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What is the “determinant” of two vectors?

Inversion of Hopf's UmlaufsatzWhat does the definition of curvature mean?Calculate the determinant when the sum of odd rows $=$ the sum of even rowsWhat does the notation $P[Xin dx]$ mean?Relations between curvature and area of simple closed plane curves.Prove the curvature of a level set equals divergence of the normalized gradientDefine a parametrized curve $beta:(a,b)rightarrowmathbb R^3$ by $beta(t)=fracdgamma(t)dt$Understanding a particular case of modifying curvature and torsion as opposed to modifying the curveIntegral of the ratio of torsion and curvatureProve that the planar curve obtained by projecting $alpha$ into its osculating plane at $P$ has the same curvature at $P$ as $alpha$.

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:

$$kappa (t) = fracdet(gamma'(t), gamma''(t)) $$

What is it supposed to mean?

linear-algebra differential-geometry notation determinant

share|cite|improve this question

asked 7 hours ago

useruser

613

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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, $colorbluedet(v,w)$ means $colorbluedet beginpmatrix v_1 & w_1 \ v_2 & w_2endpmatrix$.
  $endgroup$
  – Minus One-Twelfth
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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:

$$kappa (t) = fracdet(gamma'(t), gamma''(t)) $$

What is it supposed to mean?

linear-algebra differential-geometry notation determinant

share|cite|improve this question

asked 7 hours ago

useruser

613

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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, $colorbluedet(v,w)$ means $colorbluedet beginpmatrix v_1 & w_1 \ v_2 & w_2endpmatrix$.
  $endgroup$
  – Minus One-Twelfth
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

$$kappa (t) = fracdet(gamma'(t), gamma''(t)) $$

What is it supposed to mean?

linear-algebra differential-geometry notation determinant

share|cite|improve this question

asked 7 hours ago

useruser

613

$endgroup$

share|cite|improve this question

asked 7 hours ago

useruser

613

share|cite|improve this question

asked 7 hours ago

useruser

613

asked 7 hours ago

useruser

613

asked 7 hours ago

useruser

613

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

answered 7 hours ago

lightxbulblightxbulb

1,115311

$endgroup$

add a comment | 

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

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

3. 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

answered 6 hours ago

TaladrisTaladris

4,92431933

$endgroup$

• 
  
  
  

  
  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
  5 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Vectors in a plane $v,w$ can be written as column matrices: $$v=beginbmatrixv_1\v_2endbmatrix, w=beginbmatrixw_1\w_2endbmatrix.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=beginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$

share|cite|improve this answer

answered 1 hour ago

edmedm

3,6331425

$endgroup$

add a comment | 

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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

answered 7 hours ago

lightxbulblightxbulb

1,115311

$endgroup$

add a comment | 

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

answered 7 hours ago

lightxbulblightxbulb

1,115311

$endgroup$

add a comment | 

$begingroup$

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

answered 7 hours ago

lightxbulblightxbulb

1,115311

$endgroup$

share|cite|improve this answer

answered 7 hours ago

lightxbulblightxbulb

1,115311

answered 7 hours ago

lightxbulblightxbulb

1,115311

answered 7 hours ago

lightxbulblightxbulb

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

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

3. 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

answered 6 hours ago

TaladrisTaladris

4,92431933

$endgroup$

• 
  
  
  

  
  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
  5 hours ago
  
  

  
  
  
  

add a comment | 

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

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

3. 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

answered 6 hours ago

TaladrisTaladris

4,92431933

$endgroup$

• 
  
  
  

  
  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
  5 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

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

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

3. 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

answered 6 hours ago

TaladrisTaladris

4,92431933

$endgroup$

share|cite|improve this answer

answered 6 hours ago

TaladrisTaladris

4,92431933

answered 6 hours ago

TaladrisTaladris

4,92431933

answered 6 hours ago

TaladrisTaladris

4,92431933

1

$begingroup$

Vectors in a plane $v,w$ can be written as column matrices: $$v=beginbmatrixv_1\v_2endbmatrix, w=beginbmatrixw_1\w_2endbmatrix.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=beginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$

share|cite|improve this answer

answered 1 hour ago

edmedm

3,6331425

$endgroup$

add a comment | 

1

$begingroup$

Vectors in a plane $v,w$ can be written as column matrices: $$v=beginbmatrixv_1\v_2endbmatrix, w=beginbmatrixw_1\w_2endbmatrix.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=beginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$

share|cite|improve this answer

answered 1 hour ago

edmedm

3,6331425

$endgroup$

add a comment | 

$begingroup$

Vectors in a plane $v,w$ can be written as column matrices: $$v=beginbmatrixv_1\v_2endbmatrix, w=beginbmatrixw_1\w_2endbmatrix.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=beginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$ The determinant $det(v,w)$ is simply the determinant of this square matrix: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$

share|cite|improve this answer

answered 1 hour ago

edmedm

3,6331425

$endgroup$

share|cite|improve this answer

answered 1 hour ago

edmedm

3,6331425

answered 1 hour ago

edmedm

3,6331425

answered 1 hour ago

edmedm

3,6331425