Covariance matrix multiplied by its transpose

Question

Suppose we have a covariance matrix $A$ (it is automatically symmetric). So, $A^TA$ is also symmetric. However, do the elements of the matrix $A^TA$ keep the same pair-values as that of the matrix $A$?

Thank you in advance.

Answer 1

In the two by two case (similarly for higher dimensions), a covariance matrix is represented by:

$$A=\pmatrix{\sigma_1^2 & \sigma_{12} \\ \sigma_{12} & \sigma_2^2}$$ so that

$$A^T A = A^2 = \pmatrix{\sigma_1^2 & \sigma_{12} \\ \sigma_{12} & \sigma_2^2}\pmatrix{\sigma_1^2 & \sigma_{12} \\ \sigma_{12} & \sigma_2^2}=\pmatrix{\sigma_1^4+\sigma_{12}^2 & \sigma_{12}(\sigma_1^2 + \sigma_2^2) \\ \sigma_{12}(\sigma_1^2 + \sigma_2^2)& \sigma_2^4+ \sigma_{12}^2}.$$

We see that each term is a combination (sum of products) of elements of $A$.

Not sure exactly what you are asking, but each element of $A^TA=A^2$ is not exactly a function only of the corresponding element of $A$.