Derivative of matrix using Kronecker Product
$begingroup$
Suppose $ A(p)$ and $B(p) $ are functions which map $ mathbb{R}^{ntimes m} $ to $ mathbb{R}^{ntimes m} $ and
$$F(p)=S(I_{q}otimes A(p))(I_{q}otimes B(p))M$$
where $ S,M $ are constant matrices with dimension $ qntimes qn $ and $ qmtimes qm $ respectively. How to calculate
$$frac{dtext{vec}(F(p))}{dtext{vec}(p))}$$
Thank you very much!
derivatives matrix-calculus kronecker-product
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
$endgroup$
add a comment |
$begingroup$
Suppose $ A(p)$ and $B(p) $ are functions which map $ mathbb{R}^{ntimes m} $ to $ mathbb{R}^{ntimes m} $ and
$$F(p)=S(I_{q}otimes A(p))(I_{q}otimes B(p))M$$
where $ S,M $ are constant matrices with dimension $ qntimes qn $ and $ qmtimes qm $ respectively. How to calculate
$$frac{dtext{vec}(F(p))}{dtext{vec}(p))}$$
Thank you very much!
derivatives matrix-calculus kronecker-product
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
$endgroup$
add a comment |
$begingroup$
Suppose $ A(p)$ and $B(p) $ are functions which map $ mathbb{R}^{ntimes m} $ to $ mathbb{R}^{ntimes m} $ and
$$F(p)=S(I_{q}otimes A(p))(I_{q}otimes B(p))M$$
where $ S,M $ are constant matrices with dimension $ qntimes qn $ and $ qmtimes qm $ respectively. How to calculate
$$frac{dtext{vec}(F(p))}{dtext{vec}(p))}$$
Thank you very much!
derivatives matrix-calculus kronecker-product
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
$endgroup$
Suppose $ A(p)$ and $B(p) $ are functions which map $ mathbb{R}^{ntimes m} $ to $ mathbb{R}^{ntimes m} $ and
$$F(p)=S(I_{q}otimes A(p))(I_{q}otimes B(p))M$$
where $ S,M $ are constant matrices with dimension $ qntimes qn $ and $ qmtimes qm $ respectively. How to calculate
$$frac{dtext{vec}(F(p))}{dtext{vec}(p))}$$
Thank you very much!
derivatives matrix-calculus kronecker-product
derivatives matrix-calculus kronecker-product
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
asked Dec 5 '18 at 14:45
Yufang CuiYufang Cui
184
184
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let $,C=AB,$ and note that
$$(Iotimes A)(Iotimes B)= Iotimes C$$
Let's also use the convention where an uppercase letter denotes a matrix and a lowercase letter a vector, which are related by vectorization, e.g.
$$a={rm vec}(A),,,,b={rm vec}(B),,,,c={rm vec}(C),,,,etc$$
Finally, we'll need the SVD of the rightmost matrix in your function.
$$M = sum_k sigma_ku_kv_k^T$$
Break the function into components (corresponding to an SVD component).
$$eqalign{
F_k &= S(Iotimes C)u_kv_k^Tsigma_k cr
&= S,{rm vec}(CU_k),v_k^Tsigma_k cr
&= S(U_k^Totimes I),cv_k^Tsigma_k cr
f_k &= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big),c cr
}$$
Now find the derivative of this (vectorized) component matrix.
$$eqalign{
frac{partial f_k}{partial p}
&= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p} cr
}$$
And since $F = sum_k F_k,$ we have our answer
$$eqalign{
frac{partial f}{partial p}
&= sum_kBig(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p}
}$$
I'll leave it to you to work out the derivative on the far RHS, seeing as you told us nothing about the functional form of $A(P)$ and $B(P)$ and therefore of $C(P)$.
answered Dec 13 '18 at 2:41
greggreg
7,8101821
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $,C=AB,$ and note that
$$(Iotimes A)(Iotimes B)= Iotimes C$$
Let's also use the convention where an uppercase letter denotes a matrix and a lowercase letter a vector, which are related by vectorization, e.g.
$$a={rm vec}(A),,,,b={rm vec}(B),,,,c={rm vec}(C),,,,etc$$
Finally, we'll need the SVD of the rightmost matrix in your function.
$$M = sum_k sigma_ku_kv_k^T$$
Break the function into components (corresponding to an SVD component).
$$eqalign{
F_k &= S(Iotimes C)u_kv_k^Tsigma_k cr
&= S,{rm vec}(CU_k),v_k^Tsigma_k cr
&= S(U_k^Totimes I),cv_k^Tsigma_k cr
f_k &= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big),c cr
}$$
Now find the derivative of this (vectorized) component matrix.
$$eqalign{
frac{partial f_k}{partial p}
&= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p} cr
}$$
And since $F = sum_k F_k,$ we have our answer
$$eqalign{
frac{partial f}{partial p}
&= sum_kBig(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p}
}$$
I'll leave it to you to work out the derivative on the far RHS, seeing as you told us nothing about the functional form of $A(P)$ and $B(P)$ and therefore of $C(P)$.
answered Dec 13 '18 at 2:41
greggreg
7,8101821
$endgroup$
add a comment |
$begingroup$
Let $,C=AB,$ and note that
$$(Iotimes A)(Iotimes B)= Iotimes C$$
Let's also use the convention where an uppercase letter denotes a matrix and a lowercase letter a vector, which are related by vectorization, e.g.
$$a={rm vec}(A),,,,b={rm vec}(B),,,,c={rm vec}(C),,,,etc$$
Finally, we'll need the SVD of the rightmost matrix in your function.
$$M = sum_k sigma_ku_kv_k^T$$
Break the function into components (corresponding to an SVD component).
$$eqalign{
F_k &= S(Iotimes C)u_kv_k^Tsigma_k cr
&= S,{rm vec}(CU_k),v_k^Tsigma_k cr
&= S(U_k^Totimes I),cv_k^Tsigma_k cr
f_k &= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big),c cr
}$$
Now find the derivative of this (vectorized) component matrix.
$$eqalign{
frac{partial f_k}{partial p}
&= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p} cr
}$$
And since $F = sum_k F_k,$ we have our answer
$$eqalign{
frac{partial f}{partial p}
&= sum_kBig(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p}
}$$
I'll leave it to you to work out the derivative on the far RHS, seeing as you told us nothing about the functional form of $A(P)$ and $B(P)$ and therefore of $C(P)$.
answered Dec 13 '18 at 2:41
greggreg
7,8101821
$endgroup$
add a comment |
$begingroup$
Let $,C=AB,$ and note that
$$(Iotimes A)(Iotimes B)= Iotimes C$$
Let's also use the convention where an uppercase letter denotes a matrix and a lowercase letter a vector, which are related by vectorization, e.g.
$$a={rm vec}(A),,,,b={rm vec}(B),,,,c={rm vec}(C),,,,etc$$
Finally, we'll need the SVD of the rightmost matrix in your function.
$$M = sum_k sigma_ku_kv_k^T$$
Break the function into components (corresponding to an SVD component).
$$eqalign{
F_k &= S(Iotimes C)u_kv_k^Tsigma_k cr
&= S,{rm vec}(CU_k),v_k^Tsigma_k cr
&= S(U_k^Totimes I),cv_k^Tsigma_k cr
f_k &= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big),c cr
}$$
Now find the derivative of this (vectorized) component matrix.
$$eqalign{
frac{partial f_k}{partial p}
&= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p} cr
}$$
And since $F = sum_k F_k,$ we have our answer
$$eqalign{
frac{partial f}{partial p}
&= sum_kBig(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p}
}$$
I'll leave it to you to work out the derivative on the far RHS, seeing as you told us nothing about the functional form of $A(P)$ and $B(P)$ and therefore of $C(P)$.
answered Dec 13 '18 at 2:41
greggreg
7,8101821
$endgroup$
Let $,C=AB,$ and note that
$$(Iotimes A)(Iotimes B)= Iotimes C$$
Let's also use the convention where an uppercase letter denotes a matrix and a lowercase letter a vector, which are related by vectorization, e.g.
$$a={rm vec}(A),,,,b={rm vec}(B),,,,c={rm vec}(C),,,,etc$$
Finally, we'll need the SVD of the rightmost matrix in your function.
$$M = sum_k sigma_ku_kv_k^T$$
Break the function into components (corresponding to an SVD component).
$$eqalign{
F_k &= S(Iotimes C)u_kv_k^Tsigma_k cr
&= S,{rm vec}(CU_k),v_k^Tsigma_k cr
&= S(U_k^Totimes I),cv_k^Tsigma_k cr
f_k &= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big),c cr
}$$
Now find the derivative of this (vectorized) component matrix.
$$eqalign{
frac{partial f_k}{partial p}
&= Big(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p} cr
}$$
And since $F = sum_k F_k,$ we have our answer
$$eqalign{
frac{partial f}{partial p}
&= sum_kBig(sigma_kv_kotimesbig(S(U_k^Totimes I)big)Big)frac{partial c}{partial p}
}$$
I'll leave it to you to work out the derivative on the far RHS, seeing as you told us nothing about the functional form of $A(P)$ and $B(P)$ and therefore of $C(P)$.
answered Dec 13 '18 at 2:41
greggreg
7,8101821
edited Dec 13 '18 at 20:46
answered Dec 13 '18 at 2:41
greggreg
7,8101821
answered Dec 13 '18 at 2:41
greggreg
7,8101821
answered Dec 13 '18 at 2:41
greggreg
7,8101821
7,8101821
add a comment |
add a comment |
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