Inverse function theorem for partial derivatives of a vector function
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I have a simple vector function:
$$mathbfy = amathbfxqquad ainmathbbR,;mathbfx,mathbfyinmathbbR^n$$
The inverse is obviously:
$$mathbfy^-1 = 1over amathbfx$$
The inverse function theorem (at least for scalars) states that:
$$left[f^-1(y)right]'=frac1f'(f^-1(y))$$
But the partial derivatives of $mathbfy$ and $mathbfy^-1$ with respect to $a$ are not related that way:
$$beginalign(mathbfy)'_a &= mathbfx \ (mathbfy^-1)'_a &= -1over a^2mathbfx endalign$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
Libor
8131822
add a comment |Â
up vote
0
down vote
favorite
I have a simple vector function:
$$mathbfy = amathbfxqquad ainmathbbR,;mathbfx,mathbfyinmathbbR^n$$
The inverse is obviously:
$$mathbfy^-1 = 1over amathbfx$$
The inverse function theorem (at least for scalars) states that:
$$left[f^-1(y)right]'=frac1f'(f^-1(y))$$
But the partial derivatives of $mathbfy$ and $mathbfy^-1$ with respect to $a$ are not related that way:
$$beginalign(mathbfy)'_a &= mathbfx \ (mathbfy^-1)'_a &= -1over a^2mathbfx endalign$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
Libor
8131822
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a simple vector function:
$$mathbfy = amathbfxqquad ainmathbbR,;mathbfx,mathbfyinmathbbR^n$$
The inverse is obviously:
$$mathbfy^-1 = 1over amathbfx$$
The inverse function theorem (at least for scalars) states that:
$$left[f^-1(y)right]'=frac1f'(f^-1(y))$$
But the partial derivatives of $mathbfy$ and $mathbfy^-1$ with respect to $a$ are not related that way:
$$beginalign(mathbfy)'_a &= mathbfx \ (mathbfy^-1)'_a &= -1over a^2mathbfx endalign$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
Libor
8131822
I have a simple vector function:
$$mathbfy = amathbfxqquad ainmathbbR,;mathbfx,mathbfyinmathbbR^n$$
The inverse is obviously:
$$mathbfy^-1 = 1over amathbfx$$
The inverse function theorem (at least for scalars) states that:
$$left[f^-1(y)right]'=frac1f'(f^-1(y))$$
But the partial derivatives of $mathbfy$ and $mathbfy^-1$ with respect to $a$ are not related that way:
$$beginalign(mathbfy)'_a &= mathbfx \ (mathbfy^-1)'_a &= -1over a^2mathbfx endalign$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
Libor
8131822
asked Feb 25 '15 at 2:19
Libor
8131822
edited Feb 9 at 17:22
user99914
asked Feb 25 '15 at 2:19
Libor
8131822
asked Feb 25 '15 at 2:19
Libor
8131822
asked Feb 25 '15 at 2:19
Libor
8131822
8131822
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
0
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The derivative of a function $F:mathbbR^nto mathbbR^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = beginpmatrix
dfracpartial F_1partial x_1(x) & cdots & dfracpartial F_1partial x_n(x)\
vdots & ddots & vdots\
dfracpartial F_npartial x_1(x) & cdots & dfracpartial F_npartial x_n(x) endpmatrix$$
And the Jacobian of $F^-1$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_F^-1(x) = left(J_F(x)right)^-1$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8131822
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
– Libor
Feb 25 '15 at 2:34
The Jacobian gives you the partial derivatives...
– Tryss
Feb 25 '15 at 2:37
But the function is something like $mathbfy=amathbfx=beginpmatrixax_1 \ ax_2endpmatrix$. The only variable I care about is $a$ ($mathbfx$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
– Libor
Feb 25 '15 at 2:40
Oh, you mean your function is from $mathbbRtimes mathbbR^2 to mathbbR^2$? If it's the case, it's clearly not invertible : $$fleft(1,beginpmatrix 0 \ 0 endpmatrix right) = 1beginpmatrix 0 \ 0 endpmatrix= beginpmatrix 0 \ 0 endpmatrix = 2beginpmatrix 0 \ 0 endpmatrix = fleft(2,beginpmatrix 0 \ 0 endpmatrixright)$$
– Tryss
Feb 25 '15 at 2:43
Yes. The problem is that I already know values of $F$, $F'_a$ and $F^-1$ and only need to somehow obtain $F'_a(F^-1)$ using the reciprocal rule...
– Libor
Feb 25 '15 at 2:48
 |Â
show 1 more comment
up vote
0
down vote
The derivative w.r.t. $a$ is
$$mathbbdmathbfyover mathbbda = mathbfx$$
and the derivative of inverse is the reciproval of above, i.e.
$$mathbbdaover mathbbdmathbfy = 1over mathbfx$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
Libor
8131822
What is $1/mathbf x$ when $mathbf x$ is a vector?
– M. Winter
Jan 10 at 13:17
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The derivative of a function $F:mathbbR^nto mathbbR^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = beginpmatrix
dfracpartial F_1partial x_1(x) & cdots & dfracpartial F_1partial x_n(x)\
vdots & ddots & vdots\
dfracpartial F_npartial x_1(x) & cdots & dfracpartial F_npartial x_n(x) endpmatrix$$
And the Jacobian of $F^-1$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_F^-1(x) = left(J_F(x)right)^-1$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8131822
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
– Libor
Feb 25 '15 at 2:34
The Jacobian gives you the partial derivatives...
– Tryss
Feb 25 '15 at 2:37
But the function is something like $mathbfy=amathbfx=beginpmatrixax_1 \ ax_2endpmatrix$. The only variable I care about is $a$ ($mathbfx$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
– Libor
Feb 25 '15 at 2:40
Oh, you mean your function is from $mathbbRtimes mathbbR^2 to mathbbR^2$? If it's the case, it's clearly not invertible : $$fleft(1,beginpmatrix 0 \ 0 endpmatrix right) = 1beginpmatrix 0 \ 0 endpmatrix= beginpmatrix 0 \ 0 endpmatrix = 2beginpmatrix 0 \ 0 endpmatrix = fleft(2,beginpmatrix 0 \ 0 endpmatrixright)$$
– Tryss
Feb 25 '15 at 2:43
Yes. The problem is that I already know values of $F$, $F'_a$ and $F^-1$ and only need to somehow obtain $F'_a(F^-1)$ using the reciprocal rule...
– Libor
Feb 25 '15 at 2:48
 |Â
show 1 more comment
up vote
0
down vote
The derivative of a function $F:mathbbR^nto mathbbR^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = beginpmatrix
dfracpartial F_1partial x_1(x) & cdots & dfracpartial F_1partial x_n(x)\
vdots & ddots & vdots\
dfracpartial F_npartial x_1(x) & cdots & dfracpartial F_npartial x_n(x) endpmatrix$$
And the Jacobian of $F^-1$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_F^-1(x) = left(J_F(x)right)^-1$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8131822
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
– Libor
Feb 25 '15 at 2:34
The Jacobian gives you the partial derivatives...
– Tryss
Feb 25 '15 at 2:37
But the function is something like $mathbfy=amathbfx=beginpmatrixax_1 \ ax_2endpmatrix$. The only variable I care about is $a$ ($mathbfx$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
– Libor
Feb 25 '15 at 2:40
Oh, you mean your function is from $mathbbRtimes mathbbR^2 to mathbbR^2$? If it's the case, it's clearly not invertible : $$fleft(1,beginpmatrix 0 \ 0 endpmatrix right) = 1beginpmatrix 0 \ 0 endpmatrix= beginpmatrix 0 \ 0 endpmatrix = 2beginpmatrix 0 \ 0 endpmatrix = fleft(2,beginpmatrix 0 \ 0 endpmatrixright)$$
– Tryss
Feb 25 '15 at 2:43
Yes. The problem is that I already know values of $F$, $F'_a$ and $F^-1$ and only need to somehow obtain $F'_a(F^-1)$ using the reciprocal rule...
– Libor
Feb 25 '15 at 2:48
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
The derivative of a function $F:mathbbR^nto mathbbR^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = beginpmatrix
dfracpartial F_1partial x_1(x) & cdots & dfracpartial F_1partial x_n(x)\
vdots & ddots & vdots\
dfracpartial F_npartial x_1(x) & cdots & dfracpartial F_npartial x_n(x) endpmatrix$$
And the Jacobian of $F^-1$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_F^-1(x) = left(J_F(x)right)^-1$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8131822
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
The derivative of a function $F:mathbbR^nto mathbbR^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = beginpmatrix
dfracpartial F_1partial x_1(x) & cdots & dfracpartial F_1partial x_n(x)\
vdots & ddots & vdots\
dfracpartial F_npartial x_1(x) & cdots & dfracpartial F_npartial x_n(x) endpmatrix$$
And the Jacobian of $F^-1$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_F^-1(x) = left(J_F(x)right)^-1$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8131822
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
edited Feb 25 '15 at 16:03
Libor
8131822
edited Feb 25 '15 at 16:03
Libor
8131822
edited Feb 25 '15 at 16:03
Libor
8131822
8131822
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
answered Feb 25 '15 at 2:29
Tryss
12.9k1129
12.9k1129
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
– Libor
Feb 25 '15 at 2:34
The Jacobian gives you the partial derivatives...
– Tryss
Feb 25 '15 at 2:37
But the function is something like $mathbfy=amathbfx=beginpmatrixax_1 \ ax_2endpmatrix$. The only variable I care about is $a$ ($mathbfx$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
– Libor
Feb 25 '15 at 2:40
Oh, you mean your function is from $mathbbRtimes mathbbR^2 to mathbbR^2$? If it's the case, it's clearly not invertible : $$fleft(1,beginpmatrix 0 \ 0 endpmatrix right) = 1beginpmatrix 0 \ 0 endpmatrix= beginpmatrix 0 \ 0 endpmatrix = 2beginpmatrix 0 \ 0 endpmatrix = fleft(2,beginpmatrix 0 \ 0 endpmatrixright)$$
– Tryss
Feb 25 '15 at 2:43
Yes. The problem is that I already know values of $F$, $F'_a$ and $F^-1$ and only need to somehow obtain $F'_a(F^-1)$ using the reciprocal rule...
– Libor
Feb 25 '15 at 2:48
 |Â
show 1 more comment
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
– Libor
Feb 25 '15 at 2:34
The Jacobian gives you the partial derivatives...
– Tryss
Feb 25 '15 at 2:37
But the function is something like $mathbfy=amathbfx=beginpmatrixax_1 \ ax_2endpmatrix$. The only variable I care about is $a$ ($mathbfx$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
– Libor
Feb 25 '15 at 2:40
Oh, you mean your function is from $mathbbRtimes mathbbR^2 to mathbbR^2$? If it's the case, it's clearly not invertible : $$fleft(1,beginpmatrix 0 \ 0 endpmatrix right) = 1beginpmatrix 0 \ 0 endpmatrix= beginpmatrix 0 \ 0 endpmatrix = 2beginpmatrix 0 \ 0 endpmatrix = fleft(2,beginpmatrix 0 \ 0 endpmatrixright)$$
– Tryss
Feb 25 '15 at 2:43
Yes. The problem is that I already know values of $F$, $F'_a$ and $F^-1$ and only need to somehow obtain $F'_a(F^-1)$ using the reciprocal rule...
– Libor
Feb 25 '15 at 2:48
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
– Libor
Feb 25 '15 at 2:34
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
– Libor
Feb 25 '15 at 2:34
The Jacobian gives you the partial derivatives...
– Tryss
Feb 25 '15 at 2:37
The Jacobian gives you the partial derivatives...
– Tryss
Feb 25 '15 at 2:37
But the function is something like $mathbfy=amathbfx=beginpmatrixax_1 \ ax_2endpmatrix$. The only variable I care about is $a$ ($mathbfx$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
– Libor
Feb 25 '15 at 2:40
But the function is something like $mathbfy=amathbfx=beginpmatrixax_1 \ ax_2endpmatrix$. The only variable I care about is $a$ ($mathbfx$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
– Libor
Feb 25 '15 at 2:40
Oh, you mean your function is from $mathbbRtimes mathbbR^2 to mathbbR^2$? If it's the case, it's clearly not invertible : $$fleft(1,beginpmatrix 0 \ 0 endpmatrix right) = 1beginpmatrix 0 \ 0 endpmatrix= beginpmatrix 0 \ 0 endpmatrix = 2beginpmatrix 0 \ 0 endpmatrix = fleft(2,beginpmatrix 0 \ 0 endpmatrixright)$$
– Tryss
Feb 25 '15 at 2:43
Oh, you mean your function is from $mathbbRtimes mathbbR^2 to mathbbR^2$? If it's the case, it's clearly not invertible : $$fleft(1,beginpmatrix 0 \ 0 endpmatrix right) = 1beginpmatrix 0 \ 0 endpmatrix= beginpmatrix 0 \ 0 endpmatrix = 2beginpmatrix 0 \ 0 endpmatrix = fleft(2,beginpmatrix 0 \ 0 endpmatrixright)$$
– Tryss
Feb 25 '15 at 2:43
Yes. The problem is that I already know values of $F$, $F'_a$ and $F^-1$ and only need to somehow obtain $F'_a(F^-1)$ using the reciprocal rule...
– Libor
Feb 25 '15 at 2:48
Yes. The problem is that I already know values of $F$, $F'_a$ and $F^-1$ and only need to somehow obtain $F'_a(F^-1)$ using the reciprocal rule...
– Libor
Feb 25 '15 at 2:48
 |Â
show 1 more comment
up vote
0
down vote
The derivative w.r.t. $a$ is
$$mathbbdmathbfyover mathbbda = mathbfx$$
and the derivative of inverse is the reciproval of above, i.e.
$$mathbbdaover mathbbdmathbfy = 1over mathbfx$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
Libor
8131822
What is $1/mathbf x$ when $mathbf x$ is a vector?
– M. Winter
Jan 10 at 13:17
add a comment |Â
up vote
0
down vote
The derivative w.r.t. $a$ is
$$mathbbdmathbfyover mathbbda = mathbfx$$
and the derivative of inverse is the reciproval of above, i.e.
$$mathbbdaover mathbbdmathbfy = 1over mathbfx$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
Libor
8131822
What is $1/mathbf x$ when $mathbf x$ is a vector?
– M. Winter
Jan 10 at 13:17
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The derivative w.r.t. $a$ is
$$mathbbdmathbfyover mathbbda = mathbfx$$
and the derivative of inverse is the reciproval of above, i.e.
$$mathbbdaover mathbbdmathbfy = 1over mathbfx$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
Libor
8131822
The derivative w.r.t. $a$ is
$$mathbbdmathbfyover mathbbda = mathbfx$$
and the derivative of inverse is the reciproval of above, i.e.
$$mathbbdaover mathbbdmathbfy = 1over mathbfx$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
Libor
8131822
edited Feb 25 '15 at 18:29
answered Feb 25 '15 at 17:55
Libor
8131822
answered Feb 25 '15 at 17:55
Libor
8131822
answered Feb 25 '15 at 17:55
Libor
8131822
8131822
What is $1/mathbf x$ when $mathbf x$ is a vector?
– M. Winter
Jan 10 at 13:17
add a comment |Â
What is $1/mathbf x$ when $mathbf x$ is a vector?
– M. Winter
Jan 10 at 13:17
What is $1/mathbf x$ when $mathbf x$ is a vector?
– M. Winter
Jan 10 at 13:17
What is $1/mathbf x$ when $mathbf x$ is a vector?
– M. Winter
Jan 10 at 13:17
add a comment |Â
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