Laplacian of $ nabla^2 f( A x)$ in terms of Laplacian of $ nabla^2_y f( y)$
$begingroup$
How to find the Laplacian
begin{align}
nabla^2_x f( A x)
end{align}
where $n times m$ matrix and we know know the laplacian of
begin{align}
nabla^2_y f( y)
end{align}
This question is about chain rule for the Laplacian.
I search but could not find a formula for it.
I think that the answer is
begin{align}
nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
end{align}
matrix-calculus
asked Dec 13 '18 at 23:23
LisaLisa
643313
$endgroup$
add a comment |
$begingroup$
How to find the Laplacian
begin{align}
nabla^2_x f( A x)
end{align}
where $n times m$ matrix and we know know the laplacian of
begin{align}
nabla^2_y f( y)
end{align}
This question is about chain rule for the Laplacian.
I search but could not find a formula for it.
I think that the answer is
begin{align}
nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
end{align}
matrix-calculus
asked Dec 13 '18 at 23:23
LisaLisa
643313
$endgroup$
add a comment |
$begingroup$
How to find the Laplacian
begin{align}
nabla^2_x f( A x)
end{align}
where $n times m$ matrix and we know know the laplacian of
begin{align}
nabla^2_y f( y)
end{align}
This question is about chain rule for the Laplacian.
I search but could not find a formula for it.
I think that the answer is
begin{align}
nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
end{align}
matrix-calculus
asked Dec 13 '18 at 23:23
LisaLisa
643313
$endgroup$
How to find the Laplacian
begin{align}
nabla^2_x f( A x)
end{align}
where $n times m$ matrix and we know know the laplacian of
begin{align}
nabla^2_y f( y)
end{align}
This question is about chain rule for the Laplacian.
I search but could not find a formula for it.
I think that the answer is
begin{align}
nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
end{align}
matrix-calculus
matrix-calculus
asked Dec 13 '18 at 23:23
LisaLisa
643313
asked Dec 13 '18 at 23:23
LisaLisa
643313
edited Dec 14 '18 at 0:14
Lisa
asked Dec 13 '18 at 23:23
LisaLisa
643313
asked Dec 13 '18 at 23:23
LisaLisa
643313
asked Dec 13 '18 at 23:23
LisaLisa
643313
643313
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The gradient of the function $f(y)$ is the vector $g$ which satisfies
$$df = g^Tdy$$
and the hessian is the matrix $H$ which satisfies
$$dg = H,dy$$
and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.
Suppose a new independent vector $x$ is introduced with $,y=Ax$.
It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.
To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
$$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
Working these out is straightforward
$$eqalign{
df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
g_* &= A^Tg crcr
dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
H_* &= A^THA cr
L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
}$$
The last expression is the laplacian in terms of the new variable.
Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.
answered Dec 14 '18 at 2:45
greggreg
7,6201821
$endgroup$
$begingroup$
Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
$endgroup$
– Lisa
Dec 18 '18 at 15:32
$begingroup$
@Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
$endgroup$
– greg
Dec 18 '18 at 17:03
$begingroup$
I see now. Thank you.
$endgroup$
– Lisa
Dec 18 '18 at 17:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The gradient of the function $f(y)$ is the vector $g$ which satisfies
$$df = g^Tdy$$
and the hessian is the matrix $H$ which satisfies
$$dg = H,dy$$
and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.
Suppose a new independent vector $x$ is introduced with $,y=Ax$.
It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.
To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
$$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
Working these out is straightforward
$$eqalign{
df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
g_* &= A^Tg crcr
dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
H_* &= A^THA cr
L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
}$$
The last expression is the laplacian in terms of the new variable.
Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.
answered Dec 14 '18 at 2:45
greggreg
7,6201821
$endgroup$
$begingroup$
Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
$endgroup$
– Lisa
Dec 18 '18 at 15:32
$begingroup$
@Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
$endgroup$
– greg
Dec 18 '18 at 17:03
$begingroup$
I see now. Thank you.
$endgroup$
– Lisa
Dec 18 '18 at 17:41
add a comment |
$begingroup$
The gradient of the function $f(y)$ is the vector $g$ which satisfies
$$df = g^Tdy$$
and the hessian is the matrix $H$ which satisfies
$$dg = H,dy$$
and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.
Suppose a new independent vector $x$ is introduced with $,y=Ax$.
It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.
To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
$$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
Working these out is straightforward
$$eqalign{
df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
g_* &= A^Tg crcr
dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
H_* &= A^THA cr
L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
}$$
The last expression is the laplacian in terms of the new variable.
Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.
answered Dec 14 '18 at 2:45
greggreg
7,6201821
$endgroup$
$begingroup$
Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
$endgroup$
– Lisa
Dec 18 '18 at 15:32
$begingroup$
@Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
$endgroup$
– greg
Dec 18 '18 at 17:03
$begingroup$
I see now. Thank you.
$endgroup$
– Lisa
Dec 18 '18 at 17:41
add a comment |
$begingroup$
The gradient of the function $f(y)$ is the vector $g$ which satisfies
$$df = g^Tdy$$
and the hessian is the matrix $H$ which satisfies
$$dg = H,dy$$
and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.
Suppose a new independent vector $x$ is introduced with $,y=Ax$.
It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.
To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
$$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
Working these out is straightforward
$$eqalign{
df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
g_* &= A^Tg crcr
dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
H_* &= A^THA cr
L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
}$$
The last expression is the laplacian in terms of the new variable.
Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.
answered Dec 14 '18 at 2:45
greggreg
7,6201821
$endgroup$
The gradient of the function $f(y)$ is the vector $g$ which satisfies
$$df = g^Tdy$$
and the hessian is the matrix $H$ which satisfies
$$dg = H,dy$$
and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.
Suppose a new independent vector $x$ is introduced with $,y=Ax$.
It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.
To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
$$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
Working these out is straightforward
$$eqalign{
df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
g_* &= A^Tg crcr
dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
H_* &= A^THA cr
L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
}$$
The last expression is the laplacian in terms of the new variable.
Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.
answered Dec 14 '18 at 2:45
greggreg
7,6201821
edited Dec 14 '18 at 5:31
answered Dec 14 '18 at 2:45
greggreg
7,6201821
answered Dec 14 '18 at 2:45
greggreg
7,6201821
answered Dec 14 '18 at 2:45
greggreg
7,6201821
7,6201821
$begingroup$
Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
$endgroup$
– Lisa
Dec 18 '18 at 15:32
$begingroup$
@Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
$endgroup$
– greg
Dec 18 '18 at 17:03
$begingroup$
I see now. Thank you.
$endgroup$
– Lisa
Dec 18 '18 at 17:41
add a comment |
$begingroup$
Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
$endgroup$
– Lisa
Dec 18 '18 at 15:32
$begingroup$
@Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
$endgroup$
– greg
Dec 18 '18 at 17:03
$begingroup$
I see now. Thank you.
$endgroup$
– Lisa
Dec 18 '18 at 17:41
$begingroup$
Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
$endgroup$
– Lisa
Dec 18 '18 at 15:32
$begingroup$
Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
$endgroup$
– Lisa
Dec 18 '18 at 15:32
$begingroup$
@Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
$endgroup$
– greg
Dec 18 '18 at 17:03
$begingroup$
@Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
$endgroup$
– greg
Dec 18 '18 at 17:03
$begingroup$
I see now. Thank you.
$endgroup$
– Lisa
Dec 18 '18 at 17:41
$begingroup$
I see now. Thank you.
$endgroup$
– Lisa
Dec 18 '18 at 17:41
add a comment |
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