For a course I'm following we need to work with multivariate guassians. In this case there are four variables $x_1$ through $x_4$ with the specified covariance matrix. Even though the matrix is already given, I wish to understand it more. The number at location $(1,1)$ is 0.14 and this means the internal variance of $x_1$ right? Does the $-0.3$ next to it mean the internal variance of $x_1$ and $x_2$ together? What do the dividing lines mean?
Thanks!
$-0.3$ is the covariance between $x_1$ and $x_2$. It is $$ \operatorname{cov} (x_1,x_2) = \operatorname{E}((x_1-\operatorname{E}(x_1))(x_2 - \operatorname{E}(x_2))). $$
If $\vec x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ then the matrix $\overset{\sim} \Sigma$ is $$ \overset{\sim} \Sigma = \operatorname{E}(\ \underbrace{(\vec x - \operatorname{E}(\vec x))}_{4\times 1}\ \ \underbrace{(\vec x - \operatorname{E}(\vec x))^\top}_{1\times 4}\ ). $$
Some authors (e.g. Feller) call this non-negative-definite matrix the "variance" of the random vector $\vec x$, since it is the natural generalization to higher dimensions of the one-dimensional variance. But many call it the "covariance" because it is the matrix whose entries are the covariances between the components of $\vec x$.
If you know the spectral theorem of linear algebra then you know that a non-negative-definite symmetric matrix whose entries are real can be written as a diagonal matrix with respect to a basis found by rotating the standard basis. That means one can simply rotate the standard basis to get one with respect to which the components are uncorrelated. The diagonal entries are then the variances of those rotated components.