Derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$ w.r.t. $X$?
Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.
How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?
multivariable-calculus matrix-calculus
asked Nov 25 at 19:20
learning
275
add a comment |
Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.
How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?
multivariable-calculus matrix-calculus
asked Nov 25 at 19:20
learning
275
Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24
@user550103. yes, that's correct.
– learning
Nov 25 at 19:25
What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49
add a comment |
Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.
How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?
multivariable-calculus matrix-calculus
asked Nov 25 at 19:20
learning
275
Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.
How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?
multivariable-calculus matrix-calculus
multivariable-calculus matrix-calculus
asked Nov 25 at 19:20
learning
275
asked Nov 25 at 19:20
learning
275
asked Nov 25 at 19:20
learning
275
asked Nov 25 at 19:20
learning
275
asked Nov 25 at 19:20
learning
275
275
Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24
@user550103. yes, that's correct.
– learning
Nov 25 at 19:25
What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49
add a comment |
Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24
@user550103. yes, that's correct.
– learning
Nov 25 at 19:25
What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49
Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24
Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24
@user550103. yes, that's correct.
– learning
Nov 25 at 19:25
@user550103. yes, that's correct.
– learning
Nov 25 at 19:25
What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49
What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49
add a comment |
1 Answer
1
active
oldest
votes
Define the vectors
$$eqalign{
x &= {rm vec}(X) cr
u &= {rm vec}(U) cr
}$$
Write your function in terms of the vectors. Then find the differential and gradient.
$$eqalign{
f &= u^TBx = (B^Tu)^Tx cr
df &= (B^Tu)^T,dx cr
frac{partial f}{partial x} &= B^Tu cr
}$$
Now de-vectorize this to obtain a matrix result.
$$eqalign{
frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
}$$
answered Nov 25 at 20:19
greg
7,5001721
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
Define the vectors
$$eqalign{
x &= {rm vec}(X) cr
u &= {rm vec}(U) cr
}$$
Write your function in terms of the vectors. Then find the differential and gradient.
$$eqalign{
f &= u^TBx = (B^Tu)^Tx cr
df &= (B^Tu)^T,dx cr
frac{partial f}{partial x} &= B^Tu cr
}$$
Now de-vectorize this to obtain a matrix result.
$$eqalign{
frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
}$$
answered Nov 25 at 20:19
greg
7,5001721
add a comment |
Define the vectors
$$eqalign{
x &= {rm vec}(X) cr
u &= {rm vec}(U) cr
}$$
Write your function in terms of the vectors. Then find the differential and gradient.
$$eqalign{
f &= u^TBx = (B^Tu)^Tx cr
df &= (B^Tu)^T,dx cr
frac{partial f}{partial x} &= B^Tu cr
}$$
Now de-vectorize this to obtain a matrix result.
$$eqalign{
frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
}$$
answered Nov 25 at 20:19
greg
7,5001721
add a comment |
Define the vectors
$$eqalign{
x &= {rm vec}(X) cr
u &= {rm vec}(U) cr
}$$
Write your function in terms of the vectors. Then find the differential and gradient.
$$eqalign{
f &= u^TBx = (B^Tu)^Tx cr
df &= (B^Tu)^T,dx cr
frac{partial f}{partial x} &= B^Tu cr
}$$
Now de-vectorize this to obtain a matrix result.
$$eqalign{
frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
}$$
answered Nov 25 at 20:19
greg
7,5001721
Define the vectors
$$eqalign{
x &= {rm vec}(X) cr
u &= {rm vec}(U) cr
}$$
Write your function in terms of the vectors. Then find the differential and gradient.
$$eqalign{
f &= u^TBx = (B^Tu)^Tx cr
df &= (B^Tu)^T,dx cr
frac{partial f}{partial x} &= B^Tu cr
}$$
Now de-vectorize this to obtain a matrix result.
$$eqalign{
frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
}$$
answered Nov 25 at 20:19
greg
7,5001721
answered Nov 25 at 20:19
greg
7,5001721
answered Nov 25 at 20:19
greg
7,5001721
answered Nov 25 at 20:19
greg
7,5001721
7,5001721
add a comment |
add a comment |
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Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24
@user550103. yes, that's correct.
– learning
Nov 25 at 19:25
What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49