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Translation of 答えを知っている人はいませんでした

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How do you make a gun that shoots melee weapons and/or swords?

Writing text next to a table

If nine coins are tossed, what is the probability that the number of heads is even?

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 $beta(t)=frac{dgamma(t)}{dt}$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:

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

What is it supposed to mean?

linear-algebra differential-geometry notation determinant

share|cite|improve this question

asked 4 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, $color{blue}{det(v,w)}$ 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
  

  
  
  
  

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:

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

What is it supposed to mean?

linear-algebra differential-geometry notation determinant

share|cite|improve this question

asked 4 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, $color{blue}{det(v,w)}$ 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
  

  
  
  
  

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:

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

What is it supposed to mean?

linear-algebra differential-geometry notation determinant

share|cite|improve this question

asked 4 hours ago

useruser

613

$endgroup$

share|cite|improve this question

asked 4 hours ago

useruser

613

share|cite|improve this question

asked 4 hours ago

useruser

613

asked 4 hours ago

useruser

613

asked 4 hours ago

useruser

613

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

answered 4 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 4 hours ago

TaladrisTaladris

4,92431933

$endgroup$

• 
  
  
  

  
  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
  
  
  

  
  
  
  

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 4 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 4 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 4 hours ago

lightxbulblightxbulb

1,115311

$endgroup$

share|cite|improve this answer

answered 4 hours ago

lightxbulblightxbulb

1,115311

answered 4 hours ago

lightxbulblightxbulb

1,115311

answered 4 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 4 hours ago

TaladrisTaladris

4,92431933

$endgroup$

• 
  
  
  

  
  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
  
  
  

  
  
  
  

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 4 hours ago

TaladrisTaladris

4,92431933

$endgroup$

• 
  
  
  

  
  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
  
  
  

  
  
  
  

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 4 hours ago

TaladrisTaladris

4,92431933

$endgroup$

share|cite|improve this answer

answered 4 hours ago

TaladrisTaladris

4,92431933

answered 4 hours ago

TaladrisTaladris

4,92431933

answered 4 hours ago

TaladrisTaladris

4,92431933