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
    6 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    6 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
    6 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    6 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 6 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
    6 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    6 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
    6 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    6 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
6 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
6 hours ago














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




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












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




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










3 Answers
3






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









    $endgroup$









    • 2




      $begingroup$
      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
      4 hours ago





















    1












    $begingroup$

    Vectors in a plane $v,w$ can be written as column matrices: $$v=begin{bmatrix}v_1\v_2end{bmatrix}, w=begin{bmatrix}w_1\w_2end{bmatrix}.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=begin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbegin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      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









      $endgroup$


















        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 6 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









              $endgroup$









              • 2




                $begingroup$
                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
                4 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









              $endgroup$









              • 2




                $begingroup$
                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
                4 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 5 hours ago









              TaladrisTaladris

              4,92431933




              4,92431933








              • 2




                $begingroup$
                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
                4 hours ago
















              • 2




                $begingroup$
                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
                4 hours ago










              2




              2




              $begingroup$
              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
              4 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
              4 hours ago













              1












              $begingroup$

              Vectors in a plane $v,w$ can be written as column matrices: $$v=begin{bmatrix}v_1\v_2end{bmatrix}, w=begin{bmatrix}w_1\w_2end{bmatrix}.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=begin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbegin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Vectors in a plane $v,w$ can be written as column matrices: $$v=begin{bmatrix}v_1\v_2end{bmatrix}, w=begin{bmatrix}w_1\w_2end{bmatrix}.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=begin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbegin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Vectors in a plane $v,w$ can be written as column matrices: $$v=begin{bmatrix}v_1\v_2end{bmatrix}, w=begin{bmatrix}w_1\w_2end{bmatrix}.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=begin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbegin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Vectors in a plane $v,w$ can be written as column matrices: $$v=begin{bmatrix}v_1\v_2end{bmatrix}, w=begin{bmatrix}w_1\w_2end{bmatrix}.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=begin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbegin{bmatrix}v_1&w_1\v_2&w_2end{bmatrix}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 39 mins ago









                  edmedm

                  3,6331425




                  3,6331425






























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