Obtain an expression for the variance covariance matrix of the fitted values $\hat{Y}_{i}$ in terms of the hat matrix. Scalar trouble

Question

As the question above asks. I am attempting to obtain the variance covariance matrix of the fitted values $\hat{Y}_{i}$ in terms of the hat matrix. I essentially have completed it, but there is one step I don't understand (or maybe remember the property). Also to mention I'm doing this in matrix notation.

Recall: $\hat{Y} = \mathbf{HY}$

Applying properties of derivations for a varaince-covariance matrix:

$$\Sigma_{\hat{Y}\hat{Y}} = \mathbf{H} \mathbf{\Sigma_{YY}} \mathbf{H}^{t}$$ Where $\mathbf{H}^{t}$ is the transpose.

A solution for the problem proceeds as follows:

$$= \mathbf{H}\sigma^{2}\mathbf{I}\mathbf{H}^{t} \\ = \mathbf{H}\sigma^{2}\mathbf{I}\mathbf{H}\ \text{since H is symmetric} \\ = \sigma^{2}\mathbf{HH} \\ = \sigma^{2}\mathbf{H}$$

My only issue with any of this is treating $\sigma^{2}$ as a scalar. For me to be able to write $\mathbf{\Sigma_{YY}} = \sigma^{2}\mathbf{I}$, I have to know that there are no off diagonal terms with covariance not equal to $0$. How do I know this here? I must be missing something in regards to the theory which would allow me to express the variance-covariance matrix in this way.

Answer 1

Here is this deduction using perhaps more instructive notation,

$$cov(\hat Y)=cov(X\hat\beta)=X \cdot cov(\beta)X^T=\sigma^2 X(X^TX)^{-1}X=\sigma^2 H$$

Where the key step is,

$$cov(\beta)=\sigma^2 (X^TX)^{-1}$$

Which is true under homoskedasticity. Essentially we are saying that,

$$[Cov(\hat Y)]_{ij} = \sigma^2 \cdot h_{ij}$$

Where $h_{ij}=[X(X^TX)^{-1}X^T]_{ij}=I$.