If $X \sim \mathcal{N}(\mu, \Sigma)$ is a multivariate normal, then the quadratic $X^TAX$ has moment generating function
$$M_{X^TAX}(t)= \frac{1}{\sqrt{\det(I - 2tA\Sigma)}}\exp\left(-\frac{1}{2}\mu'[I-(I-2tA\Sigma)^{-1}]\Sigma^{-1}\mu\right), \tag{1}$$
where $A$ symmetric and $\Sigma$ positive definite.
From the answer here, a nice way of showing this is looking at the Eigendecomposition. That is,
$$X^TAX = \sum_i \lambda_iX_i^2, \quad\quad X_i \sim \mathcal{N}(\mu_i, \sigma_i^2)~~\text{iid}.$$
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
$$M_{Y_i}(t) = \frac{1}{\sqrt{1 - 2t\lambda_i\sigma_i^2}}\exp\left(\frac{\mu_i^2t\lambda_i}{1 - 2t\lambda_i\sigma_i^2}\right).$$
Hence, we arrive at the equation stated in the link, more-or-less:
$$ \begin{align} M_{X^TAX}(t) &= \prod_{i} \frac{1}{\sqrt{1 - 2t\lambda_i\sigma_i^2}}\exp\left(\frac{\mu_i^2t}{1 - 2t\lambda_i\sigma_i^2}\right) \\ &= \frac{1}{\sqrt{\det(I - 2tA\Sigma)}}\exp\left(\sum_i \frac{\mu_i^2t\lambda_i}{1 - 2t\lambda_i\sigma_i^2}\right). \tag{2} \end{align} $$
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
$$\exp\left(\sum_i \frac{\mu_i^2t\lambda_i}{1 - 2t\lambda_i\sigma_i^2}\right) = \cdots = \exp\left(-\frac{1}{2}\mu'[I-(I-2tA\Sigma)^{-1}]\Sigma^{-1}\mu\right).$$
I'm sure I'm missing some obvious linear algebra fact about the matrix decomposition here?
