Derivative of frobenius norm












0












$begingroup$


I am trying to calculate the derivative of an energy function with respect to a vector xx. The energy is given by:



$$ψ(A)=∥A−I∥_F^2.$$
Where A is a square matrix with each column as x (a column vector):



$$A=[x_1 x_2 x_3 ... x_n]$$



The aim is to find $$frac{∂ψ}{∂x}$$



[Petersen 06] gives the derivative of a Frobenius norm as $$ frac{∂∥X∥_F^2}{X}=2X$$, but I am unsure how to extend it to this case (presumably using the chain rule somehow).






derivatives norm partial-derivative







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$endgroup$












  • $begingroup$
    $psi(A)$ is the sum of squares, so to calculate partial derivatives wrt $a_{ij}$ is simple. Then you can organize those in whatever way.
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:13










  • $begingroup$
    Possible duplicate
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:14
















0












$begingroup$


I am trying to calculate the derivative of an energy function with respect to a vector xx. The energy is given by:



$$ψ(A)=∥A−I∥_F^2.$$
Where A is a square matrix with each column as x (a column vector):



$$A=[x_1 x_2 x_3 ... x_n]$$



The aim is to find $$frac{∂ψ}{∂x}$$



[Petersen 06] gives the derivative of a Frobenius norm as $$ frac{∂∥X∥_F^2}{X}=2X$$, but I am unsure how to extend it to this case (presumably using the chain rule somehow).






derivatives norm partial-derivative







share|cite|improve this question









$endgroup$












  • $begingroup$
    $psi(A)$ is the sum of squares, so to calculate partial derivatives wrt $a_{ij}$ is simple. Then you can organize those in whatever way.
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:13










  • $begingroup$
    Possible duplicate
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:14














0












0








0





$begingroup$


I am trying to calculate the derivative of an energy function with respect to a vector xx. The energy is given by:



$$ψ(A)=∥A−I∥_F^2.$$
Where A is a square matrix with each column as x (a column vector):



$$A=[x_1 x_2 x_3 ... x_n]$$



The aim is to find $$frac{∂ψ}{∂x}$$



[Petersen 06] gives the derivative of a Frobenius norm as $$ frac{∂∥X∥_F^2}{X}=2X$$, but I am unsure how to extend it to this case (presumably using the chain rule somehow).






derivatives norm partial-derivative







share|cite|improve this question









$endgroup$




I am trying to calculate the derivative of an energy function with respect to a vector xx. The energy is given by:



$$ψ(A)=∥A−I∥_F^2.$$
Where A is a square matrix with each column as x (a column vector):



$$A=[x_1 x_2 x_3 ... x_n]$$



The aim is to find $$frac{∂ψ}{∂x}$$



[Petersen 06] gives the derivative of a Frobenius norm as $$ frac{∂∥X∥_F^2}{X}=2X$$, but I am unsure how to extend it to this case (presumably using the chain rule somehow).






derivatives norm partial-derivative




derivatives norm partial-derivative






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 15 '15 at 9:04









SidSid

154




154












  • $begingroup$
    $psi(A)$ is the sum of squares, so to calculate partial derivatives wrt $a_{ij}$ is simple. Then you can organize those in whatever way.
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:13










  • $begingroup$
    Possible duplicate
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:14


















  • $begingroup$
    $psi(A)$ is the sum of squares, so to calculate partial derivatives wrt $a_{ij}$ is simple. Then you can organize those in whatever way.
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:13










  • $begingroup$
    Possible duplicate
    $endgroup$
    – A.Γ.
    Oct 15 '15 at 9:14
















$begingroup$
$psi(A)$ is the sum of squares, so to calculate partial derivatives wrt $a_{ij}$ is simple. Then you can organize those in whatever way.
$endgroup$
– A.Γ.
Oct 15 '15 at 9:13




$begingroup$
$psi(A)$ is the sum of squares, so to calculate partial derivatives wrt $a_{ij}$ is simple. Then you can organize those in whatever way.
$endgroup$
– A.Γ.
Oct 15 '15 at 9:13












$begingroup$
Possible duplicate
$endgroup$
– A.Γ.
Oct 15 '15 at 9:14




$begingroup$
Possible duplicate
$endgroup$
– A.Γ.
Oct 15 '15 at 9:14










1 Answer
1






active

oldest

votes


















0












$begingroup$

Recall that the frobeniuns norm $defnorm#1{left|#1right|_F}norm{cdot} colon mathrm{Mat}_{n,m}(mathbf R) to mathbf R$ if given by
$$ norm A = deft{mathop{rm tr}}t(A^t A)^{1/2} $$
and hence the derivative of $norm{cdot }^2$ is (we used $t(A^t H) = t(H^t A)$)
$$ D(norm{cdot}^2)(A)H = 2 t(A^t H) $$
If we denote, for given $x_1, ldots, x_{i-1}, x_{i+1}, ldots, x_m in mathbf R^n$, the map $x_i mapsto [x_1, ldots, x_n] in mathrm{Mat}_{n,m}(mathbf R)$ by $A^{hat x}$, we have by the chain rule, that the derivative of $x_i mapsto psibigl(A^hat x(x_i)bigr)$, is given by
$$ Dpsi(A^{hat x}(x_i))DA^{hat x}(x_i) $$
Now $A^{hat x}$ is affine, hence $DA^{hat x}(x_i)$ is the linear part $h mapsto [0, ldots, 0, h, 0, ldots, 0] in mathrm{Mat}_{n,m}(mathbf R)$ and $Dpsi$ is given by
$$ Dpsi(A)H = 2tbigl((A-I)^t Hbigr) $$
Hence,
$$ frac{partial psi}{partial x_i}(h)
= Dpsi(A)DA^{hat x}(x_i)h = 2tbigl((A-I)^t[0,ldots, 0, h, 0,ldots, 0])bigr) $$
Now $(A - I)^t = A^t - I^t$ has the rows $x_j^t - e_j^t$, and as
$$ (A-I)^t[0,ldots, 0, h,0,ldots, 0] = [0, ldots, (A^t - I^t)h,0,ldots, 0] $$
taking the trace leaves us with
$$ frac{partial psi}{partial x_i}(h)
= 2(x_i^t - e_i^t)h $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    From where the $$H$$ comes in the equation. Are you supposing $$H=I$$.
    $endgroup$
    – Sid
    Oct 16 '15 at 5:57










  • $begingroup$
    No. The derivative of a map $psi colon {rm Mat}_{n,m}(mathbf R)to mathbf R$ at a point $A in {rm Mat}_{n,m}(mathbf R)$ is a linear map $Dpsi(A) colon {rm Mat}_{n,m}(mathbf R) to mathbf R$, which argument I call $H$.
    $endgroup$
    – martini
    Oct 16 '15 at 6:09











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1 Answer
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1 Answer
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0












$begingroup$

Recall that the frobeniuns norm $defnorm#1{left|#1right|_F}norm{cdot} colon mathrm{Mat}_{n,m}(mathbf R) to mathbf R$ if given by
$$ norm A = deft{mathop{rm tr}}t(A^t A)^{1/2} $$
and hence the derivative of $norm{cdot }^2$ is (we used $t(A^t H) = t(H^t A)$)
$$ D(norm{cdot}^2)(A)H = 2 t(A^t H) $$
If we denote, for given $x_1, ldots, x_{i-1}, x_{i+1}, ldots, x_m in mathbf R^n$, the map $x_i mapsto [x_1, ldots, x_n] in mathrm{Mat}_{n,m}(mathbf R)$ by $A^{hat x}$, we have by the chain rule, that the derivative of $x_i mapsto psibigl(A^hat x(x_i)bigr)$, is given by
$$ Dpsi(A^{hat x}(x_i))DA^{hat x}(x_i) $$
Now $A^{hat x}$ is affine, hence $DA^{hat x}(x_i)$ is the linear part $h mapsto [0, ldots, 0, h, 0, ldots, 0] in mathrm{Mat}_{n,m}(mathbf R)$ and $Dpsi$ is given by
$$ Dpsi(A)H = 2tbigl((A-I)^t Hbigr) $$
Hence,
$$ frac{partial psi}{partial x_i}(h)
= Dpsi(A)DA^{hat x}(x_i)h = 2tbigl((A-I)^t[0,ldots, 0, h, 0,ldots, 0])bigr) $$
Now $(A - I)^t = A^t - I^t$ has the rows $x_j^t - e_j^t$, and as
$$ (A-I)^t[0,ldots, 0, h,0,ldots, 0] = [0, ldots, (A^t - I^t)h,0,ldots, 0] $$
taking the trace leaves us with
$$ frac{partial psi}{partial x_i}(h)
= 2(x_i^t - e_i^t)h $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    From where the $$H$$ comes in the equation. Are you supposing $$H=I$$.
    $endgroup$
    – Sid
    Oct 16 '15 at 5:57










  • $begingroup$
    No. The derivative of a map $psi colon {rm Mat}_{n,m}(mathbf R)to mathbf R$ at a point $A in {rm Mat}_{n,m}(mathbf R)$ is a linear map $Dpsi(A) colon {rm Mat}_{n,m}(mathbf R) to mathbf R$, which argument I call $H$.
    $endgroup$
    – martini
    Oct 16 '15 at 6:09
















0












$begingroup$

Recall that the frobeniuns norm $defnorm#1{left|#1right|_F}norm{cdot} colon mathrm{Mat}_{n,m}(mathbf R) to mathbf R$ if given by
$$ norm A = deft{mathop{rm tr}}t(A^t A)^{1/2} $$
and hence the derivative of $norm{cdot }^2$ is (we used $t(A^t H) = t(H^t A)$)
$$ D(norm{cdot}^2)(A)H = 2 t(A^t H) $$
If we denote, for given $x_1, ldots, x_{i-1}, x_{i+1}, ldots, x_m in mathbf R^n$, the map $x_i mapsto [x_1, ldots, x_n] in mathrm{Mat}_{n,m}(mathbf R)$ by $A^{hat x}$, we have by the chain rule, that the derivative of $x_i mapsto psibigl(A^hat x(x_i)bigr)$, is given by
$$ Dpsi(A^{hat x}(x_i))DA^{hat x}(x_i) $$
Now $A^{hat x}$ is affine, hence $DA^{hat x}(x_i)$ is the linear part $h mapsto [0, ldots, 0, h, 0, ldots, 0] in mathrm{Mat}_{n,m}(mathbf R)$ and $Dpsi$ is given by
$$ Dpsi(A)H = 2tbigl((A-I)^t Hbigr) $$
Hence,
$$ frac{partial psi}{partial x_i}(h)
= Dpsi(A)DA^{hat x}(x_i)h = 2tbigl((A-I)^t[0,ldots, 0, h, 0,ldots, 0])bigr) $$
Now $(A - I)^t = A^t - I^t$ has the rows $x_j^t - e_j^t$, and as
$$ (A-I)^t[0,ldots, 0, h,0,ldots, 0] = [0, ldots, (A^t - I^t)h,0,ldots, 0] $$
taking the trace leaves us with
$$ frac{partial psi}{partial x_i}(h)
= 2(x_i^t - e_i^t)h $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    From where the $$H$$ comes in the equation. Are you supposing $$H=I$$.
    $endgroup$
    – Sid
    Oct 16 '15 at 5:57










  • $begingroup$
    No. The derivative of a map $psi colon {rm Mat}_{n,m}(mathbf R)to mathbf R$ at a point $A in {rm Mat}_{n,m}(mathbf R)$ is a linear map $Dpsi(A) colon {rm Mat}_{n,m}(mathbf R) to mathbf R$, which argument I call $H$.
    $endgroup$
    – martini
    Oct 16 '15 at 6:09














0












0








0





$begingroup$

Recall that the frobeniuns norm $defnorm#1{left|#1right|_F}norm{cdot} colon mathrm{Mat}_{n,m}(mathbf R) to mathbf R$ if given by
$$ norm A = deft{mathop{rm tr}}t(A^t A)^{1/2} $$
and hence the derivative of $norm{cdot }^2$ is (we used $t(A^t H) = t(H^t A)$)
$$ D(norm{cdot}^2)(A)H = 2 t(A^t H) $$
If we denote, for given $x_1, ldots, x_{i-1}, x_{i+1}, ldots, x_m in mathbf R^n$, the map $x_i mapsto [x_1, ldots, x_n] in mathrm{Mat}_{n,m}(mathbf R)$ by $A^{hat x}$, we have by the chain rule, that the derivative of $x_i mapsto psibigl(A^hat x(x_i)bigr)$, is given by
$$ Dpsi(A^{hat x}(x_i))DA^{hat x}(x_i) $$
Now $A^{hat x}$ is affine, hence $DA^{hat x}(x_i)$ is the linear part $h mapsto [0, ldots, 0, h, 0, ldots, 0] in mathrm{Mat}_{n,m}(mathbf R)$ and $Dpsi$ is given by
$$ Dpsi(A)H = 2tbigl((A-I)^t Hbigr) $$
Hence,
$$ frac{partial psi}{partial x_i}(h)
= Dpsi(A)DA^{hat x}(x_i)h = 2tbigl((A-I)^t[0,ldots, 0, h, 0,ldots, 0])bigr) $$
Now $(A - I)^t = A^t - I^t$ has the rows $x_j^t - e_j^t$, and as
$$ (A-I)^t[0,ldots, 0, h,0,ldots, 0] = [0, ldots, (A^t - I^t)h,0,ldots, 0] $$
taking the trace leaves us with
$$ frac{partial psi}{partial x_i}(h)
= 2(x_i^t - e_i^t)h $$






share|cite|improve this answer









$endgroup$



Recall that the frobeniuns norm $defnorm#1{left|#1right|_F}norm{cdot} colon mathrm{Mat}_{n,m}(mathbf R) to mathbf R$ if given by
$$ norm A = deft{mathop{rm tr}}t(A^t A)^{1/2} $$
and hence the derivative of $norm{cdot }^2$ is (we used $t(A^t H) = t(H^t A)$)
$$ D(norm{cdot}^2)(A)H = 2 t(A^t H) $$
If we denote, for given $x_1, ldots, x_{i-1}, x_{i+1}, ldots, x_m in mathbf R^n$, the map $x_i mapsto [x_1, ldots, x_n] in mathrm{Mat}_{n,m}(mathbf R)$ by $A^{hat x}$, we have by the chain rule, that the derivative of $x_i mapsto psibigl(A^hat x(x_i)bigr)$, is given by
$$ Dpsi(A^{hat x}(x_i))DA^{hat x}(x_i) $$
Now $A^{hat x}$ is affine, hence $DA^{hat x}(x_i)$ is the linear part $h mapsto [0, ldots, 0, h, 0, ldots, 0] in mathrm{Mat}_{n,m}(mathbf R)$ and $Dpsi$ is given by
$$ Dpsi(A)H = 2tbigl((A-I)^t Hbigr) $$
Hence,
$$ frac{partial psi}{partial x_i}(h)
= Dpsi(A)DA^{hat x}(x_i)h = 2tbigl((A-I)^t[0,ldots, 0, h, 0,ldots, 0])bigr) $$
Now $(A - I)^t = A^t - I^t$ has the rows $x_j^t - e_j^t$, and as
$$ (A-I)^t[0,ldots, 0, h,0,ldots, 0] = [0, ldots, (A^t - I^t)h,0,ldots, 0] $$
taking the trace leaves us with
$$ frac{partial psi}{partial x_i}(h)
= 2(x_i^t - e_i^t)h $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 15 '15 at 9:29









martinimartini

70.4k45991




70.4k45991












  • $begingroup$
    From where the $$H$$ comes in the equation. Are you supposing $$H=I$$.
    $endgroup$
    – Sid
    Oct 16 '15 at 5:57










  • $begingroup$
    No. The derivative of a map $psi colon {rm Mat}_{n,m}(mathbf R)to mathbf R$ at a point $A in {rm Mat}_{n,m}(mathbf R)$ is a linear map $Dpsi(A) colon {rm Mat}_{n,m}(mathbf R) to mathbf R$, which argument I call $H$.
    $endgroup$
    – martini
    Oct 16 '15 at 6:09


















  • $begingroup$
    From where the $$H$$ comes in the equation. Are you supposing $$H=I$$.
    $endgroup$
    – Sid
    Oct 16 '15 at 5:57










  • $begingroup$
    No. The derivative of a map $psi colon {rm Mat}_{n,m}(mathbf R)to mathbf R$ at a point $A in {rm Mat}_{n,m}(mathbf R)$ is a linear map $Dpsi(A) colon {rm Mat}_{n,m}(mathbf R) to mathbf R$, which argument I call $H$.
    $endgroup$
    – martini
    Oct 16 '15 at 6:09
















$begingroup$
From where the $$H$$ comes in the equation. Are you supposing $$H=I$$.
$endgroup$
– Sid
Oct 16 '15 at 5:57




$begingroup$
From where the $$H$$ comes in the equation. Are you supposing $$H=I$$.
$endgroup$
– Sid
Oct 16 '15 at 5:57












$begingroup$
No. The derivative of a map $psi colon {rm Mat}_{n,m}(mathbf R)to mathbf R$ at a point $A in {rm Mat}_{n,m}(mathbf R)$ is a linear map $Dpsi(A) colon {rm Mat}_{n,m}(mathbf R) to mathbf R$, which argument I call $H$.
$endgroup$
– martini
Oct 16 '15 at 6:09




$begingroup$
No. The derivative of a map $psi colon {rm Mat}_{n,m}(mathbf R)to mathbf R$ at a point $A in {rm Mat}_{n,m}(mathbf R)$ is a linear map $Dpsi(A) colon {rm Mat}_{n,m}(mathbf R) to mathbf R$, which argument I call $H$.
$endgroup$
– martini
Oct 16 '15 at 6:09


















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