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







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$endgroup$












  • $begingroup$
    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
    5 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    5 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
    5 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    5 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 5 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
    5 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    5 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
    5 hours ago












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










  • $begingroup$
    That is the determinant of their components.
    $endgroup$
    – Bernard
    5 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
5 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
5 hours ago














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




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












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




$begingroup$
That is the determinant of their components.
$endgroup$
– Bernard
5 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
      3 hours ago





















    0












    $begingroup$

    Sometimes determinant of $n$ by $n$ matrix is defined as an alternating $n$-linear form; it assigns a number to each $n$-tuple of vectors $(v_1,v_2,cdots, v_n)$ following the axioms that




    1. For each $i=1,2,cdots, n$, the map $v_imapsto det(v_1,v_2,cdots, v_n)$ is linear.


    2. $det(v_1,cdots,v_i,cdots,v_j,cdots,v_n)=-det(v_1,cdots,v_j,cdots,v_i,cdots,v_n)$ for each $ine j$


    3. $det(e_1,e_2,cdots,e_n)=1$ where $(e_i)_{i=1}^n$ is the standard basis.



    Following this notation, we have that
    $$begin{align*}
    det(v,w)&=det(v_1e_1+v_2e_2,w_1e_1+w_2e_2)
    \=&v_1w_1det(e_1,e_1)+v_1w_2det(e_1,e_2)\&+v_2w_1det(e_2,e_1)+v_2w_2det(e_2,e_2)\=&v_1w_2det(e_1,e_2)-v_2w_1det(e_1,e_2)\=&v_1w_2-v_2w_1\=&left|begin{array}{cc}v_1 &w_1\v_2&w_2end{array}right|,
    end{align*}$$     so it is not different from the determinant of the matrix $(v, w)$ formed by letting $v,w$ as its columns.






    share|cite









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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 5 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
                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









              $endgroup$









              • 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
                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








              • 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
                3 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
                3 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
              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













              0












              $begingroup$

              Sometimes determinant of $n$ by $n$ matrix is defined as an alternating $n$-linear form; it assigns a number to each $n$-tuple of vectors $(v_1,v_2,cdots, v_n)$ following the axioms that




              1. For each $i=1,2,cdots, n$, the map $v_imapsto det(v_1,v_2,cdots, v_n)$ is linear.


              2. $det(v_1,cdots,v_i,cdots,v_j,cdots,v_n)=-det(v_1,cdots,v_j,cdots,v_i,cdots,v_n)$ for each $ine j$


              3. $det(e_1,e_2,cdots,e_n)=1$ where $(e_i)_{i=1}^n$ is the standard basis.



              Following this notation, we have that
              $$begin{align*}
              det(v,w)&=det(v_1e_1+v_2e_2,w_1e_1+w_2e_2)
              \=&v_1w_1det(e_1,e_1)+v_1w_2det(e_1,e_2)\&+v_2w_1det(e_2,e_1)+v_2w_2det(e_2,e_2)\=&v_1w_2det(e_1,e_2)-v_2w_1det(e_1,e_2)\=&v_1w_2-v_2w_1\=&left|begin{array}{cc}v_1 &w_1\v_2&w_2end{array}right|,
              end{align*}$$               so it is not different from the determinant of the matrix $(v, w)$ formed by letting $v,w$ as its columns.






              share|cite









              $endgroup$


















                0












                $begingroup$

                Sometimes determinant of $n$ by $n$ matrix is defined as an alternating $n$-linear form; it assigns a number to each $n$-tuple of vectors $(v_1,v_2,cdots, v_n)$ following the axioms that




                1. For each $i=1,2,cdots, n$, the map $v_imapsto det(v_1,v_2,cdots, v_n)$ is linear.


                2. $det(v_1,cdots,v_i,cdots,v_j,cdots,v_n)=-det(v_1,cdots,v_j,cdots,v_i,cdots,v_n)$ for each $ine j$


                3. $det(e_1,e_2,cdots,e_n)=1$ where $(e_i)_{i=1}^n$ is the standard basis.



                Following this notation, we have that
                $$begin{align*}
                det(v,w)&=det(v_1e_1+v_2e_2,w_1e_1+w_2e_2)
                \=&v_1w_1det(e_1,e_1)+v_1w_2det(e_1,e_2)\&+v_2w_1det(e_2,e_1)+v_2w_2det(e_2,e_2)\=&v_1w_2det(e_1,e_2)-v_2w_1det(e_1,e_2)\=&v_1w_2-v_2w_1\=&left|begin{array}{cc}v_1 &w_1\v_2&w_2end{array}right|,
                end{align*}$$                 so it is not different from the determinant of the matrix $(v, w)$ formed by letting $v,w$ as its columns.






                share|cite









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Sometimes determinant of $n$ by $n$ matrix is defined as an alternating $n$-linear form; it assigns a number to each $n$-tuple of vectors $(v_1,v_2,cdots, v_n)$ following the axioms that




                  1. For each $i=1,2,cdots, n$, the map $v_imapsto det(v_1,v_2,cdots, v_n)$ is linear.


                  2. $det(v_1,cdots,v_i,cdots,v_j,cdots,v_n)=-det(v_1,cdots,v_j,cdots,v_i,cdots,v_n)$ for each $ine j$


                  3. $det(e_1,e_2,cdots,e_n)=1$ where $(e_i)_{i=1}^n$ is the standard basis.



                  Following this notation, we have that
                  $$begin{align*}
                  det(v,w)&=det(v_1e_1+v_2e_2,w_1e_1+w_2e_2)
                  \=&v_1w_1det(e_1,e_1)+v_1w_2det(e_1,e_2)\&+v_2w_1det(e_2,e_1)+v_2w_2det(e_2,e_2)\=&v_1w_2det(e_1,e_2)-v_2w_1det(e_1,e_2)\=&v_1w_2-v_2w_1\=&left|begin{array}{cc}v_1 &w_1\v_2&w_2end{array}right|,
                  end{align*}$$                   so it is not different from the determinant of the matrix $(v, w)$ formed by letting $v,w$ as its columns.






                  share|cite









                  $endgroup$



                  Sometimes determinant of $n$ by $n$ matrix is defined as an alternating $n$-linear form; it assigns a number to each $n$-tuple of vectors $(v_1,v_2,cdots, v_n)$ following the axioms that




                  1. For each $i=1,2,cdots, n$, the map $v_imapsto det(v_1,v_2,cdots, v_n)$ is linear.


                  2. $det(v_1,cdots,v_i,cdots,v_j,cdots,v_n)=-det(v_1,cdots,v_j,cdots,v_i,cdots,v_n)$ for each $ine j$


                  3. $det(e_1,e_2,cdots,e_n)=1$ where $(e_i)_{i=1}^n$ is the standard basis.



                  Following this notation, we have that
                  $$begin{align*}
                  det(v,w)&=det(v_1e_1+v_2e_2,w_1e_1+w_2e_2)
                  \=&v_1w_1det(e_1,e_1)+v_1w_2det(e_1,e_2)\&+v_2w_1det(e_2,e_1)+v_2w_2det(e_2,e_2)\=&v_1w_2det(e_1,e_2)-v_2w_1det(e_1,e_2)\=&v_1w_2-v_2w_1\=&left|begin{array}{cc}v_1 &w_1\v_2&w_2end{array}right|,
                  end{align*}$$                   so it is not different from the determinant of the matrix $(v, w)$ formed by letting $v,w$ as its columns.







                  share|cite












                  share|cite



                  share|cite










                  answered 1 min ago









                  SongSong

                  16.8k21145




                  16.8k21145






























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