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Gradient and hessian of $log(x^TAx)$
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I am working on a optimization problem which involves the gradient and hessian of $log(x^TAx)$, where $x$ is an unknown vector and $A$ is a positive definite matrix. How can I derive them? Thanks!
matrix-calculus
asked Jan 24 at 23:11
BayesBayes
277
$endgroup$
add a comment |
$begingroup$
I am working on a optimization problem which involves the gradient and hessian of $log(x^TAx)$, where $x$ is an unknown vector and $A$ is a positive definite matrix. How can I derive them? Thanks!
matrix-calculus
asked Jan 24 at 23:11
BayesBayes
277
$endgroup$
add a comment |
$begingroup$
I am working on a optimization problem which involves the gradient and hessian of $log(x^TAx)$, where $x$ is an unknown vector and $A$ is a positive definite matrix. How can I derive them? Thanks!
matrix-calculus
asked Jan 24 at 23:11
BayesBayes
277
$endgroup$
I am working on a optimization problem which involves the gradient and hessian of $log(x^TAx)$, where $x$ is an unknown vector and $A$ is a positive definite matrix. How can I derive them? Thanks!
matrix-calculus
matrix-calculus
asked Jan 24 at 23:11
BayesBayes
277
asked Jan 24 at 23:11
BayesBayes
277
asked Jan 24 at 23:11
BayesBayes
277
asked Jan 24 at 23:11
BayesBayes
277
asked Jan 24 at 23:11
BayesBayes
277
277
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
First, let's define a quadratic form and calculate its differential.
$$eqalign{
alpha &= x^TAx cr
dalpha &= 2(Ax)^T dx cr
}$$
The objective function is the log of the preceeding, so its differential and gradient are easy to calculate.
$$eqalign{
lambda &= logalpha cr
dlambda
&= alpha^{-1}dalpha cr
&= 2alpha^{-1}(Ax)^T dx cr
g = frac{partiallambda}{partial x} &= 2alpha^{-1}Ax cr
}$$
Now let's calculate the differential of $g$ and thence the hessian.
$$eqalign{
dg
&= 2alpha^{-1}A,dx - 2Ax,,alpha^{-2},dalpha cr
&= 2alpha^{-1}A,dx - 4alpha^{-2}Ax(Ax)^T,dx cr
H = frac{partial g}{partial x}
&= 2alpha^{-1}A - 4alpha^{-2}Ax(Ax)^T cr
&= 2alpha^{-1}A - gg^T cr
}$$
answered Jan 25 at 0:15
greggreg
8,8401824
$endgroup$
$begingroup$
Thank you very much. BTW, A is not necessarily symmetric. Your solution is enough for me to derive the final answer.
$endgroup$
– Bayes
Jan 25 at 0:32
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
First, let's define a quadratic form and calculate its differential.
$$eqalign{
alpha &= x^TAx cr
dalpha &= 2(Ax)^T dx cr
}$$
The objective function is the log of the preceeding, so its differential and gradient are easy to calculate.
$$eqalign{
lambda &= logalpha cr
dlambda
&= alpha^{-1}dalpha cr
&= 2alpha^{-1}(Ax)^T dx cr
g = frac{partiallambda}{partial x} &= 2alpha^{-1}Ax cr
}$$
Now let's calculate the differential of $g$ and thence the hessian.
$$eqalign{
dg
&= 2alpha^{-1}A,dx - 2Ax,,alpha^{-2},dalpha cr
&= 2alpha^{-1}A,dx - 4alpha^{-2}Ax(Ax)^T,dx cr
H = frac{partial g}{partial x}
&= 2alpha^{-1}A - 4alpha^{-2}Ax(Ax)^T cr
&= 2alpha^{-1}A - gg^T cr
}$$
answered Jan 25 at 0:15
greggreg
8,8401824
$endgroup$
$begingroup$
Thank you very much. BTW, A is not necessarily symmetric. Your solution is enough for me to derive the final answer.
$endgroup$
– Bayes
Jan 25 at 0:32
add a comment |
$begingroup$
First, let's define a quadratic form and calculate its differential.
$$eqalign{
alpha &= x^TAx cr
dalpha &= 2(Ax)^T dx cr
}$$
The objective function is the log of the preceeding, so its differential and gradient are easy to calculate.
$$eqalign{
lambda &= logalpha cr
dlambda
&= alpha^{-1}dalpha cr
&= 2alpha^{-1}(Ax)^T dx cr
g = frac{partiallambda}{partial x} &= 2alpha^{-1}Ax cr
}$$
Now let's calculate the differential of $g$ and thence the hessian.
$$eqalign{
dg
&= 2alpha^{-1}A,dx - 2Ax,,alpha^{-2},dalpha cr
&= 2alpha^{-1}A,dx - 4alpha^{-2}Ax(Ax)^T,dx cr
H = frac{partial g}{partial x}
&= 2alpha^{-1}A - 4alpha^{-2}Ax(Ax)^T cr
&= 2alpha^{-1}A - gg^T cr
}$$
answered Jan 25 at 0:15
greggreg
8,8401824
$endgroup$
$begingroup$
Thank you very much. BTW, A is not necessarily symmetric. Your solution is enough for me to derive the final answer.
$endgroup$
– Bayes
Jan 25 at 0:32
add a comment |
$begingroup$
First, let's define a quadratic form and calculate its differential.
$$eqalign{
alpha &= x^TAx cr
dalpha &= 2(Ax)^T dx cr
}$$
The objective function is the log of the preceeding, so its differential and gradient are easy to calculate.
$$eqalign{
lambda &= logalpha cr
dlambda
&= alpha^{-1}dalpha cr
&= 2alpha^{-1}(Ax)^T dx cr
g = frac{partiallambda}{partial x} &= 2alpha^{-1}Ax cr
}$$
Now let's calculate the differential of $g$ and thence the hessian.
$$eqalign{
dg
&= 2alpha^{-1}A,dx - 2Ax,,alpha^{-2},dalpha cr
&= 2alpha^{-1}A,dx - 4alpha^{-2}Ax(Ax)^T,dx cr
H = frac{partial g}{partial x}
&= 2alpha^{-1}A - 4alpha^{-2}Ax(Ax)^T cr
&= 2alpha^{-1}A - gg^T cr
}$$
answered Jan 25 at 0:15
greggreg
8,8401824
$endgroup$
First, let's define a quadratic form and calculate its differential.
$$eqalign{
alpha &= x^TAx cr
dalpha &= 2(Ax)^T dx cr
}$$
The objective function is the log of the preceeding, so its differential and gradient are easy to calculate.
$$eqalign{
lambda &= logalpha cr
dlambda
&= alpha^{-1}dalpha cr
&= 2alpha^{-1}(Ax)^T dx cr
g = frac{partiallambda}{partial x} &= 2alpha^{-1}Ax cr
}$$
Now let's calculate the differential of $g$ and thence the hessian.
$$eqalign{
dg
&= 2alpha^{-1}A,dx - 2Ax,,alpha^{-2},dalpha cr
&= 2alpha^{-1}A,dx - 4alpha^{-2}Ax(Ax)^T,dx cr
H = frac{partial g}{partial x}
&= 2alpha^{-1}A - 4alpha^{-2}Ax(Ax)^T cr
&= 2alpha^{-1}A - gg^T cr
}$$
answered Jan 25 at 0:15
greggreg
8,8401824
edited Jan 25 at 0:21
answered Jan 25 at 0:15
greggreg
8,8401824
answered Jan 25 at 0:15
greggreg
8,8401824
answered Jan 25 at 0:15
greggreg
8,8401824
8,8401824
$begingroup$
Thank you very much. BTW, A is not necessarily symmetric. Your solution is enough for me to derive the final answer.
$endgroup$
– Bayes
Jan 25 at 0:32
add a comment |
$begingroup$
Thank you very much. BTW, A is not necessarily symmetric. Your solution is enough for me to derive the final answer.
$endgroup$
– Bayes
Jan 25 at 0:32
$begingroup$
Thank you very much. BTW, A is not necessarily symmetric. Your solution is enough for me to derive the final answer.
$endgroup$
– Bayes
Jan 25 at 0:32
$begingroup$
Thank you very much. BTW, A is not necessarily symmetric. Your solution is enough for me to derive the final answer.
$endgroup$
– Bayes
Jan 25 at 0:32
add a comment |
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