Transpose the answer matrix to the right dimension after partial differentiation the product of matrices

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I would like to understand about transposing matrix after partial differentiation for the right dimension of answer



For a function of the product of matrices $W$ and $H$ with dimension $i*k$ and $k*j$, respectively:



$$eqalign
Omega = sum_kW_ikK_kj
$$



A colon represents Frobenius product. $1$ is matrix of ones in $i*j$ dimension. So,
$$eqalign
Omega &= 1:WH cr
$$



Differentiation function $Omega$ wrt $H$:



$$eqalign
partial Omega &= 1:Wpartial H cr
fracpartial Omegapartial H &= 1:Wcr
fracpartial Omegapartial H &= W^T1cr
$$



I wonder how to resolve the last two lines? I know the answer should be in $k*j$ dimension and $W^T 1$ is in the right dimension for the answer. But is there the reference/rule to do this? Or is it from the observation that the answer must be $kj$ and we just make a product between $1$ and $W$ to be $kj$?




matrices matrix-calculus transpose




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  • I believe you meant to define $Omega$ as the scalar quantity $$Omega = sum_isum_jsum_k W_ikH_kj$$ The cyclic property of the trace allows you to rearrange the terms in a Frobenius product. For example $$eqalignA:BC&=rm tr(A(BC)^T)cr&=rm tr(AC^TB^T)cr&=rm tr(B^TAC^T) =B^TA:Ccr$$
    – greg
    Aug 24 at 4:40















up vote
0
down vote

favorite












I would like to understand about transposing matrix after partial differentiation for the right dimension of answer



For a function of the product of matrices $W$ and $H$ with dimension $i*k$ and $k*j$, respectively:



$$eqalign
Omega = sum_kW_ikK_kj
$$



A colon represents Frobenius product. $1$ is matrix of ones in $i*j$ dimension. So,
$$eqalign
Omega &= 1:WH cr
$$



Differentiation function $Omega$ wrt $H$:



$$eqalign
partial Omega &= 1:Wpartial H cr
fracpartial Omegapartial H &= 1:Wcr
fracpartial Omegapartial H &= W^T1cr
$$



I wonder how to resolve the last two lines? I know the answer should be in $k*j$ dimension and $W^T 1$ is in the right dimension for the answer. But is there the reference/rule to do this? Or is it from the observation that the answer must be $kj$ and we just make a product between $1$ and $W$ to be $kj$?




matrices matrix-calculus transpose




share|cite|improve this question




















  • I believe you meant to define $Omega$ as the scalar quantity $$Omega = sum_isum_jsum_k W_ikH_kj$$ The cyclic property of the trace allows you to rearrange the terms in a Frobenius product. For example $$eqalignA:BC&=rm tr(A(BC)^T)cr&=rm tr(AC^TB^T)cr&=rm tr(B^TAC^T) =B^TA:Ccr$$
    – greg
    Aug 24 at 4:40













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I would like to understand about transposing matrix after partial differentiation for the right dimension of answer



For a function of the product of matrices $W$ and $H$ with dimension $i*k$ and $k*j$, respectively:



$$eqalign
Omega = sum_kW_ikK_kj
$$



A colon represents Frobenius product. $1$ is matrix of ones in $i*j$ dimension. So,
$$eqalign
Omega &= 1:WH cr
$$



Differentiation function $Omega$ wrt $H$:



$$eqalign
partial Omega &= 1:Wpartial H cr
fracpartial Omegapartial H &= 1:Wcr
fracpartial Omegapartial H &= W^T1cr
$$



I wonder how to resolve the last two lines? I know the answer should be in $k*j$ dimension and $W^T 1$ is in the right dimension for the answer. But is there the reference/rule to do this? Or is it from the observation that the answer must be $kj$ and we just make a product between $1$ and $W$ to be $kj$?




matrices matrix-calculus transpose




share|cite|improve this question












I would like to understand about transposing matrix after partial differentiation for the right dimension of answer



For a function of the product of matrices $W$ and $H$ with dimension $i*k$ and $k*j$, respectively:



$$eqalign
Omega = sum_kW_ikK_kj
$$



A colon represents Frobenius product. $1$ is matrix of ones in $i*j$ dimension. So,
$$eqalign
Omega &= 1:WH cr
$$



Differentiation function $Omega$ wrt $H$:



$$eqalign
partial Omega &= 1:Wpartial H cr
fracpartial Omegapartial H &= 1:Wcr
fracpartial Omegapartial H &= W^T1cr
$$



I wonder how to resolve the last two lines? I know the answer should be in $k*j$ dimension and $W^T 1$ is in the right dimension for the answer. But is there the reference/rule to do this? Or is it from the observation that the answer must be $kj$ and we just make a product between $1$ and $W$ to be $kj$?





matrices matrix-calculus transpose





share|cite|improve this question











share|cite|improve this question




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asked Aug 16 at 7:08









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  • I believe you meant to define $Omega$ as the scalar quantity $$Omega = sum_isum_jsum_k W_ikH_kj$$ The cyclic property of the trace allows you to rearrange the terms in a Frobenius product. For example $$eqalignA:BC&=rm tr(A(BC)^T)cr&=rm tr(AC^TB^T)cr&=rm tr(B^TAC^T) =B^TA:Ccr$$
    – greg
    Aug 24 at 4:40

















  • I believe you meant to define $Omega$ as the scalar quantity $$Omega = sum_isum_jsum_k W_ikH_kj$$ The cyclic property of the trace allows you to rearrange the terms in a Frobenius product. For example $$eqalignA:BC&=rm tr(A(BC)^T)cr&=rm tr(AC^TB^T)cr&=rm tr(B^TAC^T) =B^TA:Ccr$$
    – greg
    Aug 24 at 4:40
















I believe you meant to define $Omega$ as the scalar quantity $$Omega = sum_isum_jsum_k W_ikH_kj$$ The cyclic property of the trace allows you to rearrange the terms in a Frobenius product. For example $$eqalignA:BC&=rm tr(A(BC)^T)cr&=rm tr(AC^TB^T)cr&=rm tr(B^TAC^T) =B^TA:Ccr$$
– greg
Aug 24 at 4:40





I believe you meant to define $Omega$ as the scalar quantity $$Omega = sum_isum_jsum_k W_ikH_kj$$ The cyclic property of the trace allows you to rearrange the terms in a Frobenius product. For example $$eqalignA:BC&=rm tr(A(BC)^T)cr&=rm tr(AC^TB^T)cr&=rm tr(B^TAC^T) =B^TA:Ccr$$
– greg
Aug 24 at 4:40
















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