Why are the non-diagonals in a Jacobian zeros?












1












$begingroup$


From the Matrix Calculus for Deep Learning, in the "Derivatives of vector element-wise binary operators" section, it says




Any time the general function is a vector, we know that $f_i(w)$ reduces to $f_i(w_i) = w_i$.




enter image description here



Why is it that $f_i(w)$ reduces to $f_i(w_i) = w_i$ for $y=w+x$ ?



Can someone show an example of why when i != j, the partial derivatives are zero?




Is this answer satisfactory? Something like this?



$frac{d}{dw_i}f_i(w_j)$ when i != j, the scalar derivative is 0






matrix-calculus jacobian







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migrated from stats.stackexchange.com Jan 15 at 14:13


This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.


















  • $begingroup$
    In the quoted passage I don't see anything about $ineq j.$ It's $w_i$ and $x_i.$ Also, the passage assumes at the start that $f(mathbf w) = mathbf w,$ so that simplifies a lot of things right away.
    $endgroup$
    – David K
    Jan 15 at 14:18


















1












$begingroup$


From the Matrix Calculus for Deep Learning, in the "Derivatives of vector element-wise binary operators" section, it says




Any time the general function is a vector, we know that $f_i(w)$ reduces to $f_i(w_i) = w_i$.




enter image description here



Why is it that $f_i(w)$ reduces to $f_i(w_i) = w_i$ for $y=w+x$ ?



Can someone show an example of why when i != j, the partial derivatives are zero?




Is this answer satisfactory? Something like this?



$frac{d}{dw_i}f_i(w_j)$ when i != j, the scalar derivative is 0






matrix-calculus jacobian







share|cite|improve this question









$endgroup$



migrated from stats.stackexchange.com Jan 15 at 14:13


This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.


















  • $begingroup$
    In the quoted passage I don't see anything about $ineq j.$ It's $w_i$ and $x_i.$ Also, the passage assumes at the start that $f(mathbf w) = mathbf w,$ so that simplifies a lot of things right away.
    $endgroup$
    – David K
    Jan 15 at 14:18
















1












1








1





$begingroup$


From the Matrix Calculus for Deep Learning, in the "Derivatives of vector element-wise binary operators" section, it says




Any time the general function is a vector, we know that $f_i(w)$ reduces to $f_i(w_i) = w_i$.




enter image description here



Why is it that $f_i(w)$ reduces to $f_i(w_i) = w_i$ for $y=w+x$ ?



Can someone show an example of why when i != j, the partial derivatives are zero?




Is this answer satisfactory? Something like this?



$frac{d}{dw_i}f_i(w_j)$ when i != j, the scalar derivative is 0






matrix-calculus jacobian







share|cite|improve this question









$endgroup$




From the Matrix Calculus for Deep Learning, in the "Derivatives of vector element-wise binary operators" section, it says




Any time the general function is a vector, we know that $f_i(w)$ reduces to $f_i(w_i) = w_i$.




enter image description here



Why is it that $f_i(w)$ reduces to $f_i(w_i) = w_i$ for $y=w+x$ ?



Can someone show an example of why when i != j, the partial derivatives are zero?




Is this answer satisfactory? Something like this?



$frac{d}{dw_i}f_i(w_j)$ when i != j, the scalar derivative is 0






matrix-calculus jacobian




matrix-calculus jacobian






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 13:37









alvasalvas

679




679




migrated from stats.stackexchange.com Jan 15 at 14:13


This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.









migrated from stats.stackexchange.com Jan 15 at 14:13


This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.














  • $begingroup$
    In the quoted passage I don't see anything about $ineq j.$ It's $w_i$ and $x_i.$ Also, the passage assumes at the start that $f(mathbf w) = mathbf w,$ so that simplifies a lot of things right away.
    $endgroup$
    – David K
    Jan 15 at 14:18




















  • $begingroup$
    In the quoted passage I don't see anything about $ineq j.$ It's $w_i$ and $x_i.$ Also, the passage assumes at the start that $f(mathbf w) = mathbf w,$ so that simplifies a lot of things right away.
    $endgroup$
    – David K
    Jan 15 at 14:18


















$begingroup$
In the quoted passage I don't see anything about $ineq j.$ It's $w_i$ and $x_i.$ Also, the passage assumes at the start that $f(mathbf w) = mathbf w,$ so that simplifies a lot of things right away.
$endgroup$
– David K
Jan 15 at 14:18






$begingroup$
In the quoted passage I don't see anything about $ineq j.$ It's $w_i$ and $x_i.$ Also, the passage assumes at the start that $f(mathbf w) = mathbf w,$ so that simplifies a lot of things right away.
$endgroup$
– David K
Jan 15 at 14:18












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$begingroup$

The reason that the off-diagonal elements are 0 is because the derivative of a constant is 0. The key bit is that $w_i$ looks like a constant if we take the derivative with respect to $w_j$ for $j ne i$.



Here is more of the matrix calculus article I co-authored:



enter image description here






share|cite|improve this answer









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    $begingroup$

    The reason that the off-diagonal elements are 0 is because the derivative of a constant is 0. The key bit is that $w_i$ looks like a constant if we take the derivative with respect to $w_j$ for $j ne i$.



    Here is more of the matrix calculus article I co-authored:



    enter image description here






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The reason that the off-diagonal elements are 0 is because the derivative of a constant is 0. The key bit is that $w_i$ looks like a constant if we take the derivative with respect to $w_j$ for $j ne i$.



      Here is more of the matrix calculus article I co-authored:



      enter image description here






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The reason that the off-diagonal elements are 0 is because the derivative of a constant is 0. The key bit is that $w_i$ looks like a constant if we take the derivative with respect to $w_j$ for $j ne i$.



        Here is more of the matrix calculus article I co-authored:



        enter image description here






        share|cite|improve this answer









        $endgroup$



        The reason that the off-diagonal elements are 0 is because the derivative of a constant is 0. The key bit is that $w_i$ looks like a constant if we take the derivative with respect to $w_j$ for $j ne i$.



        Here is more of the matrix calculus article I co-authored:



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 17:56









        Terence ParrTerence Parr

        1093




        1093






























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