I was just wondering if someone could explain the following series of equalities: The Gaussian integral may be evaluated using an orthogonal transformation $R$ to diagonalise the real symmetric operator $K$ $$\int d\phi' e^{-\frac{1}{2} \phi' K \phi'} = \int d \phi' e^{-\frac{1}{2} \phi' R^TR K R^TR \phi'} = \int \frac{d \phi''}{\text{det R}} e^{-\phi'' RKR^T \phi''} = \prod_{\lambda} \int_{-\infty}^{\infty}d \phi_{\lambda}e^{-\frac{1}{2} \lambda \phi_{\lambda}^2} = \prod_{\lambda} \frac{1}{\sqrt{2 \pi \lambda}} \propto \frac{1}{\sqrt{\text{det K}}}$$
It's the second and third equalities which I don't quite understand. I am not sure where the factor of $1/2$ went in the second equality (perhaps a typo) or why $\phi' R^T = \phi''$ given that $R\phi' = \phi''$ and I think we are expressing $RKR^T = D$, some diagonal matrix but I am not sure how this gives the third equality.
Thanks!
A preliminary important modification: I have added "transpose" operators to some $\phi$s, $\phi'$s $\phi''$s occurring on the left of expressions:
$$\int d\phi' e^{-\frac{1}{2} \phi'^T K \phi'} \overset{(1)}{=} \int d \phi' e^{-\frac{1}{2} \phi'^T R^TR K R^TR \phi'} \overset{(2)}{=} \int \frac{d \phi''}{\text{det R}} e^{-\frac{1}{2}\phi''^T RKR^T \phi''} \overset{(3)}{=} \prod_{\lambda} \int_{-\infty}^{\infty}d \phi_{\lambda}e^{-\frac{1}{2} \lambda \phi_{\lambda}^2} \overset{(4)}{=} \prod_{\lambda} \frac{1}{\sqrt{2 \pi \lambda}} \propto \frac{1}{\sqrt{\text{det K}}} \ \ \ (I)$$
(A number has been given to the different "equal" signs).
with $R$ an orthogonal transform, i.e.,
$$R^TR=RR^T=I \Leftrightarrow R^{-1}=R^T \ \ \ (II)$$
You ask about the change of variables, why $R\phi' = \phi'' \Rightarrow \phi' R^T = \phi''$. Due to the "important modification" mentionned at the beginning, this question can be rewritten in this way: why $R\phi' = \phi'' \Rightarrow \phi'^T R^T = \phi''^T$, and the answer is "simply by the transposition rule of a product".
The third equality is due to the fact that $R$ has been chosen so as to diagonalize matrix $K$, i.e., be such that $RKR^T=diag(\lambda_1,\cdots,\lambda_n)$. (it is always possible to diagonalize a symmetric matrix using an orthogonal matrix).
In this way, the quadratic form can be decomposed (for the sake of definiteness, we restrict the dimension to 2):
$$\begin{bmatrix} x_1 & x_2 \end{bmatrix}\begin{bmatrix} \lambda_1 & 0\\0&\lambda_2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\lambda_1 x_1^2 +\lambda_2 x_2^2$$
Thus:
$$exp(\lambda_1 x_1^2 +\lambda_2 x_2^2)=exp(\lambda_1 x_1^2) \times exp(\lambda_2 x_2^2)$$
In this way, the integral over $\mathbb{R}^2$ is separable into the product of two integrals over $\mathbb{R}$, and in the case of $\mathbb{R}^n$, into the product of $n$ 1D integrals. Each of these integrals is easily transformable into the classical integral:
$$\int_{-\infty}^{+\infty}exp\left(-\frac{1}{2}\frac{x^2}{\sigma^2}\right)dx=\sigma \sqrt{2 \pi}$$
whence the result.
Remark: I find inapropriate in (I) the indexing by $\lambda$; it should be an indexing by a certain $k \ (k=1\cdots n))$.
For examples, see : http://education.illinois.edu/courses/edpsy494/lectures/MoreLinearAlgebra-online.pdf