Fourier Transform of $e^{-}$, $A$ is a symmetric, positive definite matrix
$begingroup$
I would like to understand how to fourier transform the function $f:mathbb R^ntomathbb R$,
$f(x):=e^{-lt x,Axgt}$
with $A$ being a positive definite, symmetric matrix.
I understand that $lt x,Axgt$ will end up being something like $sum_{i=1}^nsum_{j=1}^n a_{ij}x_ix_j$ with a couple of summarizable summands due to the symmetry of $A$:
$frac 1{{(2pi)}^{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,xgt}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.
integration matrices complex-analysis fourier-transform positive-definite
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
$endgroup$
add a comment |
$begingroup$
I would like to understand how to fourier transform the function $f:mathbb R^ntomathbb R$,
$f(x):=e^{-lt x,Axgt}$
with $A$ being a positive definite, symmetric matrix.
I understand that $lt x,Axgt$ will end up being something like $sum_{i=1}^nsum_{j=1}^n a_{ij}x_ix_j$ with a couple of summarizable summands due to the symmetry of $A$:
$frac 1{{(2pi)}^{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,xgt}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.
integration matrices complex-analysis fourier-transform positive-definite
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
$endgroup$
$begingroup$
Let $h(x) = e^{-pi |x|^2}$. (With the $xi$ normalization) it is its own Fourier transform. Iff $A$ is positive definite then $A =pi B^top B$ with $B in GL_n(mathbb{R})$ then $e^{-langle x,Axrangle } = h(B x)$ whose Fourier transform is $|det(B)|^{-1} h(B^{-top} xi)$ (change of variable $y = Bx, dy= d(Bx)=det(B) dx$ in the Fourier integral)
$endgroup$
– reuns
Jan 19 at 1:43
$begingroup$
(I will look into this further in the morning, too late in the night for me right now, think this will help though, thanks already)
$endgroup$
– DepressedUnicorn
Jan 19 at 2:09
add a comment |
$begingroup$
I would like to understand how to fourier transform the function $f:mathbb R^ntomathbb R$,
$f(x):=e^{-lt x,Axgt}$
with $A$ being a positive definite, symmetric matrix.
I understand that $lt x,Axgt$ will end up being something like $sum_{i=1}^nsum_{j=1}^n a_{ij}x_ix_j$ with a couple of summarizable summands due to the symmetry of $A$:
$frac 1{{(2pi)}^{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,xgt}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.
integration matrices complex-analysis fourier-transform positive-definite
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
$endgroup$
I would like to understand how to fourier transform the function $f:mathbb R^ntomathbb R$,
$f(x):=e^{-lt x,Axgt}$
with $A$ being a positive definite, symmetric matrix.
I understand that $lt x,Axgt$ will end up being something like $sum_{i=1}^nsum_{j=1}^n a_{ij}x_ix_j$ with a couple of summarizable summands due to the symmetry of $A$:
$frac 1{{(2pi)}^{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,xgt}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.
integration matrices complex-analysis fourier-transform positive-definite
integration matrices complex-analysis fourier-transform positive-definite
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
edited Jan 19 at 6:36
Andrews
4081317
edited Jan 19 at 6:36
Andrews
4081317
edited Jan 19 at 6:36
Andrews
4081317
4081317
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
asked Jan 19 at 1:40
DepressedUnicornDepressedUnicorn
83
83
$begingroup$
Let $h(x) = e^{-pi |x|^2}$. (With the $xi$ normalization) it is its own Fourier transform. Iff $A$ is positive definite then $A =pi B^top B$ with $B in GL_n(mathbb{R})$ then $e^{-langle x,Axrangle } = h(B x)$ whose Fourier transform is $|det(B)|^{-1} h(B^{-top} xi)$ (change of variable $y = Bx, dy= d(Bx)=det(B) dx$ in the Fourier integral)
$endgroup$
– reuns
Jan 19 at 1:43
$begingroup$
(I will look into this further in the morning, too late in the night for me right now, think this will help though, thanks already)
$endgroup$
– DepressedUnicorn
Jan 19 at 2:09
add a comment |
$begingroup$
Let $h(x) = e^{-pi |x|^2}$. (With the $xi$ normalization) it is its own Fourier transform. Iff $A$ is positive definite then $A =pi B^top B$ with $B in GL_n(mathbb{R})$ then $e^{-langle x,Axrangle } = h(B x)$ whose Fourier transform is $|det(B)|^{-1} h(B^{-top} xi)$ (change of variable $y = Bx, dy= d(Bx)=det(B) dx$ in the Fourier integral)
$endgroup$
– reuns
Jan 19 at 1:43
$begingroup$
(I will look into this further in the morning, too late in the night for me right now, think this will help though, thanks already)
$endgroup$
– DepressedUnicorn
Jan 19 at 2:09
$begingroup$
Let $h(x) = e^{-pi |x|^2}$. (With the $xi$ normalization) it is its own Fourier transform. Iff $A$ is positive definite then $A =pi B^top B$ with $B in GL_n(mathbb{R})$ then $e^{-langle x,Axrangle } = h(B x)$ whose Fourier transform is $|det(B)|^{-1} h(B^{-top} xi)$ (change of variable $y = Bx, dy= d(Bx)=det(B) dx$ in the Fourier integral)
$endgroup$
– reuns
Jan 19 at 1:43
$begingroup$
Let $h(x) = e^{-pi |x|^2}$. (With the $xi$ normalization) it is its own Fourier transform. Iff $A$ is positive definite then $A =pi B^top B$ with $B in GL_n(mathbb{R})$ then $e^{-langle x,Axrangle } = h(B x)$ whose Fourier transform is $|det(B)|^{-1} h(B^{-top} xi)$ (change of variable $y = Bx, dy= d(Bx)=det(B) dx$ in the Fourier integral)
$endgroup$
– reuns
Jan 19 at 1:43
$begingroup$
(I will look into this further in the morning, too late in the night for me right now, think this will help though, thanks already)
$endgroup$
– DepressedUnicorn
Jan 19 at 2:09
$begingroup$
(I will look into this further in the morning, too late in the night for me right now, think this will help though, thanks already)
$endgroup$
– DepressedUnicorn
Jan 19 at 2:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$A=U^TLambda U$$
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 $mathbb R^n$, $langle t, xrangle = langle Ut, Ux rangle$.
With that, the Fourier transform of $f$ now becomes
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt Ux, Lambda Ux gt}e^{-i lt Ut,Uxgt}d^nx\
end{split}$$
By changing the variable $y=Ux$, we get
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt Ut,ygt}|det U|d^nx\
end{split}$$
Noting that $|det U|=1$ and $U^TU=I$
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt t,ygt}d^ny$$
If $lambda_1,...,lambda_n$ denote the diagonal elements of $Lambda $ (all strictly positive),
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} prod_{k=1}^ne^{-lambda_ky_k^2}e^{-i t_ky_k}d^ny=prod_{k=1}^n frac 1 {2pi}int_{mathbb R}e^{-lambda_ky_k^2}e^{-i t_ky_k}dy_k$$
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.
edited Jan 19 at 6:27
reuns
20.2k21148
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
$endgroup$
$begingroup$
Thanks, this makes a lot of sense now.
$endgroup$
– DepressedUnicorn
Jan 19 at 13:39
add a comment |
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1 Answer
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$begingroup$
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
$$A=U^TLambda U$$
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 $mathbb R^n$, $langle t, xrangle = langle Ut, Ux rangle$.
With that, the Fourier transform of $f$ now becomes
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt Ux, Lambda Ux gt}e^{-i lt Ut,Uxgt}d^nx\
end{split}$$
By changing the variable $y=Ux$, we get
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt Ut,ygt}|det U|d^nx\
end{split}$$
Noting that $|det U|=1$ and $U^TU=I$
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt t,ygt}d^ny$$
If $lambda_1,...,lambda_n$ denote the diagonal elements of $Lambda $ (all strictly positive),
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} prod_{k=1}^ne^{-lambda_ky_k^2}e^{-i t_ky_k}d^ny=prod_{k=1}^n frac 1 {2pi}int_{mathbb R}e^{-lambda_ky_k^2}e^{-i t_ky_k}dy_k$$
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.
edited Jan 19 at 6:27
reuns
20.2k21148
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
$endgroup$
$begingroup$
Thanks, this makes a lot of sense now.
$endgroup$
– DepressedUnicorn
Jan 19 at 13:39
add a comment |
$begingroup$
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
$$A=U^TLambda U$$
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 $mathbb R^n$, $langle t, xrangle = langle Ut, Ux rangle$.
With that, the Fourier transform of $f$ now becomes
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt Ux, Lambda Ux gt}e^{-i lt Ut,Uxgt}d^nx\
end{split}$$
By changing the variable $y=Ux$, we get
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt Ut,ygt}|det U|d^nx\
end{split}$$
Noting that $|det U|=1$ and $U^TU=I$
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt t,ygt}d^ny$$
If $lambda_1,...,lambda_n$ denote the diagonal elements of $Lambda $ (all strictly positive),
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} prod_{k=1}^ne^{-lambda_ky_k^2}e^{-i t_ky_k}d^ny=prod_{k=1}^n frac 1 {2pi}int_{mathbb R}e^{-lambda_ky_k^2}e^{-i t_ky_k}dy_k$$
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.
edited Jan 19 at 6:27
reuns
20.2k21148
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
$endgroup$
$begingroup$
Thanks, this makes a lot of sense now.
$endgroup$
– DepressedUnicorn
Jan 19 at 13:39
add a comment |
$begingroup$
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
$$A=U^TLambda U$$
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 $mathbb R^n$, $langle t, xrangle = langle Ut, Ux rangle$.
With that, the Fourier transform of $f$ now becomes
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt Ux, Lambda Ux gt}e^{-i lt Ut,Uxgt}d^nx\
end{split}$$
By changing the variable $y=Ux$, we get
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt Ut,ygt}|det U|d^nx\
end{split}$$
Noting that $|det U|=1$ and $U^TU=I$
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt t,ygt}d^ny$$
If $lambda_1,...,lambda_n$ denote the diagonal elements of $Lambda $ (all strictly positive),
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} prod_{k=1}^ne^{-lambda_ky_k^2}e^{-i t_ky_k}d^ny=prod_{k=1}^n frac 1 {2pi}int_{mathbb R}e^{-lambda_ky_k^2}e^{-i t_ky_k}dy_k$$
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.
edited Jan 19 at 6:27
reuns
20.2k21148
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
$endgroup$
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
$$A=U^TLambda U$$
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 $mathbb R^n$, $langle t, xrangle = langle Ut, Ux rangle$.
With that, the Fourier transform of $f$ now becomes
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt Ux, Lambda Ux gt}e^{-i lt Ut,Uxgt}d^nx\
end{split}$$
By changing the variable $y=Ux$, we get
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt Ut,ygt}|det U|d^nx\
end{split}$$
Noting that $|det U|=1$ and $U^TU=I$
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt t,ygt}d^ny$$
If $lambda_1,...,lambda_n$ denote the diagonal elements of $Lambda $ (all strictly positive),
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} prod_{k=1}^ne^{-lambda_ky_k^2}e^{-i t_ky_k}d^ny=prod_{k=1}^n frac 1 {2pi}int_{mathbb R}e^{-lambda_ky_k^2}e^{-i t_ky_k}dy_k$$
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.
edited Jan 19 at 6:27
reuns
20.2k21148
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
edited Jan 19 at 6:27
reuns
20.2k21148
edited Jan 19 at 6:27
reuns
20.2k21148
edited Jan 19 at 6:27
reuns
20.2k21148
20.2k21148
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
answered Jan 19 at 4:15
Stefan LafonStefan Lafon
1,86618
1,86618
$begingroup$
Thanks, this makes a lot of sense now.
$endgroup$
– DepressedUnicorn
Jan 19 at 13:39
add a comment |
$begingroup$
Thanks, this makes a lot of sense now.
$endgroup$
– DepressedUnicorn
Jan 19 at 13:39
$begingroup$
Thanks, this makes a lot of sense now.
$endgroup$
– DepressedUnicorn
Jan 19 at 13:39
$begingroup$
Thanks, this makes a lot of sense now.
$endgroup$
– DepressedUnicorn
Jan 19 at 13:39
add a comment |
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$begingroup$
Let $h(x) = e^{-pi |x|^2}$. (With the $xi$ normalization) it is its own Fourier transform. Iff $A$ is positive definite then $A =pi B^top B$ with $B in GL_n(mathbb{R})$ then $e^{-langle x,Axrangle } = h(B x)$ whose Fourier transform is $|det(B)|^{-1} h(B^{-top} xi)$ (change of variable $y = Bx, dy= d(Bx)=det(B) dx$ in the Fourier integral)
$endgroup$
– reuns
Jan 19 at 1:43
$begingroup$
(I will look into this further in the morning, too late in the night for me right now, think this will help though, thanks already)
$endgroup$
– DepressedUnicorn
Jan 19 at 2:09