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 $alpha$.
Smooth vector fields on a surface modulo diffeomorphisms
Why aren't there more Gauls like Obelix?
"If + would" conditional in present perfect tense
How do we create new idioms and use them in a novel?
How to educate team mate to take screenshots for bugs with out unwanted stuff
Was it really inappropriate to write a pull request for the company I interviewed with?
cannot log in to the server after changing SSH port
Locked Away- What am I?
One circle's diameter is different from others within a series of circles
Short scifi story where reproductive organs are converted to produce "materials", pregnant protagonist is "found fit" to be a mother
What would be the most expensive material to an intergalactic society?
Writing text next to a table
Why restrict private health insurance?
Giving a career talk in my old university, how prominently should I tell students my salary?
The (Easy) Road to Code
Has a sovereign Communist government ever run, and conceded loss, on a fair election?
Traveling to heavily polluted city, what practical measures can I take to minimize impact?
Playing a 7-string guitar song on a 6-string guitar
Does the US political system, in principle, allow for a no-party system?
How is it possible to drive VGA displays at such high pixel clock frequencies?
What does *dead* mean in *What do you mean, dead?*?
How to write a chaotic neutral protagonist and prevent my readers from thinking they are evil?
Converting from "matrix" data into "coordinate" data
I am the person who abides by rules, but breaks the rules. Who am I?
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 $alpha$.
$begingroup$
I came across the notation $det(v,w)$ where $v$ and $w$ are vectors. Specifically, it was about the curvature of a plane curve:
$$kappa (t) = fracdet(gamma'(t), gamma''(t)) ^3$$
What is it supposed to mean?
linear-algebra differential-geometry notation determinant
asked 6 hours ago
useruser
613
$endgroup$
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:
$$kappa (t) = fracdet(gamma'(t), gamma''(t)) ^3$$
What is it supposed to mean?
linear-algebra differential-geometry notation determinant
asked 6 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
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
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:
$$kappa (t) = fracdet(gamma'(t), gamma''(t)) ^3$$
What is it supposed to mean?
linear-algebra differential-geometry notation determinant
asked 6 hours ago
useruser
613
$endgroup$
I came across the notation $det(v,w)$ where $v$ and $w$ are vectors. Specifically, it was about the curvature of a plane curve:
$$kappa (t) = fracdet(gamma'(t), gamma''(t)) ^3$$
What is it supposed to mean?
linear-algebra differential-geometry notation determinant
linear-algebra differential-geometry notation determinant
asked 6 hours ago
useruser
613
asked 6 hours ago
useruser
613
asked 6 hours ago
useruser
613
asked 6 hours ago
useruser
613
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, $colorbluedet(v,w)$ means $colorbluedet beginpmatrix v_1 & w_1 \ v_2 & w_2endpmatrix$.
$endgroup$
– 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
add a comment |
$begingroup$
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
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, $colorbluedet(v,w)$ means $colorbluedet beginpmatrix v_1 & w_1 \ v_2 & w_2endpmatrix$.
$endgroup$
– 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, $colorbluedet(v,w)$ means $colorbluedet beginpmatrix v_1 & w_1 \ v_2 & w_2endpmatrix$.
$endgroup$
– 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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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).
answered 5 hours ago
lightxbulblightxbulb
1,115311
$endgroup$
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:
multilinear, that is linear in each variable:
$$det(v_1,dots, a v_j+b w_j,dots,v_n)
=
a det(v_1,dots, v_j,dots,v_n)
+ bdet(v_1,dots, w_j,dots,v_n)$$alternating: switching two vectors transforms the determinant in its opposite
$$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$
- 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.
answered 5 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
4 hours ago
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: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$
answered 17 mins ago
edmedm
3,6231425
$endgroup$
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
);
);
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
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
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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).
answered 5 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).
answered 5 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).
answered 5 hours ago
lightxbulblightxbulb
1,115311
$endgroup$
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).
answered 5 hours ago
lightxbulblightxbulb
1,115311
answered 5 hours ago
lightxbulblightxbulb
1,115311
answered 5 hours ago
lightxbulblightxbulb
1,115311
answered 5 hours ago
lightxbulblightxbulb
1,115311
1,115311
add a comment |
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:
multilinear, that is linear in each variable:
$$det(v_1,dots, a v_j+b w_j,dots,v_n)
=
a det(v_1,dots, v_j,dots,v_n)
+ bdet(v_1,dots, w_j,dots,v_n)$$alternating: switching two vectors transforms the determinant in its opposite
$$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$
- 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.
answered 5 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
4 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:
multilinear, that is linear in each variable:
$$det(v_1,dots, a v_j+b w_j,dots,v_n)
=
a det(v_1,dots, v_j,dots,v_n)
+ bdet(v_1,dots, w_j,dots,v_n)$$alternating: switching two vectors transforms the determinant in its opposite
$$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$
- 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.
answered 5 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
4 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:
multilinear, that is linear in each variable:
$$det(v_1,dots, a v_j+b w_j,dots,v_n)
=
a det(v_1,dots, v_j,dots,v_n)
+ bdet(v_1,dots, w_j,dots,v_n)$$alternating: switching two vectors transforms the determinant in its opposite
$$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$
- 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.
answered 5 hours ago
TaladrisTaladris
4,92431933
$endgroup$
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:
multilinear, that is linear in each variable:
$$det(v_1,dots, a v_j+b w_j,dots,v_n)
=
a det(v_1,dots, v_j,dots,v_n)
+ bdet(v_1,dots, w_j,dots,v_n)$$alternating: switching two vectors transforms the determinant in its opposite
$$det(v_1,dots, v_i, dots, v_j,dots,v_n) = det(v_1,dots, v_j, dots, v_i,dots,v_n)$$
- 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.
answered 5 hours ago
TaladrisTaladris
4,92431933
answered 5 hours ago
TaladrisTaladris
4,92431933
answered 5 hours ago
TaladrisTaladris
4,92431933
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
add a comment |
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
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: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$
answered 17 mins ago
edmedm
3,6231425
$endgroup$
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: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$
answered 17 mins ago
edmedm
3,6231425
$endgroup$
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: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$
answered 17 mins ago
edmedm
3,6231425
$endgroup$
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: $$det(v,w)=detbeginbmatrixv_1&w_1\v_2&w_2endbmatrix.$$
answered 17 mins ago
edmedm
3,6231425
answered 17 mins ago
edmedm
3,6231425
answered 17 mins ago
edmedm
3,6231425
answered 17 mins ago
edmedm
3,6231425
3,6231425
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
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