Ryfujk

Why do phishing e-mails use faked e-mail addresses instead of the real one?

Is there a way to make cleveref distinguish two environments with the same counter?

How exactly does an Ethernet collision happen in the cable, since nodes use different circuits for Tx and Rx?

Use Mercury as quenching liquid for swords?

Is it a Cyclops number? "Nobody" knows!

What is better: yes / no radio, or simple checkbox?

Will expression retain the same definition if particle is changed?

Computation logic of Partway in TikZ

Smooth vector fields on a surface modulo diffeomorphisms

How to copy the rest of lines of a file to another file

Are these two graphs isomorphic? Why/Why not?

Traveling to heavily polluted city, what practical measures can I take to minimize impact?

Help! My Character is too much for her story!

"If + would" conditional in present perfect tense

Logistic regression BIC: what's the right N?

What would be the most expensive material to an intergalactic society?

Does an unused member variable take up memory?

Can the Witch Sight warlock invocation see through the Mirror Image spell?

How to educate team mate to take screenshots for bugs with out unwanted stuff

Should we avoid writing fiction about historical events without extensive research?

What do you call someone who likes to pick fights?

What will happen if my luggage gets delayed?

What can I do if someone tampers with my SSH public key?

How do I increase the number of TTY consoles?

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 5 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
  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
  

  
  
  
  

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 5 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
  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
  

  
  
  
  

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 5 hours ago

useruser

613

$endgroup$

share|cite|improve this question

asked 5 hours ago

useruser

613

share|cite|improve this question

asked 5 hours ago

useruser

613

asked 5 hours ago

useruser

613

asked 5 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 | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3141770%2fwhat-is-the-determinant-of-two-vectors%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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