I stumbled on the following text on Wikipedia:
Suppose the random column vectors X, Y live in $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, and the vector $(X, Y)$ in $\mathbb{R}^{n+m}$ has a multivariate normal distribution whose covariance is the symmetric positive-definite matrix source: https://en.wikipedia.org/wiki/Schur_complement
This is probably very basic - but when I visualize a $3$-dimensional vector $X$ ($\mathbb{R}^3$) and a $3$-dimensional vector $Y$ ($\mathbb{R}^3$), then I naively would think that the vector $(X,Y)$ lives in a $3$-dimensional space as well (and not in a $\mathbb{R}^6$ space). what is the thinking error that I am making?
Maybe it's easier if you go to smaller dimensions, where you can still imagine the full space.
Let's first look at the absolute simplest case, $m=n=1$. A vector in $\mathbb R^1$ is, of course, just a real number. So your two vectors in $\mathbb R^1$ are two real numbers $x$ and $y$. Thus your combined vector is $(x,y)$ which clearly describes a point in $\mathbb R^2$.
The point is, in $(x,y)$ the first vector, $x$, lives in another $\mathbb R^1$ (namely, the $x$ axis) than the second vector, $y$, which lives on the $y$-axis. Together, the two one-dimensional spaces ($x$-axis, $y$-axis) span a two-dimensional space (the $x$-$y$-plane).
Similarly, you can consider the case $n=2$, $m=1$. Then $X$ is a point i the plane, and $y$ is a point on the line. Now $(X,y)$ is a point in space whose base point is given by $X$, and whose height is given by $y$. Again, the space $\mathbb R^2$ (the base plane) in which $X$ lives, and the space $\mathbb R^1$ (the vertical axis) in which $y$ lives, lie in different directions, and the dimension of the total space is $n+m=2+1=3$.
The same happens with your two $\mathbb R^3$ vectors: They lie in two different three-dimensional subspaces of $\mathbb R^6$ which lie in completely different directions.