Doubt about proof of positive semi-definite matrix implies covariance matrix

Question

I have a doubt about the proof of the fact that a positive semi-definite matrix is a covariance matrix. The professor do the following proof:

Let $\Sigma$ be a positive semi-definite $p \times p$ matrix with range $r \leq p$. Then, exists a $p \times r$ matrix $C$ (with range $r$) such that $$\Sigma = C\cdot C^T$$ Then, let $X_1,\ldots,X_r$ be $r$ independent random variables wiht expected value 0 and covariance matrix $I_r$. So, we have that \begin{align*} \text{Cov}(CXC^T) &= C\cdot \text{Cov}(X)\cdot C^T \\ &= C \cdot I_r \cdot C^T \\ &= C \cdot C^T = \Sigma \end{align*} where $X=(X_1,\ldots,X_r)$.

My question is ¿Do the r.v's $X_1,\ldots,X_r$ need to be independent? I think that the only requirement is that $\text{Cov}(X) = I_r$, but my proffesor told me that they HAVE to be independent. I just don't get it.

Thanks for the help.

Answer 1

If I was doing this, I would explicitly use the spectral theorem to say $$ \Sigma = \tilde C \tilde \Lambda\tilde C^T $$ where $\tilde C \in \mathbb R^{p\times p}$ and $\tilde \Lambda$ is a $p\times p$ diagonal matrix with the last $p-r$ entries being exactly zero. Letting $C$ be the first $r$ columns of $C$, and $\Lambda = \text{diag}(\lambda_1, \dots, \lambda_r)$ we then have $$ \Sigma = C\Lambda C^T $$ as the low rank factorization. We can then take $X \sim \mathcal N(\mathbf 0, \Lambda)$ so $$ \text{Var}(CXC^T) = C\Lambda C^T = \Sigma. $$

In this case the elements of $X$ happen to be independent because $\text{Cov}(X_i,X_j) = 0 \iff X_i\perp X_j$ for a multivariate Gaussian, but independence isn't required, just non-correlation.