I talked to my advisor about this (twice actually), but I simply do not understand his reasoning unfortunately. Given the following matrix (which is a covariance matrix in the broader context, but I guess that's not important here) $$ \Sigma = (\sigma_{ij}) = \begin{pmatrix} \omega_{11} & 0 & 0 \\ 0 & \omega_{22} & \omega_{23} + \lambda_{23}\omega_{22} \\ 0 & \omega_{23} + \lambda_{23}\omega_{22} & \omega_{33} + 2\omega_{23}\lambda_{23} + \lambda_{23}^2\omega_{22} \end{pmatrix}, $$ I want to show that there is an infinite amount of choices for the entries of $\Lambda$ and / or $\Omega$ such that the (covariance) matrix stays the same, so $|\mathcal{F}(\Lambda, \Omega)| = \infty$, which denotes the fiber of a function that simply maps $(\Lambda, \Omega)$ onto the (covariance) matrix. The entries of $\Lambda$ and $\Omega$ are unknown coefficients. Now he told me something the lines of treating every entry as an equation with $\omega_{22}$ as a free variable but not the other $\omega_{ij}$, so something like
$\begin{equation} \sigma_{22} = \omega_{22} \\ \sigma_{23} = \omega_{23} + \lambda_{23}\omega_{22} \Leftrightarrow \omega_{23} = \sigma_{23} - \lambda_{23}\omega_{22}\\ \sigma_{33} = \omega_{33} + 2\omega_{23}\lambda_{23} + \lambda_{23}^2\omega_{22} \Leftrightarrow \omega_{33} = \sigma_{33} - 2\omega_{23}\lambda_{23} - \lambda_{23}^2\omega_{22} \end{equation}$
I don't understand the point of that. If I treat $\omega_{22}$ as a free variable, wouldn't this always yield a different matrix since $\omega_{22}$ stands alone on the diagonal of $\Sigma$?
I am guessing your advisor meant to say you fix $\omega_{11}=a$ and $\omega_{22}=b$ and then show that the following system has infinitely many solutions: $$ \begin{cases} \omega_{23} + \lambda_{23}b = \sigma_{23} \\ \omega_{33} + 2\omega_{23}\lambda_{23} + b^2\lambda_{23}^2 = \sigma_{33} \end{cases}$$ for given $\sigma_{ij}$ values. In the above two equations, the variables are $\omega_{23}, \lambda_{23}, \omega_{33}$ and so it should have infinitely many solutions because you only have two equations.