How to calculate the $frac{partial det(mathbf X)}{partial mathbf X}$ and $frac{partial tr(mathbf...
$begingroup$
How to calculate the $frac{partial det(mathbf X)}{partial mathbf X}$ and $frac{partial tr(mathbf X^n)}{partial mathbf X}$ by using Frobenius product?i tried to begin the calculation,but i stuck here at once
$ det(mathbf X)=I_{mn}:X$ and $ tr(mathbf X^n)=I_{mn}:X^n$,but i don't know how to calculate it.
matrices partial-derivative frobenius-method
asked Dec 1 '18 at 7:46
shineeleshineele
377
$endgroup$
add a comment |
$begingroup$
How to calculate the $frac{partial det(mathbf X)}{partial mathbf X}$ and $frac{partial tr(mathbf X^n)}{partial mathbf X}$ by using Frobenius product?i tried to begin the calculation,but i stuck here at once
$ det(mathbf X)=I_{mn}:X$ and $ tr(mathbf X^n)=I_{mn}:X^n$,but i don't know how to calculate it.
matrices partial-derivative frobenius-method
asked Dec 1 '18 at 7:46
shineeleshineele
377
$endgroup$
add a comment |
$begingroup$
How to calculate the $frac{partial det(mathbf X)}{partial mathbf X}$ and $frac{partial tr(mathbf X^n)}{partial mathbf X}$ by using Frobenius product?i tried to begin the calculation,but i stuck here at once
$ det(mathbf X)=I_{mn}:X$ and $ tr(mathbf X^n)=I_{mn}:X^n$,but i don't know how to calculate it.
matrices partial-derivative frobenius-method
asked Dec 1 '18 at 7:46
shineeleshineele
377
$endgroup$
How to calculate the $frac{partial det(mathbf X)}{partial mathbf X}$ and $frac{partial tr(mathbf X^n)}{partial mathbf X}$ by using Frobenius product?i tried to begin the calculation,but i stuck here at once
$ det(mathbf X)=I_{mn}:X$ and $ tr(mathbf X^n)=I_{mn}:X^n$,but i don't know how to calculate it.
matrices partial-derivative frobenius-method
matrices partial-derivative frobenius-method
asked Dec 1 '18 at 7:46
shineeleshineele
377
asked Dec 1 '18 at 7:46
shineeleshineele
377
asked Dec 1 '18 at 7:46
shineeleshineele
377
asked Dec 1 '18 at 7:46
shineeleshineele
377
asked Dec 1 '18 at 7:46
shineeleshineele
377
377
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let's start by finding the gradient of a small concrete example.
$$eqalign{
phi &= {rm tr}(X^3) = I:X^3 cr
dphi
&= I:(dX,X,X+X,dX,X+X,X,dX) cr
&= X^TX^T:dX + X^TX^T:dX + X^TX^T:dX cr
&= (3X^2)^T:dX cr
frac{partialphi}{partial X} &= (3X^{2})^T cr
}$$
This can be immediately extended to higher powers and scalar coefficients.
$$eqalign{
phi &= {rm tr}(alpha X^k) cr
dphi &= (nalpha X^{k-1})^T:dX cr
frac{partialphi}{partial X} &= (kalpha X^{k-1})^T cr
}$$
Then extended again to any function expressed as a power series.
$$eqalign{
F &= sum_kalpha_kX^k implies
F' &= sum_kkalpha_kX^{k-1} cr
phi &= {rm tr}(F) cr
dphi &= (F')^T:dX cr
frac{partialphi}{partial X} &= (F')^T crcr
}$$
To handle the determinant, consider the following
$$eqalign{
exp({rm tr}(X)) &=
expBig(sum_klambda_kBig) =
prod_kexp(lambda_k) =
det(exp(X)) cr
{rm tr}(X) &= log(det(exp(X))) cr
}$$
Now assume $X=log(Y)$ in which case $Y=exp(X)$ and
$$
{rm tr}(log(Y)) = log(det(Y))
$$
This allows us to use the above trace-trick to find the gradient of
$$eqalign{
phi &= log(det(X)) = {rm tr}(log(X)) cr
dphi &= (X^{-1})^T:dX cr
frac{partialphi}{partial X} &= (X^{-1})^T cr
}$$
To find the gradient of the determinant, note that the preceding was a logarithmic derivative, so
$$eqalign{
frac{partialphi}{partial X}
&= frac{tfrac{partial,det(X)}{partial X}}{det(X)} cr
frac{partial,det(X)}{partial X} &= det(X),(X^{-1})^T cr
}$$
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
$endgroup$
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
$begingroup$
Let's start by finding the gradient of a small concrete example.
$$eqalign{
phi &= {rm tr}(X^3) = I:X^3 cr
dphi
&= I:(dX,X,X+X,dX,X+X,X,dX) cr
&= X^TX^T:dX + X^TX^T:dX + X^TX^T:dX cr
&= (3X^2)^T:dX cr
frac{partialphi}{partial X} &= (3X^{2})^T cr
}$$
This can be immediately extended to higher powers and scalar coefficients.
$$eqalign{
phi &= {rm tr}(alpha X^k) cr
dphi &= (nalpha X^{k-1})^T:dX cr
frac{partialphi}{partial X} &= (kalpha X^{k-1})^T cr
}$$
Then extended again to any function expressed as a power series.
$$eqalign{
F &= sum_kalpha_kX^k implies
F' &= sum_kkalpha_kX^{k-1} cr
phi &= {rm tr}(F) cr
dphi &= (F')^T:dX cr
frac{partialphi}{partial X} &= (F')^T crcr
}$$
To handle the determinant, consider the following
$$eqalign{
exp({rm tr}(X)) &=
expBig(sum_klambda_kBig) =
prod_kexp(lambda_k) =
det(exp(X)) cr
{rm tr}(X) &= log(det(exp(X))) cr
}$$
Now assume $X=log(Y)$ in which case $Y=exp(X)$ and
$$
{rm tr}(log(Y)) = log(det(Y))
$$
This allows us to use the above trace-trick to find the gradient of
$$eqalign{
phi &= log(det(X)) = {rm tr}(log(X)) cr
dphi &= (X^{-1})^T:dX cr
frac{partialphi}{partial X} &= (X^{-1})^T cr
}$$
To find the gradient of the determinant, note that the preceding was a logarithmic derivative, so
$$eqalign{
frac{partialphi}{partial X}
&= frac{tfrac{partial,det(X)}{partial X}}{det(X)} cr
frac{partial,det(X)}{partial X} &= det(X),(X^{-1})^T cr
}$$
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
$endgroup$
add a comment |
$begingroup$
Let's start by finding the gradient of a small concrete example.
$$eqalign{
phi &= {rm tr}(X^3) = I:X^3 cr
dphi
&= I:(dX,X,X+X,dX,X+X,X,dX) cr
&= X^TX^T:dX + X^TX^T:dX + X^TX^T:dX cr
&= (3X^2)^T:dX cr
frac{partialphi}{partial X} &= (3X^{2})^T cr
}$$
This can be immediately extended to higher powers and scalar coefficients.
$$eqalign{
phi &= {rm tr}(alpha X^k) cr
dphi &= (nalpha X^{k-1})^T:dX cr
frac{partialphi}{partial X} &= (kalpha X^{k-1})^T cr
}$$
Then extended again to any function expressed as a power series.
$$eqalign{
F &= sum_kalpha_kX^k implies
F' &= sum_kkalpha_kX^{k-1} cr
phi &= {rm tr}(F) cr
dphi &= (F')^T:dX cr
frac{partialphi}{partial X} &= (F')^T crcr
}$$
To handle the determinant, consider the following
$$eqalign{
exp({rm tr}(X)) &=
expBig(sum_klambda_kBig) =
prod_kexp(lambda_k) =
det(exp(X)) cr
{rm tr}(X) &= log(det(exp(X))) cr
}$$
Now assume $X=log(Y)$ in which case $Y=exp(X)$ and
$$
{rm tr}(log(Y)) = log(det(Y))
$$
This allows us to use the above trace-trick to find the gradient of
$$eqalign{
phi &= log(det(X)) = {rm tr}(log(X)) cr
dphi &= (X^{-1})^T:dX cr
frac{partialphi}{partial X} &= (X^{-1})^T cr
}$$
To find the gradient of the determinant, note that the preceding was a logarithmic derivative, so
$$eqalign{
frac{partialphi}{partial X}
&= frac{tfrac{partial,det(X)}{partial X}}{det(X)} cr
frac{partial,det(X)}{partial X} &= det(X),(X^{-1})^T cr
}$$
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
$endgroup$
add a comment |
$begingroup$
Let's start by finding the gradient of a small concrete example.
$$eqalign{
phi &= {rm tr}(X^3) = I:X^3 cr
dphi
&= I:(dX,X,X+X,dX,X+X,X,dX) cr
&= X^TX^T:dX + X^TX^T:dX + X^TX^T:dX cr
&= (3X^2)^T:dX cr
frac{partialphi}{partial X} &= (3X^{2})^T cr
}$$
This can be immediately extended to higher powers and scalar coefficients.
$$eqalign{
phi &= {rm tr}(alpha X^k) cr
dphi &= (nalpha X^{k-1})^T:dX cr
frac{partialphi}{partial X} &= (kalpha X^{k-1})^T cr
}$$
Then extended again to any function expressed as a power series.
$$eqalign{
F &= sum_kalpha_kX^k implies
F' &= sum_kkalpha_kX^{k-1} cr
phi &= {rm tr}(F) cr
dphi &= (F')^T:dX cr
frac{partialphi}{partial X} &= (F')^T crcr
}$$
To handle the determinant, consider the following
$$eqalign{
exp({rm tr}(X)) &=
expBig(sum_klambda_kBig) =
prod_kexp(lambda_k) =
det(exp(X)) cr
{rm tr}(X) &= log(det(exp(X))) cr
}$$
Now assume $X=log(Y)$ in which case $Y=exp(X)$ and
$$
{rm tr}(log(Y)) = log(det(Y))
$$
This allows us to use the above trace-trick to find the gradient of
$$eqalign{
phi &= log(det(X)) = {rm tr}(log(X)) cr
dphi &= (X^{-1})^T:dX cr
frac{partialphi}{partial X} &= (X^{-1})^T cr
}$$
To find the gradient of the determinant, note that the preceding was a logarithmic derivative, so
$$eqalign{
frac{partialphi}{partial X}
&= frac{tfrac{partial,det(X)}{partial X}}{det(X)} cr
frac{partial,det(X)}{partial X} &= det(X),(X^{-1})^T cr
}$$
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
$endgroup$
Let's start by finding the gradient of a small concrete example.
$$eqalign{
phi &= {rm tr}(X^3) = I:X^3 cr
dphi
&= I:(dX,X,X+X,dX,X+X,X,dX) cr
&= X^TX^T:dX + X^TX^T:dX + X^TX^T:dX cr
&= (3X^2)^T:dX cr
frac{partialphi}{partial X} &= (3X^{2})^T cr
}$$
This can be immediately extended to higher powers and scalar coefficients.
$$eqalign{
phi &= {rm tr}(alpha X^k) cr
dphi &= (nalpha X^{k-1})^T:dX cr
frac{partialphi}{partial X} &= (kalpha X^{k-1})^T cr
}$$
Then extended again to any function expressed as a power series.
$$eqalign{
F &= sum_kalpha_kX^k implies
F' &= sum_kkalpha_kX^{k-1} cr
phi &= {rm tr}(F) cr
dphi &= (F')^T:dX cr
frac{partialphi}{partial X} &= (F')^T crcr
}$$
To handle the determinant, consider the following
$$eqalign{
exp({rm tr}(X)) &=
expBig(sum_klambda_kBig) =
prod_kexp(lambda_k) =
det(exp(X)) cr
{rm tr}(X) &= log(det(exp(X))) cr
}$$
Now assume $X=log(Y)$ in which case $Y=exp(X)$ and
$$
{rm tr}(log(Y)) = log(det(Y))
$$
This allows us to use the above trace-trick to find the gradient of
$$eqalign{
phi &= log(det(X)) = {rm tr}(log(X)) cr
dphi &= (X^{-1})^T:dX cr
frac{partialphi}{partial X} &= (X^{-1})^T cr
}$$
To find the gradient of the determinant, note that the preceding was a logarithmic derivative, so
$$eqalign{
frac{partialphi}{partial X}
&= frac{tfrac{partial,det(X)}{partial X}}{det(X)} cr
frac{partial,det(X)}{partial X} &= det(X),(X^{-1})^T cr
}$$
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
edited Dec 1 '18 at 18:03
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
answered Dec 1 '18 at 17:37
lynnlynn
1,766177
1,766177
add a comment |
add a comment |
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