Why is the Jacobi Matrix not a matrix?












1












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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







share|cite|improve this question









$endgroup$








  • 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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










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