What is the “determinant” of two vectors?












5












$begingroup$


I came across the notation $det(v,w)$ where $v$ and $w$ are vectors. Specifically, it was about the curvature of a plane curve:



$$kappa (t) = frac{det(gamma'(t), gamma''(t)) }{|gamma'(t)|^3}$$



What is it supposed to mean?






linear-algebra differential-geometry notation determinant







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are they 2D vectors? You may want to see this section of the Wikipedia article on Curvature. As you should be able to see from there, $color{blue}{det(v,w)}$ means $color{blue}{det begin{pmatrix} v_1 & w_1 \ v_2 & w_2end{pmatrix}}$.
    $endgroup$
    – Minus One-Twelfth
    4 hours ago












  • $begingroup$
    @MinusOne-Twelfth yes
    $endgroup$
    – user
    4 hours ago










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    4 hours ago
















5












$begingroup$


I came across the notation $det(v,w)$ where $v$ and $w$ are vectors. Specifically, it was about the curvature of a plane curve:



$$kappa (t) = frac{det(gamma'(t), gamma''(t)) }{|gamma'(t)|^3}$$



What is it supposed to mean?






linear-algebra differential-geometry notation determinant







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are they 2D vectors? You may want to see this section of the Wikipedia article on Curvature. As you should be able to see from there, $color{blue}{det(v,w)}$ means $color{blue}{det begin{pmatrix} v_1 & w_1 \ v_2 & w_2end{pmatrix}}$.
    $endgroup$
    – Minus One-Twelfth
    4 hours ago












  • $begingroup$
    @MinusOne-Twelfth yes
    $endgroup$
    – user
    4 hours ago










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    4 hours ago














5












5








5


1



$begingroup$


I came across the notation $det(v,w)$ where $v$ and $w$ are vectors. Specifically, it was about the curvature of a plane curve:



$$kappa (t) = frac{det(gamma'(t), gamma''(t)) }{|gamma'(t)|^3}$$



What is it supposed to mean?






linear-algebra differential-geometry notation determinant







share|cite|improve this question









$endgroup$




I came across the notation $det(v,w)$ where $v$ and $w$ are vectors. Specifically, it was about the curvature of a plane curve:



$$kappa (t) = frac{det(gamma'(t), gamma''(t)) }{|gamma'(t)|^3}$$



What is it supposed to mean?






linear-algebra differential-geometry notation determinant




linear-algebra differential-geometry notation determinant






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









useruser

613




613












  • $begingroup$
    Are they 2D vectors? You may want to see this section of the Wikipedia article on Curvature. As you should be able to see from there, $color{blue}{det(v,w)}$ means $color{blue}{det begin{pmatrix} v_1 & w_1 \ v_2 & w_2end{pmatrix}}$.
    $endgroup$
    – Minus One-Twelfth
    4 hours ago












  • $begingroup$
    @MinusOne-Twelfth yes
    $endgroup$
    – user
    4 hours ago










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    4 hours ago


















  • $begingroup$
    Are they 2D vectors? You may want to see this section of the Wikipedia article on Curvature. As you should be able to see from there, $color{blue}{det(v,w)}$ means $color{blue}{det begin{pmatrix} v_1 & w_1 \ v_2 & w_2end{pmatrix}}$.
    $endgroup$
    – Minus One-Twelfth
    4 hours ago












  • $begingroup$
    @MinusOne-Twelfth yes
    $endgroup$
    – user
    4 hours ago










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    4 hours ago
















$begingroup$
Are they 2D vectors? You may want to see this section of the Wikipedia article on Curvature. As you should be able to see from there, $color{blue}{det(v,w)}$ means $color{blue}{det begin{pmatrix} v_1 & w_1 \ v_2 & w_2end{pmatrix}}$.
$endgroup$
– Minus One-Twelfth
4 hours ago






$begingroup$
Are they 2D vectors? You may want to see this section of the Wikipedia article on Curvature. As you should be able to see from there, $color{blue}{det(v,w)}$ means $color{blue}{det begin{pmatrix} v_1 & w_1 \ v_2 & w_2end{pmatrix}}$.
$endgroup$
– Minus One-Twelfth
4 hours ago














$begingroup$
@MinusOne-Twelfth yes
$endgroup$
– user
4 hours ago




$begingroup$
@MinusOne-Twelfth yes
$endgroup$
– user
4 hours ago












$begingroup$
That is the determinant of their components.
$endgroup$
– Bernard
4 hours ago




$begingroup$
That is the determinant of their components.
$endgroup$
– Bernard
4 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

They formed a matrix by stacking $gamma'(t)$ and $gamma''(t)$ next to each other as column vectors. You can also regard it as the cross product of the two vectors if you extend both with a $z=0$ coordinate and take the z component of the resulting vector (that way you can relate it to the 3d formula in a way).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    In general, the determinant of $n$ vectors $v_1$, $v_2$, $dots$, $v_n$ in $mathbb R^n$ is the determinant of the matrix whose columns are $v_1$, $dots$, $v_n$ (in that order).



    Seen as an application whose inputs are vectors, the determinant has nice properties:




    1. multilinear, that is linear in each variable:
      $$det(v_1,dots, a v_j+b w_j,dots,v_n)
      =
      a det(v_1,dots, v_j,dots,v_n)
      + bdet(v_1,dots, w_j,dots,v_n)$$


    2. alternating: switching two vectors transforms the determinant in its opposite



    $$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$




    1. The value on the canonical basis $(e_1,dots,e_n)$ of $mathbb R^n$ is $1$.


    $$det(e_1,dots,e_n) = 1 $$



    Actually, it can be proved that the determinant is the unique alternating multilinear form whose value on the canonical basis is $1$. Many (most?) Linear Algebra books use this as a definition of the determinant (before extending the definition to matrices and then linear applications). I think it is equivalent but more satisfying than introducing the determinant by the strange "expansion along a row" formula most PreCalculus textbook use.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix.
      $endgroup$
      – lightxbulb
      3 hours ago













    Your Answer





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    2 Answers
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    2 Answers
    2






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    active

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    $begingroup$

    They formed a matrix by stacking $gamma'(t)$ and $gamma''(t)$ next to each other as column vectors. You can also regard it as the cross product of the two vectors if you extend both with a $z=0$ coordinate and take the z component of the resulting vector (that way you can relate it to the 3d formula in a way).






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      They formed a matrix by stacking $gamma'(t)$ and $gamma''(t)$ next to each other as column vectors. You can also regard it as the cross product of the two vectors if you extend both with a $z=0$ coordinate and take the z component of the resulting vector (that way you can relate it to the 3d formula in a way).






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        They formed a matrix by stacking $gamma'(t)$ and $gamma''(t)$ next to each other as column vectors. You can also regard it as the cross product of the two vectors if you extend both with a $z=0$ coordinate and take the z component of the resulting vector (that way you can relate it to the 3d formula in a way).






        share|cite|improve this answer









        $endgroup$



        They formed a matrix by stacking $gamma'(t)$ and $gamma''(t)$ next to each other as column vectors. You can also regard it as the cross product of the two vectors if you extend both with a $z=0$ coordinate and take the z component of the resulting vector (that way you can relate it to the 3d formula in a way).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        lightxbulblightxbulb

        1,115311




        1,115311























            1












            $begingroup$

            In general, the determinant of $n$ vectors $v_1$, $v_2$, $dots$, $v_n$ in $mathbb R^n$ is the determinant of the matrix whose columns are $v_1$, $dots$, $v_n$ (in that order).



            Seen as an application whose inputs are vectors, the determinant has nice properties:




            1. multilinear, that is linear in each variable:
              $$det(v_1,dots, a v_j+b w_j,dots,v_n)
              =
              a det(v_1,dots, v_j,dots,v_n)
              + bdet(v_1,dots, w_j,dots,v_n)$$


            2. alternating: switching two vectors transforms the determinant in its opposite



            $$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$




            1. The value on the canonical basis $(e_1,dots,e_n)$ of $mathbb R^n$ is $1$.


            $$det(e_1,dots,e_n) = 1 $$



            Actually, it can be proved that the determinant is the unique alternating multilinear form whose value on the canonical basis is $1$. Many (most?) Linear Algebra books use this as a definition of the determinant (before extending the definition to matrices and then linear applications). I think it is equivalent but more satisfying than introducing the determinant by the strange "expansion along a row" formula most PreCalculus textbook use.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix.
              $endgroup$
              – lightxbulb
              3 hours ago


















            1












            $begingroup$

            In general, the determinant of $n$ vectors $v_1$, $v_2$, $dots$, $v_n$ in $mathbb R^n$ is the determinant of the matrix whose columns are $v_1$, $dots$, $v_n$ (in that order).



            Seen as an application whose inputs are vectors, the determinant has nice properties:




            1. multilinear, that is linear in each variable:
              $$det(v_1,dots, a v_j+b w_j,dots,v_n)
              =
              a det(v_1,dots, v_j,dots,v_n)
              + bdet(v_1,dots, w_j,dots,v_n)$$


            2. alternating: switching two vectors transforms the determinant in its opposite



            $$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$




            1. The value on the canonical basis $(e_1,dots,e_n)$ of $mathbb R^n$ is $1$.


            $$det(e_1,dots,e_n) = 1 $$



            Actually, it can be proved that the determinant is the unique alternating multilinear form whose value on the canonical basis is $1$. Many (most?) Linear Algebra books use this as a definition of the determinant (before extending the definition to matrices and then linear applications). I think it is equivalent but more satisfying than introducing the determinant by the strange "expansion along a row" formula most PreCalculus textbook use.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix.
              $endgroup$
              – lightxbulb
              3 hours ago
















            1












            1








            1





            $begingroup$

            In general, the determinant of $n$ vectors $v_1$, $v_2$, $dots$, $v_n$ in $mathbb R^n$ is the determinant of the matrix whose columns are $v_1$, $dots$, $v_n$ (in that order).



            Seen as an application whose inputs are vectors, the determinant has nice properties:




            1. multilinear, that is linear in each variable:
              $$det(v_1,dots, a v_j+b w_j,dots,v_n)
              =
              a det(v_1,dots, v_j,dots,v_n)
              + bdet(v_1,dots, w_j,dots,v_n)$$


            2. alternating: switching two vectors transforms the determinant in its opposite



            $$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$




            1. The value on the canonical basis $(e_1,dots,e_n)$ of $mathbb R^n$ is $1$.


            $$det(e_1,dots,e_n) = 1 $$



            Actually, it can be proved that the determinant is the unique alternating multilinear form whose value on the canonical basis is $1$. Many (most?) Linear Algebra books use this as a definition of the determinant (before extending the definition to matrices and then linear applications). I think it is equivalent but more satisfying than introducing the determinant by the strange "expansion along a row" formula most PreCalculus textbook use.






            share|cite|improve this answer









            $endgroup$



            In general, the determinant of $n$ vectors $v_1$, $v_2$, $dots$, $v_n$ in $mathbb R^n$ is the determinant of the matrix whose columns are $v_1$, $dots$, $v_n$ (in that order).



            Seen as an application whose inputs are vectors, the determinant has nice properties:




            1. multilinear, that is linear in each variable:
              $$det(v_1,dots, a v_j+b w_j,dots,v_n)
              =
              a det(v_1,dots, v_j,dots,v_n)
              + bdet(v_1,dots, w_j,dots,v_n)$$


            2. alternating: switching two vectors transforms the determinant in its opposite



            $$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$




            1. The value on the canonical basis $(e_1,dots,e_n)$ of $mathbb R^n$ is $1$.


            $$det(e_1,dots,e_n) = 1 $$



            Actually, it can be proved that the determinant is the unique alternating multilinear form whose value on the canonical basis is $1$. Many (most?) Linear Algebra books use this as a definition of the determinant (before extending the definition to matrices and then linear applications). I think it is equivalent but more satisfying than introducing the determinant by the strange "expansion along a row" formula most PreCalculus textbook use.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            TaladrisTaladris

            4,92431933




            4,92431933








            • 1




              $begingroup$
              I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix.
              $endgroup$
              – lightxbulb
              3 hours ago
















            • 1




              $begingroup$
              I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix.
              $endgroup$
              – lightxbulb
              3 hours ago










            1




            1




            $begingroup$
            I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix.
            $endgroup$
            – lightxbulb
            3 hours ago






            $begingroup$
            I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix.
            $endgroup$
            – lightxbulb
            3 hours ago




















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