Spectral decomposition and quadratic forms 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

The point is that
$$z^\prime L^{-1}LAL^\prime \left(L^\prime\right)^{-1}z = (L^\prime x)^\prime L^{-1}LAL^\prime \left(L^\prime\right)^{-1}(L^\prime x) = x^\prime \left(LAL^\prime\right) x.$$
By definition, $\lambda^{*}$ is an eigenvalue of $A\Sigma$ if and only if there is a nonzero vector $e$ for which $\lambda^{*}e = (A\Sigma)e.$  Left-multiply this equation by $L$ and expand:
$$\lambda^{*}(L e) = L(\lambda^{*}e)= L (A\Sigma e) = L(AL^\prime L e) = (LAL^\prime)(Le).$$
Because the notation implies $L^{-1}$ exists, $Le\ne 0$ and therefore (again by the very definition) $Le$ is an eigenvector of $LAL^\prime$ with eigenvector $\lambda^{*}.$  Thus, the eigenvalues of $A\Sigma$ are those of $LAL^\prime$ (and $L$ maps the eigenvectors of $A\Sigma$ into those of $LAL^\prime$).  The paper's claim is immediate.
