why is $frac{partial y_{mn}}{partial mathbf X}= mathbf B mathbf I_{mn}^Tmathbf A$
I have find some reason why $frac{partial y_{mn}}{partial mathbf X}= mathbf B mathbf I_{mn}^Tmathbf A$ when $mathbf Y=mathbf A mathbf X^T mathbf B$ in some books,but i can't still understand why,can anyone teach me how to calculate it in detail?
$mathbf A$ is $m*k$ matrix,$mathbf B$ is $i*n$ matrix,$mathbf X$ is $i*k$ matrix
matrices derivatives
asked Nov 29 '18 at 12:30
shineele
377
|
show 2 more comments
I have find some reason why $frac{partial y_{mn}}{partial mathbf X}= mathbf B mathbf I_{mn}^Tmathbf A$ when $mathbf Y=mathbf A mathbf X^T mathbf B$ in some books,but i can't still understand why,can anyone teach me how to calculate it in detail?
$mathbf A$ is $m*k$ matrix,$mathbf B$ is $i*n$ matrix,$mathbf X$ is $i*k$ matrix
matrices derivatives
asked Nov 29 '18 at 12:30
shineele
377
What are the orders of $A,B,X$?
– Mostafa Ayaz
Nov 29 '18 at 12:34
@MostafaAyaz i have edited the question
– shineele
Nov 29 '18 at 12:38
One more thing: how do you define $X^TB$ when $X$ is $ktimes i$ and $B$ is $itimes n$? Shouldn't it be $XB$?
– Mostafa Ayaz
Nov 29 '18 at 12:40
sry,terrible mistake
– shineele
Nov 29 '18 at 12:45
@MostafaAyaz i improved it
– shineele
Nov 29 '18 at 12:57
|
show 2 more comments
I have find some reason why $frac{partial y_{mn}}{partial mathbf X}= mathbf B mathbf I_{mn}^Tmathbf A$ when $mathbf Y=mathbf A mathbf X^T mathbf B$ in some books,but i can't still understand why,can anyone teach me how to calculate it in detail?
$mathbf A$ is $m*k$ matrix,$mathbf B$ is $i*n$ matrix,$mathbf X$ is $i*k$ matrix
matrices derivatives
asked Nov 29 '18 at 12:30
shineele
377
I have find some reason why $frac{partial y_{mn}}{partial mathbf X}= mathbf B mathbf I_{mn}^Tmathbf A$ when $mathbf Y=mathbf A mathbf X^T mathbf B$ in some books,but i can't still understand why,can anyone teach me how to calculate it in detail?
$mathbf A$ is $m*k$ matrix,$mathbf B$ is $i*n$ matrix,$mathbf X$ is $i*k$ matrix
matrices derivatives
matrices derivatives
asked Nov 29 '18 at 12:30
shineele
377
asked Nov 29 '18 at 12:30
shineele
377
edited Nov 29 '18 at 12:45
asked Nov 29 '18 at 12:30
shineele
377
asked Nov 29 '18 at 12:30
shineele
377
asked Nov 29 '18 at 12:30
shineele
377
377
What are the orders of $A,B,X$?
– Mostafa Ayaz
Nov 29 '18 at 12:34
@MostafaAyaz i have edited the question
– shineele
Nov 29 '18 at 12:38
One more thing: how do you define $X^TB$ when $X$ is $ktimes i$ and $B$ is $itimes n$? Shouldn't it be $XB$?
– Mostafa Ayaz
Nov 29 '18 at 12:40
sry,terrible mistake
– shineele
Nov 29 '18 at 12:45
@MostafaAyaz i improved it
– shineele
Nov 29 '18 at 12:57
|
show 2 more comments
What are the orders of $A,B,X$?
– Mostafa Ayaz
Nov 29 '18 at 12:34
@MostafaAyaz i have edited the question
– shineele
Nov 29 '18 at 12:38
One more thing: how do you define $X^TB$ when $X$ is $ktimes i$ and $B$ is $itimes n$? Shouldn't it be $XB$?
– Mostafa Ayaz
Nov 29 '18 at 12:40
sry,terrible mistake
– shineele
Nov 29 '18 at 12:45
@MostafaAyaz i improved it
– shineele
Nov 29 '18 at 12:57
What are the orders of $A,B,X$?
– Mostafa Ayaz
Nov 29 '18 at 12:34
What are the orders of $A,B,X$?
– Mostafa Ayaz
Nov 29 '18 at 12:34
@MostafaAyaz i have edited the question
– shineele
Nov 29 '18 at 12:38
@MostafaAyaz i have edited the question
– shineele
Nov 29 '18 at 12:38
One more thing: how do you define $X^TB$ when $X$ is $ktimes i$ and $B$ is $itimes n$? Shouldn't it be $XB$?
– Mostafa Ayaz
Nov 29 '18 at 12:40
One more thing: how do you define $X^TB$ when $X$ is $ktimes i$ and $B$ is $itimes n$? Shouldn't it be $XB$?
– Mostafa Ayaz
Nov 29 '18 at 12:40
sry,terrible mistake
– shineele
Nov 29 '18 at 12:45
sry,terrible mistake
– shineele
Nov 29 '18 at 12:45
@MostafaAyaz i improved it
– shineele
Nov 29 '18 at 12:57
@MostafaAyaz i improved it
– shineele
Nov 29 '18 at 12:57
|
show 2 more comments
1 Answer
1
active
oldest
votes
Starting with
$$eqalign{
Y &= AX^TB cr
}$$
Take the Frobenius product with the single-entry matrix to isolate the $(m,n)$ element.
$$eqalign{
y_{mn} &= E_{mn}:Y cr
&= E_{mn}:AX^TB cr
&= E_{mn}^T:B^TXA^T cr
&= BE_{mn}^TA:X cr
}$$
Now find the differential and gradient of this element wrt $X$
$$eqalign{
dy_{mn} &= BE_{mn}^TA:dX cr
frac{partial y_{mn}}{partial X} &= BE^T_{mn}A cr
}$$
There is no standard notation for the single-entry matrix. I use the letter ${mathbf E}$ others use ${mathbf J}$, but ${mathbf I}$ is a confusing choice since it usually denotes the identity matrix.
NB: The Frobenius product is just a convenient way to write the trace function
$$A:B = {rm Tr}(A^TB)$$
answered Nov 29 '18 at 14:42
greg
7,5451821
so in this calculation,$BB^T=I$?
– shineele
Nov 29 '18 at 15:09
Why would you think that? This calculation is agnostic about the properties of the $(A,B,X)$ matrices other than they must have compatible dimensions to allow the product $AX^TB$.
– greg
Nov 29 '18 at 15:19
because here $E_{mn}^T:B^TXA^T = BE_{mn}^TA:X $
– shineele
Nov 30 '18 at 1:18
You can prove this using the properties of the trace $$eqalign{ E^T:B^TXA^T &={rm Tr}(EB^TXA^T)cr &={rm Tr}(A^TEB^TX)cr &={rm Tr}big((BE^TA)^TXbig)cr &=BE^TA:X }$$
– greg
Nov 30 '18 at 4:16
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
Starting with
$$eqalign{
Y &= AX^TB cr
}$$
Take the Frobenius product with the single-entry matrix to isolate the $(m,n)$ element.
$$eqalign{
y_{mn} &= E_{mn}:Y cr
&= E_{mn}:AX^TB cr
&= E_{mn}^T:B^TXA^T cr
&= BE_{mn}^TA:X cr
}$$
Now find the differential and gradient of this element wrt $X$
$$eqalign{
dy_{mn} &= BE_{mn}^TA:dX cr
frac{partial y_{mn}}{partial X} &= BE^T_{mn}A cr
}$$
There is no standard notation for the single-entry matrix. I use the letter ${mathbf E}$ others use ${mathbf J}$, but ${mathbf I}$ is a confusing choice since it usually denotes the identity matrix.
NB: The Frobenius product is just a convenient way to write the trace function
$$A:B = {rm Tr}(A^TB)$$
answered Nov 29 '18 at 14:42
greg
7,5451821
so in this calculation,$BB^T=I$?
– shineele
Nov 29 '18 at 15:09
Why would you think that? This calculation is agnostic about the properties of the $(A,B,X)$ matrices other than they must have compatible dimensions to allow the product $AX^TB$.
– greg
Nov 29 '18 at 15:19
because here $E_{mn}^T:B^TXA^T = BE_{mn}^TA:X $
– shineele
Nov 30 '18 at 1:18
You can prove this using the properties of the trace $$eqalign{ E^T:B^TXA^T &={rm Tr}(EB^TXA^T)cr &={rm Tr}(A^TEB^TX)cr &={rm Tr}big((BE^TA)^TXbig)cr &=BE^TA:X }$$
– greg
Nov 30 '18 at 4:16
add a comment |
Starting with
$$eqalign{
Y &= AX^TB cr
}$$
Take the Frobenius product with the single-entry matrix to isolate the $(m,n)$ element.
$$eqalign{
y_{mn} &= E_{mn}:Y cr
&= E_{mn}:AX^TB cr
&= E_{mn}^T:B^TXA^T cr
&= BE_{mn}^TA:X cr
}$$
Now find the differential and gradient of this element wrt $X$
$$eqalign{
dy_{mn} &= BE_{mn}^TA:dX cr
frac{partial y_{mn}}{partial X} &= BE^T_{mn}A cr
}$$
There is no standard notation for the single-entry matrix. I use the letter ${mathbf E}$ others use ${mathbf J}$, but ${mathbf I}$ is a confusing choice since it usually denotes the identity matrix.
NB: The Frobenius product is just a convenient way to write the trace function
$$A:B = {rm Tr}(A^TB)$$
answered Nov 29 '18 at 14:42
greg
7,5451821
so in this calculation,$BB^T=I$?
– shineele
Nov 29 '18 at 15:09
Why would you think that? This calculation is agnostic about the properties of the $(A,B,X)$ matrices other than they must have compatible dimensions to allow the product $AX^TB$.
– greg
Nov 29 '18 at 15:19
because here $E_{mn}^T:B^TXA^T = BE_{mn}^TA:X $
– shineele
Nov 30 '18 at 1:18
You can prove this using the properties of the trace $$eqalign{ E^T:B^TXA^T &={rm Tr}(EB^TXA^T)cr &={rm Tr}(A^TEB^TX)cr &={rm Tr}big((BE^TA)^TXbig)cr &=BE^TA:X }$$
– greg
Nov 30 '18 at 4:16
add a comment |
Starting with
$$eqalign{
Y &= AX^TB cr
}$$
Take the Frobenius product with the single-entry matrix to isolate the $(m,n)$ element.
$$eqalign{
y_{mn} &= E_{mn}:Y cr
&= E_{mn}:AX^TB cr
&= E_{mn}^T:B^TXA^T cr
&= BE_{mn}^TA:X cr
}$$
Now find the differential and gradient of this element wrt $X$
$$eqalign{
dy_{mn} &= BE_{mn}^TA:dX cr
frac{partial y_{mn}}{partial X} &= BE^T_{mn}A cr
}$$
There is no standard notation for the single-entry matrix. I use the letter ${mathbf E}$ others use ${mathbf J}$, but ${mathbf I}$ is a confusing choice since it usually denotes the identity matrix.
NB: The Frobenius product is just a convenient way to write the trace function
$$A:B = {rm Tr}(A^TB)$$
answered Nov 29 '18 at 14:42
greg
7,5451821
Starting with
$$eqalign{
Y &= AX^TB cr
}$$
Take the Frobenius product with the single-entry matrix to isolate the $(m,n)$ element.
$$eqalign{
y_{mn} &= E_{mn}:Y cr
&= E_{mn}:AX^TB cr
&= E_{mn}^T:B^TXA^T cr
&= BE_{mn}^TA:X cr
}$$
Now find the differential and gradient of this element wrt $X$
$$eqalign{
dy_{mn} &= BE_{mn}^TA:dX cr
frac{partial y_{mn}}{partial X} &= BE^T_{mn}A cr
}$$
There is no standard notation for the single-entry matrix. I use the letter ${mathbf E}$ others use ${mathbf J}$, but ${mathbf I}$ is a confusing choice since it usually denotes the identity matrix.
NB: The Frobenius product is just a convenient way to write the trace function
$$A:B = {rm Tr}(A^TB)$$
answered Nov 29 '18 at 14:42
greg
7,5451821
answered Nov 29 '18 at 14:42
greg
7,5451821
answered Nov 29 '18 at 14:42
greg
7,5451821
answered Nov 29 '18 at 14:42
greg
7,5451821
7,5451821
so in this calculation,$BB^T=I$?
– shineele
Nov 29 '18 at 15:09
Why would you think that? This calculation is agnostic about the properties of the $(A,B,X)$ matrices other than they must have compatible dimensions to allow the product $AX^TB$.
– greg
Nov 29 '18 at 15:19
because here $E_{mn}^T:B^TXA^T = BE_{mn}^TA:X $
– shineele
Nov 30 '18 at 1:18
You can prove this using the properties of the trace $$eqalign{ E^T:B^TXA^T &={rm Tr}(EB^TXA^T)cr &={rm Tr}(A^TEB^TX)cr &={rm Tr}big((BE^TA)^TXbig)cr &=BE^TA:X }$$
– greg
Nov 30 '18 at 4:16
add a comment |
so in this calculation,$BB^T=I$?
– shineele
Nov 29 '18 at 15:09
Why would you think that? This calculation is agnostic about the properties of the $(A,B,X)$ matrices other than they must have compatible dimensions to allow the product $AX^TB$.
– greg
Nov 29 '18 at 15:19
because here $E_{mn}^T:B^TXA^T = BE_{mn}^TA:X $
– shineele
Nov 30 '18 at 1:18
You can prove this using the properties of the trace $$eqalign{ E^T:B^TXA^T &={rm Tr}(EB^TXA^T)cr &={rm Tr}(A^TEB^TX)cr &={rm Tr}big((BE^TA)^TXbig)cr &=BE^TA:X }$$
– greg
Nov 30 '18 at 4:16
so in this calculation,$BB^T=I$?
– shineele
Nov 29 '18 at 15:09
so in this calculation,$BB^T=I$?
– shineele
Nov 29 '18 at 15:09
Why would you think that? This calculation is agnostic about the properties of the $(A,B,X)$ matrices other than they must have compatible dimensions to allow the product $AX^TB$.
– greg
Nov 29 '18 at 15:19
Why would you think that? This calculation is agnostic about the properties of the $(A,B,X)$ matrices other than they must have compatible dimensions to allow the product $AX^TB$.
– greg
Nov 29 '18 at 15:19
because here $E_{mn}^T:B^TXA^T = BE_{mn}^TA:X $
– shineele
Nov 30 '18 at 1:18
because here $E_{mn}^T:B^TXA^T = BE_{mn}^TA:X $
– shineele
Nov 30 '18 at 1:18
You can prove this using the properties of the trace $$eqalign{ E^T:B^TXA^T &={rm Tr}(EB^TXA^T)cr &={rm Tr}(A^TEB^TX)cr &={rm Tr}big((BE^TA)^TXbig)cr &=BE^TA:X }$$
– greg
Nov 30 '18 at 4:16
You can prove this using the properties of the trace $$eqalign{ E^T:B^TXA^T &={rm Tr}(EB^TXA^T)cr &={rm Tr}(A^TEB^TX)cr &={rm Tr}big((BE^TA)^TXbig)cr &=BE^TA:X }$$
– greg
Nov 30 '18 at 4:16
add a comment |
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What are the orders of $A,B,X$?
– Mostafa Ayaz
Nov 29 '18 at 12:34
@MostafaAyaz i have edited the question
– shineele
Nov 29 '18 at 12:38
One more thing: how do you define $X^TB$ when $X$ is $ktimes i$ and $B$ is $itimes n$? Shouldn't it be $XB$?
– Mostafa Ayaz
Nov 29 '18 at 12:40
sry,terrible mistake
– shineele
Nov 29 '18 at 12:45
@MostafaAyaz i improved it
– shineele
Nov 29 '18 at 12:57