Inverse function theorem for partial derivatives of a vector function
$begingroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
$endgroup$
add a comment |
$begingroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
$endgroup$
add a comment |
$begingroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
$endgroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
asked Feb 25 '15 at 2:19
LiborLibor
8781822
edited Feb 9 '18 at 17:22
user99914
asked Feb 25 '15 at 2:19
LiborLibor
8781822
asked Feb 25 '15 at 2:19
LiborLibor
8781822
asked Feb 25 '15 at 2:19
LiborLibor
8781822
8781822
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
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
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
$endgroup$
$begingroup$
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.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
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...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
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
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
$endgroup$
$begingroup$
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.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
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...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
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
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
$endgroup$
$begingroup$
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.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
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...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
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
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
$endgroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
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
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
edited Feb 25 '15 at 16:03
Libor
8781822
edited Feb 25 '15 at 16:03
Libor
8781822
edited Feb 25 '15 at 16:03
Libor
8781822
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
answered Feb 25 '15 at 2:29
TryssTryss
13k1229
13k1229
$begingroup$
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.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
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...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
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.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
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...
$endgroup$
– Libor
Feb 25 '15 at 2:48
$begingroup$
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.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
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.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
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...
$endgroup$
– Libor
Feb 25 '15 at 2:48
$begingroup$
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...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
edited Feb 25 '15 at 18:29
answered Feb 25 '15 at 17:55
LiborLibor
8781822
answered Feb 25 '15 at 17:55
LiborLibor
8781822
answered Feb 25 '15 at 17:55
LiborLibor
8781822
8781822
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
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