Trouble solving for the Jacobian update formula in Broyden's method
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I'm having trouble understanding how to update this formula (it's Broyden's method in multiple dimensions) by solving the following equations:
$$A^{(m)}(x^{(m)}-x^{(m-1)}) = f(x^{(m)})-f(x^{(m-1)})$$ and
$$A^{(m)}v=A^{(m-1)}v,$$ where $A^{(m)}$ is a matrix, $x^{(m)}$ is the $m^{th}$ iteration, and for all vectors v orthogonal to $(x^{(m)}-x^{(m-1)})$.
These two equations should lead to $$A^{(m)}=$$ $$A^{(m-1)} + frac{f(x^{(m)})-f(x^{(m-1)})-A^{(m-1)}(x^{(m)}-x^{(m-1)})}{(x^{(m)}-x^{(m-1)})^T(x^{(m)}-x^{(m-1)})}(x^{(m)}-x^{(m-1)})^T$$.
How does one find the last equation?
numerical-methods numerical-linear-algebra
asked Dec 13 '18 at 14:56
platypus17platypus17
296
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding how to update this formula (it's Broyden's method in multiple dimensions) by solving the following equations:
$$A^{(m)}(x^{(m)}-x^{(m-1)}) = f(x^{(m)})-f(x^{(m-1)})$$ and
$$A^{(m)}v=A^{(m-1)}v,$$ where $A^{(m)}$ is a matrix, $x^{(m)}$ is the $m^{th}$ iteration, and for all vectors v orthogonal to $(x^{(m)}-x^{(m-1)})$.
These two equations should lead to $$A^{(m)}=$$ $$A^{(m-1)} + frac{f(x^{(m)})-f(x^{(m-1)})-A^{(m-1)}(x^{(m)}-x^{(m-1)})}{(x^{(m)}-x^{(m-1)})^T(x^{(m)}-x^{(m-1)})}(x^{(m)}-x^{(m-1)})^T$$.
How does one find the last equation?
numerical-methods numerical-linear-algebra
asked Dec 13 '18 at 14:56
platypus17platypus17
296
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding how to update this formula (it's Broyden's method in multiple dimensions) by solving the following equations:
$$A^{(m)}(x^{(m)}-x^{(m-1)}) = f(x^{(m)})-f(x^{(m-1)})$$ and
$$A^{(m)}v=A^{(m-1)}v,$$ where $A^{(m)}$ is a matrix, $x^{(m)}$ is the $m^{th}$ iteration, and for all vectors v orthogonal to $(x^{(m)}-x^{(m-1)})$.
These two equations should lead to $$A^{(m)}=$$ $$A^{(m-1)} + frac{f(x^{(m)})-f(x^{(m-1)})-A^{(m-1)}(x^{(m)}-x^{(m-1)})}{(x^{(m)}-x^{(m-1)})^T(x^{(m)}-x^{(m-1)})}(x^{(m)}-x^{(m-1)})^T$$.
How does one find the last equation?
numerical-methods numerical-linear-algebra
asked Dec 13 '18 at 14:56
platypus17platypus17
296
$endgroup$
I'm having trouble understanding how to update this formula (it's Broyden's method in multiple dimensions) by solving the following equations:
$$A^{(m)}(x^{(m)}-x^{(m-1)}) = f(x^{(m)})-f(x^{(m-1)})$$ and
$$A^{(m)}v=A^{(m-1)}v,$$ where $A^{(m)}$ is a matrix, $x^{(m)}$ is the $m^{th}$ iteration, and for all vectors v orthogonal to $(x^{(m)}-x^{(m-1)})$.
These two equations should lead to $$A^{(m)}=$$ $$A^{(m-1)} + frac{f(x^{(m)})-f(x^{(m-1)})-A^{(m-1)}(x^{(m)}-x^{(m-1)})}{(x^{(m)}-x^{(m-1)})^T(x^{(m)}-x^{(m-1)})}(x^{(m)}-x^{(m-1)})^T$$.
How does one find the last equation?
numerical-methods numerical-linear-algebra
numerical-methods numerical-linear-algebra
asked Dec 13 '18 at 14:56
platypus17platypus17
296
asked Dec 13 '18 at 14:56
platypus17platypus17
296
edited Dec 13 '18 at 15:46
platypus17
asked Dec 13 '18 at 14:56
platypus17platypus17
296
asked Dec 13 '18 at 14:56
platypus17platypus17
296
asked Dec 13 '18 at 14:56
platypus17platypus17
296
296
add a comment |
add a comment |
1 Answer
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$begingroup$
You want $A_{+1}s=d$ while $A_{+1}v=Av$ for $vperp s$. The second condition means that $A_{+1}$ is a rank-one update of $A$,
$$
A_{+1}=A+ws^top.
$$
Now insert this into the first condition
$$
d=A_{+1}s=As+ws^top simplies w = frac{d-As}{s^top s}.
$$
The orthogonality comes from the minimization in the Frobenius norm, $A_{+1}$ is there the closest matrix to $A$ such that the first condition holds. As Lagrange function this gives (using the notation $A:B={rm trace}(AB^T)$)
begin{align}
L(X,w)&=frac12|X-A|_F^2-w^top(Xs-d)
\
&=frac12(X-A):(X-A)-(ws^top):X+w^top d
\
&=frac12|X-A-ws^top|^2-(ws^top):A-frac12(w^top s)_F^2+w^top d
end{align}
with the stationarity condition
$$
0=(X-A)-ws^top
$$
which implies the invariance on the space orthogonal to $s$.
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You want $A_{+1}s=d$ while $A_{+1}v=Av$ for $vperp s$. The second condition means that $A_{+1}$ is a rank-one update of $A$,
$$
A_{+1}=A+ws^top.
$$
Now insert this into the first condition
$$
d=A_{+1}s=As+ws^top simplies w = frac{d-As}{s^top s}.
$$
The orthogonality comes from the minimization in the Frobenius norm, $A_{+1}$ is there the closest matrix to $A$ such that the first condition holds. As Lagrange function this gives (using the notation $A:B={rm trace}(AB^T)$)
begin{align}
L(X,w)&=frac12|X-A|_F^2-w^top(Xs-d)
\
&=frac12(X-A):(X-A)-(ws^top):X+w^top d
\
&=frac12|X-A-ws^top|^2-(ws^top):A-frac12(w^top s)_F^2+w^top d
end{align}
with the stationarity condition
$$
0=(X-A)-ws^top
$$
which implies the invariance on the space orthogonal to $s$.
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
$endgroup$
add a comment |
$begingroup$
You want $A_{+1}s=d$ while $A_{+1}v=Av$ for $vperp s$. The second condition means that $A_{+1}$ is a rank-one update of $A$,
$$
A_{+1}=A+ws^top.
$$
Now insert this into the first condition
$$
d=A_{+1}s=As+ws^top simplies w = frac{d-As}{s^top s}.
$$
The orthogonality comes from the minimization in the Frobenius norm, $A_{+1}$ is there the closest matrix to $A$ such that the first condition holds. As Lagrange function this gives (using the notation $A:B={rm trace}(AB^T)$)
begin{align}
L(X,w)&=frac12|X-A|_F^2-w^top(Xs-d)
\
&=frac12(X-A):(X-A)-(ws^top):X+w^top d
\
&=frac12|X-A-ws^top|^2-(ws^top):A-frac12(w^top s)_F^2+w^top d
end{align}
with the stationarity condition
$$
0=(X-A)-ws^top
$$
which implies the invariance on the space orthogonal to $s$.
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
$endgroup$
add a comment |
$begingroup$
You want $A_{+1}s=d$ while $A_{+1}v=Av$ for $vperp s$. The second condition means that $A_{+1}$ is a rank-one update of $A$,
$$
A_{+1}=A+ws^top.
$$
Now insert this into the first condition
$$
d=A_{+1}s=As+ws^top simplies w = frac{d-As}{s^top s}.
$$
The orthogonality comes from the minimization in the Frobenius norm, $A_{+1}$ is there the closest matrix to $A$ such that the first condition holds. As Lagrange function this gives (using the notation $A:B={rm trace}(AB^T)$)
begin{align}
L(X,w)&=frac12|X-A|_F^2-w^top(Xs-d)
\
&=frac12(X-A):(X-A)-(ws^top):X+w^top d
\
&=frac12|X-A-ws^top|^2-(ws^top):A-frac12(w^top s)_F^2+w^top d
end{align}
with the stationarity condition
$$
0=(X-A)-ws^top
$$
which implies the invariance on the space orthogonal to $s$.
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
$endgroup$
You want $A_{+1}s=d$ while $A_{+1}v=Av$ for $vperp s$. The second condition means that $A_{+1}$ is a rank-one update of $A$,
$$
A_{+1}=A+ws^top.
$$
Now insert this into the first condition
$$
d=A_{+1}s=As+ws^top simplies w = frac{d-As}{s^top s}.
$$
The orthogonality comes from the minimization in the Frobenius norm, $A_{+1}$ is there the closest matrix to $A$ such that the first condition holds. As Lagrange function this gives (using the notation $A:B={rm trace}(AB^T)$)
begin{align}
L(X,w)&=frac12|X-A|_F^2-w^top(Xs-d)
\
&=frac12(X-A):(X-A)-(ws^top):X+w^top d
\
&=frac12|X-A-ws^top|^2-(ws^top):A-frac12(w^top s)_F^2+w^top d
end{align}
with the stationarity condition
$$
0=(X-A)-ws^top
$$
which implies the invariance on the space orthogonal to $s$.
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
edited Dec 13 '18 at 22:47
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
answered Dec 13 '18 at 22:34
LutzLLutzL
59.1k42056
59.1k42056
add a comment |
add a comment |
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