Fourier Transform of $e^{-}$, $A$ is a symmetric, positive definite matrix I would like to understand how to fourier transform the function $f:\mathbb R^n\to\mathbb R$,
$f(x):=e^{-\lt x,Ax\gt}$
with $A$ being a positive definite, symmetric matrix.
I understand that $\lt x,Ax\gt$ will end up being something like $\sum_{i=1}^n\sum_{j=1}^n a_{ij}x_ix_j$ with a couple of summarizable summands due to the symmetry of $A$:
$\frac 1{{(2\pi)}^{\frac 1 n}}\int_{\mathbb R^n} e^{-(a_{11}x_1^2+a_{22}x_2^2+...+a_{nn}x_n^2+2a_{12}x_1x_2+...+2a_{23}x_2x_3+...+2a_{n,n-1}x_nx_{n-1})}e^{-i \lt t,x\gt}d^nx $
(Is that even correct?..)
But I am having trouble actually calculating the integral, as I can't see anything I already know I could substitute in this, not even for $n=1$.
I'd appreciate any help with this.
 A: You can simplify the Fourier transform expression if you express everything in the basis of eigenvectors of $A$.
Remember that, by the spectral theorem, $A$ can be diagonalized in an orthonormal basis. That means there is a unitary transform $U$ and a diagonal matrix $\Lambda$ such that
$$A=U^T\Lambda U$$
And because $A$ is positive definite, $\Lambda$'s diagonal elements are all strictly positive. Also, because $U$ is unitary, for any $t$ and $x$ in $\mathbb R^n$, $\langle t, x\rangle = \langle Ut, Ux \rangle$.
With that, the Fourier transform of $f$ now becomes
$$\begin{split}
\hat{f}(t) &=
\frac 1{(2\pi)^n}\int_{\mathbb R^n} e^{-\lt Ux, \Lambda Ux \gt}e^{-i \lt Ut,Ux\gt}d^nx\\
\end{split}$$
By changing the variable $y=Ux$, we get
$$\begin{split}
\hat{f}(t) &=
\frac 1{(2\pi)^n}\int_{\mathbb R^n} e^{-\lt y, \Lambda y \gt}e^{-i \lt Ut,y\gt}|\det U|d^nx\\
\end{split}$$
Noting that $|\det U|=1$ and $U^TU=I$
$$\hat{f}(U^Tt)=\frac 1{(2\pi)^n}\int_{\mathbb R^n} e^{-\lt y, \Lambda y \gt}e^{-i \lt t,y\gt}d^ny$$
If $\lambda_1,...,\lambda_n$ denote the diagonal elements of $\Lambda $ (all strictly positive),
$$\hat{f}(U^Tt)=\frac 1{(2\pi)^n}\int_{\mathbb R^n} \prod_{k=1}^ne^{-\lambda_ky_k^2}e^{-i t_ky_k}d^ny=\prod_{k=1}^n \frac 1 {2\pi}\int_{\mathbb R}e^{-\lambda_ky_k^2}e^{-i t_ky_k}dy_k$$
In other words, if you rotate the axes to align with the eigenvectors of $A$, the $n$-dimensional Fourier transform is the product of $1$-dimensional Fourier transforms of the univariate Gaussian function. And you know the expression for the Fourier transform of a $1$-dimensional Gaussian function.
