m)\times(n+m)}$) as the distribution of $X$ and $Y$.  Now apply the theorem mentioned in the first paragraph above.
