Generating correlated random numbers: Why does Cholesky decomposition work?

The co-variance matrix of any random vector $Y$ is given as $\mathbb{E} \left(YY^T \right)$, where $Y$ is a random column vector of size $n \times 1$. Now take a random vector, $X$, consisting of uncorrelated random variables with each random variable, $X_i$, having zero mean and unit variance $1$. Since $X_i$'s are uncorrelated random variables with zero mean and unit variance, we have $\mathbb{E} \left(X_iX_j\right) = \delta_{ij}$. Hence, $$\mathbb{E} \left( X X^T \right) = I$$ To generate a random vector with a given covariance matrix $Q$, look at the Cholesky decomposition of $Q$ i.e. $Q = LL^T$. Note that it is possible to obtain a Cholesky decomposition of $Q$ since by definition the co-variance matrix $Q$ is symmetric and positive definite.

Now look at the random vector $Z = LX$. We have $$\mathbb{E} \left(ZZ^T\right) = \mathbb{E} \left((LX)(LX)^T \right) = \underbrace{\mathbb{E} \left(LX X^T L^T\right) = L \mathbb{E} \left(XX^T \right) L^T}_{\text{ Since expectation is a linear operator}} = LIL^T = LL^T = Q$$ Hence, the random vector $Z$ has the desired co-variance matrix, $Q$.