Why is the Jacobi Matrix not a matrix?
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In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
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show 2 more comments
$begingroup$
In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
$endgroup$
1
$begingroup$
No, a point $p$ has already been plugged in.
$endgroup$
– Randall
Nov 28 '18 at 12:22
1
$begingroup$
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
$endgroup$
– Math_QED
Nov 28 '18 at 12:25
1
$begingroup$
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
$endgroup$
– Jakobian
Nov 28 '18 at 14:25
1
$begingroup$
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
$endgroup$
– Randall
Nov 28 '18 at 18:03
1
$begingroup$
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
$endgroup$
– Arjihad
Nov 28 '18 at 18:08
|
show 2 more comments
$begingroup$
In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
$endgroup$
In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
asked Nov 28 '18 at 12:18
ArjihadArjihad
390111
390111
1
$begingroup$
No, a point $p$ has already been plugged in.
$endgroup$
– Randall
Nov 28 '18 at 12:22
1
$begingroup$
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
$endgroup$
– Math_QED
Nov 28 '18 at 12:25
1
$begingroup$
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
$endgroup$
– Jakobian
Nov 28 '18 at 14:25
1
$begingroup$
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
$endgroup$
– Randall
Nov 28 '18 at 18:03
1
$begingroup$
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
$endgroup$
– Arjihad
Nov 28 '18 at 18:08
|
show 2 more comments
1
$begingroup$
No, a point $p$ has already been plugged in.
$endgroup$
– Randall
Nov 28 '18 at 12:22
1
$begingroup$
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
$endgroup$
– Math_QED
Nov 28 '18 at 12:25
1
$begingroup$
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
$endgroup$
– Jakobian
Nov 28 '18 at 14:25
1
$begingroup$
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
$endgroup$
– Randall
Nov 28 '18 at 18:03
1
$begingroup$
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
$endgroup$
– Arjihad
Nov 28 '18 at 18:08
1
1
$begingroup$
No, a point $p$ has already been plugged in.
$endgroup$
– Randall
Nov 28 '18 at 12:22
$begingroup$
No, a point $p$ has already been plugged in.
$endgroup$
– Randall
Nov 28 '18 at 12:22
1
1
$begingroup$
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
$endgroup$
– Math_QED
Nov 28 '18 at 12:25
$begingroup$
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
$endgroup$
– Math_QED
Nov 28 '18 at 12:25
1
1
$begingroup$
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
$endgroup$
– Jakobian
Nov 28 '18 at 14:25
$begingroup$
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
$endgroup$
– Jakobian
Nov 28 '18 at 14:25
1
1
$begingroup$
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
$endgroup$
– Randall
Nov 28 '18 at 18:03
$begingroup$
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
$endgroup$
– Randall
Nov 28 '18 at 18:03
1
1
$begingroup$
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
$endgroup$
– Arjihad
Nov 28 '18 at 18:08
$begingroup$
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
$endgroup$
– Arjihad
Nov 28 '18 at 18:08
|
show 2 more comments
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1
$begingroup$
No, a point $p$ has already been plugged in.
$endgroup$
– Randall
Nov 28 '18 at 12:22
1
$begingroup$
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
$endgroup$
– Math_QED
Nov 28 '18 at 12:25
1
$begingroup$
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
$endgroup$
– Jakobian
Nov 28 '18 at 14:25
1
$begingroup$
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
$endgroup$
– Randall
Nov 28 '18 at 18:03
1
$begingroup$
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
$endgroup$
– Arjihad
Nov 28 '18 at 18:08