Let $\mathbf{x}\sim N(0,I)$ and $A$ a real-valued square matrix. The spectral decomposition allows us to rewrite a quadratic form $\mathbf{x}^\top A \mathbf{x}$ as a sum of iid chi-squared random variables weighted by the eigenvalues of the square matrix $A$, $$ \mathbf{x}^\top A \mathbf{x} = \mathbf{x}^\top Q^\top\Lambda Q\mathbf{x} = (Q\mathbf{x})^\top \Lambda Q\mathbf{x} = \sum_{i=1}^n \lambda_i(A)\chi^2_i. $$
Now let $\Sigma$ be a positive-definite matrix which we can decompose as $\Sigma=L^\top L$, $\mathbf{z}\sim N(0,\Sigma)$ and $(L^\top)^{-1}Z=X\sim N(0,I)$. Using the same rationale as above,
$$ \begin{align} \mathbf{z}^\top L^{-1} L A L^\top(L^\top)^{-1}\mathbf{z} &= ((L^\top)^{-1} \mathbf{z})^\top L A L^\top ((L^\top)^{-1}\mathbf{z}) \\ &= \mathbf{x}^\top L A L^\top \mathbf{x} \\ &= \mathbf{x}^\top L Q \Lambda Q L^\top \mathbf{x} \\ \end{align} $$ The paper I'm reading says that this is equal to $$ \sum_{i=1}^n \lambda^*_i(A\Sigma) \chi^2_i $$ such that $\lambda^*_i(A\Sigma)$ is the i-th eigenvalue of $A\Sigma$. I can't seem to find out the in-between steps. I'd appreciate some light on this issue.
References: Related question about quadratic forms and the Chi-squared distribution
