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How can I derive the matrix?
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I have questions about matrix differentiation problem.
x∈ℜn, A∈ℜm∗n
x is an n*1 vector, A is an m*n matrix.
How to derive the following equation?
$$fracdAxdA$$
not the equation $$fracdAxdx$$
Thanks for reading my questions.
matrices derivatives matrix-calculus
asked Jul 30 at 7:59
ljh1830
61
add a comment |Â
up vote
0
down vote
favorite
I have questions about matrix differentiation problem.
x∈ℜn, A∈ℜm∗n
x is an n*1 vector, A is an m*n matrix.
How to derive the following equation?
$$fracdAxdA$$
not the equation $$fracdAxdx$$
Thanks for reading my questions.
matrices derivatives matrix-calculus
asked Jul 30 at 7:59
ljh1830
61
Welcome to Mathematics Stack Exchange! You should be using MathJax to format your mathematics. math.stackexchange.com/help/notation
– Miksu
Jul 30 at 8:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have questions about matrix differentiation problem.
x∈ℜn, A∈ℜm∗n
x is an n*1 vector, A is an m*n matrix.
How to derive the following equation?
$$fracdAxdA$$
not the equation $$fracdAxdx$$
Thanks for reading my questions.
matrices derivatives matrix-calculus
asked Jul 30 at 7:59
ljh1830
61
I have questions about matrix differentiation problem.
x∈ℜn, A∈ℜm∗n
x is an n*1 vector, A is an m*n matrix.
How to derive the following equation?
$$fracdAxdA$$
not the equation $$fracdAxdx$$
Thanks for reading my questions.
matrices derivatives matrix-calculus
asked Jul 30 at 7:59
ljh1830
61
edited Jul 30 at 8:09
asked Jul 30 at 7:59
ljh1830
61
asked Jul 30 at 7:59
ljh1830
61
asked Jul 30 at 7:59
ljh1830
61
61
Welcome to Mathematics Stack Exchange! You should be using MathJax to format your mathematics. math.stackexchange.com/help/notation
– Miksu
Jul 30 at 8:02
add a comment |Â
Welcome to Mathematics Stack Exchange! You should be using MathJax to format your mathematics. math.stackexchange.com/help/notation
– Miksu
Jul 30 at 8:02
Welcome to Mathematics Stack Exchange! You should be using MathJax to format your mathematics. math.stackexchange.com/help/notation
– Miksu
Jul 30 at 8:02
Welcome to Mathematics Stack Exchange! You should be using MathJax to format your mathematics. math.stackexchange.com/help/notation
– Miksu
Jul 30 at 8:02
add a comment |Â
2 Answers
2
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oldest
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up vote
1
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The function
$$P_x: Alongmapsto Ax$$
is obviously linear, so the Freched differential of $P_x$ in any point is again $P_x$:
$$DP(A_0) = P_x.$$
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
add a comment |Â
up vote
1
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You have a function
$$ f: mathbbR^m times n rightarrow mathbbR^n : A mapsto Ax $$
with fixed $x$. Hence, your derivative must be a mapping
$$ Df: mathbbR^m times n times mathbbR^m times n rightarrow mathbbR^n , (A_0, H) mapsto Df(A_0)[H], $$
which is linear in $H$, and that satisfies the 1st order Taylor equation
$$
f(A_0 + dA) = f(A_0) + Df(A_0)[dA] + o(dA) .
$$
If we insert our function we get
$$ f(A_0 + dA) = A_0 x + dA x = f(A_0) + dA x. $$
So $Df(A_0)[dA] := dA x$ does the job. If you are looking for the derivative written as tensor you can write the last formula via Einstein summation:
$$ Df(A_0)[dA] _i := dA_i,k x_k = delta_i,j dA_j,k x_k =
(delta_i,j x_k) dA_j,k $$
So the derivative $Df(A_0)$ can be identified with the 3-dimensional tensor $T in mathbbR^m times m times n$ given by
$$T_i,j,k = delta_i,j x_k. $$
Hereby $$delta_i,j:= begincases 1, &i=j\ 0, &ineq j endcases.$$
answered Aug 1 at 11:32
til
694
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The function
$$P_x: Alongmapsto Ax$$
is obviously linear, so the Freched differential of $P_x$ in any point is again $P_x$:
$$DP(A_0) = P_x.$$
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
add a comment |Â
up vote
1
down vote
The function
$$P_x: Alongmapsto Ax$$
is obviously linear, so the Freched differential of $P_x$ in any point is again $P_x$:
$$DP(A_0) = P_x.$$
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The function
$$P_x: Alongmapsto Ax$$
is obviously linear, so the Freched differential of $P_x$ in any point is again $P_x$:
$$DP(A_0) = P_x.$$
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
The function
$$P_x: Alongmapsto Ax$$
is obviously linear, so the Freched differential of $P_x$ in any point is again $P_x$:
$$DP(A_0) = P_x.$$
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Jul 30 at 8:20
MartÃn-Blas Pérez Pinilla
33.4k42570
33.4k42570
add a comment |Â
add a comment |Â
up vote
1
down vote
You have a function
$$ f: mathbbR^m times n rightarrow mathbbR^n : A mapsto Ax $$
with fixed $x$. Hence, your derivative must be a mapping
$$ Df: mathbbR^m times n times mathbbR^m times n rightarrow mathbbR^n , (A_0, H) mapsto Df(A_0)[H], $$
which is linear in $H$, and that satisfies the 1st order Taylor equation
$$
f(A_0 + dA) = f(A_0) + Df(A_0)[dA] + o(dA) .
$$
If we insert our function we get
$$ f(A_0 + dA) = A_0 x + dA x = f(A_0) + dA x. $$
So $Df(A_0)[dA] := dA x$ does the job. If you are looking for the derivative written as tensor you can write the last formula via Einstein summation:
$$ Df(A_0)[dA] _i := dA_i,k x_k = delta_i,j dA_j,k x_k =
(delta_i,j x_k) dA_j,k $$
So the derivative $Df(A_0)$ can be identified with the 3-dimensional tensor $T in mathbbR^m times m times n$ given by
$$T_i,j,k = delta_i,j x_k. $$
Hereby $$delta_i,j:= begincases 1, &i=j\ 0, &ineq j endcases.$$
answered Aug 1 at 11:32
til
694
add a comment |Â
up vote
1
down vote
You have a function
$$ f: mathbbR^m times n rightarrow mathbbR^n : A mapsto Ax $$
with fixed $x$. Hence, your derivative must be a mapping
$$ Df: mathbbR^m times n times mathbbR^m times n rightarrow mathbbR^n , (A_0, H) mapsto Df(A_0)[H], $$
which is linear in $H$, and that satisfies the 1st order Taylor equation
$$
f(A_0 + dA) = f(A_0) + Df(A_0)[dA] + o(dA) .
$$
If we insert our function we get
$$ f(A_0 + dA) = A_0 x + dA x = f(A_0) + dA x. $$
So $Df(A_0)[dA] := dA x$ does the job. If you are looking for the derivative written as tensor you can write the last formula via Einstein summation:
$$ Df(A_0)[dA] _i := dA_i,k x_k = delta_i,j dA_j,k x_k =
(delta_i,j x_k) dA_j,k $$
So the derivative $Df(A_0)$ can be identified with the 3-dimensional tensor $T in mathbbR^m times m times n$ given by
$$T_i,j,k = delta_i,j x_k. $$
Hereby $$delta_i,j:= begincases 1, &i=j\ 0, &ineq j endcases.$$
answered Aug 1 at 11:32
til
694
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have a function
$$ f: mathbbR^m times n rightarrow mathbbR^n : A mapsto Ax $$
with fixed $x$. Hence, your derivative must be a mapping
$$ Df: mathbbR^m times n times mathbbR^m times n rightarrow mathbbR^n , (A_0, H) mapsto Df(A_0)[H], $$
which is linear in $H$, and that satisfies the 1st order Taylor equation
$$
f(A_0 + dA) = f(A_0) + Df(A_0)[dA] + o(dA) .
$$
If we insert our function we get
$$ f(A_0 + dA) = A_0 x + dA x = f(A_0) + dA x. $$
So $Df(A_0)[dA] := dA x$ does the job. If you are looking for the derivative written as tensor you can write the last formula via Einstein summation:
$$ Df(A_0)[dA] _i := dA_i,k x_k = delta_i,j dA_j,k x_k =
(delta_i,j x_k) dA_j,k $$
So the derivative $Df(A_0)$ can be identified with the 3-dimensional tensor $T in mathbbR^m times m times n$ given by
$$T_i,j,k = delta_i,j x_k. $$
Hereby $$delta_i,j:= begincases 1, &i=j\ 0, &ineq j endcases.$$
answered Aug 1 at 11:32
til
694
You have a function
$$ f: mathbbR^m times n rightarrow mathbbR^n : A mapsto Ax $$
with fixed $x$. Hence, your derivative must be a mapping
$$ Df: mathbbR^m times n times mathbbR^m times n rightarrow mathbbR^n , (A_0, H) mapsto Df(A_0)[H], $$
which is linear in $H$, and that satisfies the 1st order Taylor equation
$$
f(A_0 + dA) = f(A_0) + Df(A_0)[dA] + o(dA) .
$$
If we insert our function we get
$$ f(A_0 + dA) = A_0 x + dA x = f(A_0) + dA x. $$
So $Df(A_0)[dA] := dA x$ does the job. If you are looking for the derivative written as tensor you can write the last formula via Einstein summation:
$$ Df(A_0)[dA] _i := dA_i,k x_k = delta_i,j dA_j,k x_k =
(delta_i,j x_k) dA_j,k $$
So the derivative $Df(A_0)$ can be identified with the 3-dimensional tensor $T in mathbbR^m times m times n$ given by
$$T_i,j,k = delta_i,j x_k. $$
Hereby $$delta_i,j:= begincases 1, &i=j\ 0, &ineq j endcases.$$
answered Aug 1 at 11:32
til
694
answered Aug 1 at 11:32
til
694
answered Aug 1 at 11:32
til
694
answered Aug 1 at 11:32
til
694
694
add a comment |Â
add a comment |Â
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Welcome to Mathematics Stack Exchange! You should be using MathJax to format your mathematics. math.stackexchange.com/help/notation
– Miksu
Jul 30 at 8:02