Multivariate Normal Quadratic MGF: Eigendecomposition to Matrix form

If $X sim mathcal{N}(mu, Sigma)$ is a multivariate normal, then the quadratic $X^TAX$ has moment generating function

$$M_{X^TAX}(t)= frac{1}{sqrt{det(I – 2tASigma)}}expleft(-frac{1}{2}mu'[I-(I-2tASigma)^{-1}]Sigma^{-1}muright), tag{1}$$

where $A$ symmetric and $Sigma$ positive definite.

From the answer here, a nice way of showing this is looking at the Eigendecomposition. That is,

$$X^TAX = sum_i lambda_iX_i^2, quadquad X_i sim mathcal{N}(mu_i, sigma_i^2)~~text{iid}.$$

Then, letting $Y_i := mu_i + sigma_iZ_i$, where $Z_i sim mathcal{N}(0, 1)$, we get that for $t < 1/2$ the m.g.f. for any $i$ is

$$M_{Y_i}(t) = frac{1}{sqrt{1 – 2tlambda_isigma_i^2}}expleft(frac{mu_i^2tlambda_i}{1 – 2tlambda_isigma_i^2}right).$$

Hence, we arrive at the equation stated in the link, more-or-less:

$$
begin{align}
M_{X^TAX}(t) &= prod_{i} frac{1}{sqrt{1 – 2tlambda_isigma_i^2}}expleft(frac{mu_i^2t}{1 – 2tlambda_isigma_i^2}right) \
&= frac{1}{sqrt{det(I – 2tASigma)}}expleft(sum_i frac{mu_i^2tlambda_i}{1 – 2tlambda_isigma_i^2}right). tag{2}
end{align}
$$

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My question: I can’t seem to work out how to go from the exponential in $(2)$ back to the exponential in $(1)$, i.e. how do we get

$$expleft(sum_i frac{mu_i^2tlambda_i}{1 – 2tlambda_isigma_i^2}right) = cdots = expleft(-frac{1}{2}mu'[I-(I-2tASigma)^{-1}]Sigma^{-1}muright).$$

I’m sure I’m missing some obvious linear algebra fact about the matrix decomposition here?