why determinant is volume of parallelepiped in any dimensions
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for $n = 2,$ I can visualize that the determinant $n times n$ matrix is the area of the parallelograms by actually calculate the area by coordinates. But how can one easily realize that it is true for any dimensions?
geometry matrices determinant
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add a comment |
$begingroup$
for $n = 2,$ I can visualize that the determinant $n times n$ matrix is the area of the parallelograms by actually calculate the area by coordinates. But how can one easily realize that it is true for any dimensions?
geometry matrices determinant
$endgroup$
add a comment |
$begingroup$
for $n = 2,$ I can visualize that the determinant $n times n$ matrix is the area of the parallelograms by actually calculate the area by coordinates. But how can one easily realize that it is true for any dimensions?
geometry matrices determinant
$endgroup$
for $n = 2,$ I can visualize that the determinant $n times n$ matrix is the area of the parallelograms by actually calculate the area by coordinates. But how can one easily realize that it is true for any dimensions?
geometry matrices determinant
geometry matrices determinant
edited Jun 23 '13 at 13:47
Sujaan Kunalan
7,151123972
7,151123972
asked Jun 23 '13 at 13:30
ahalaahala
1,06811023
1,06811023
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6 Answers
6
active
oldest
votes
$begingroup$
If the column vectors are linearly dependent, both the determinant and the volume are zero.
So assume linear independence.
The determinant remains unchanged when adding multiples of one column to another. This corresponds to a skew translation of the parallelepiped, which does not affect its volume.
By a finite sequence of such operations, you can transform your matrix to diagonal form, where the relation between determinant (=product of diagonal entries) and volume of a "rectangle" (=product of side lengths) is apparent.
$endgroup$
1
$begingroup$
thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant?
$endgroup$
– ahala
Jun 23 '13 at 15:34
1
$begingroup$
Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional
$endgroup$
– Hagen von Eitzen
May 2 '14 at 11:58
$begingroup$
I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear.
$endgroup$
– Mitch
Oct 19 '18 at 19:52
add a comment |
$begingroup$
Here is the same argument as Muphrid's, perhaps written in an elementary way.
Apply Gram-Schmidt orthogonalization to ${v_{1},ldots,v_{n}}$, so that
begin{eqnarray*}
v_{1} & = & v_{1}\
v_{2} & = & c_{12}v_{1}+v_{2}^{perp}\
v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp}\
& vdots
end{eqnarray*}
where $v_{2}^{perp}$ is orthogonal to $v_{1}$; and $v_{3}^{perp}$
is orthogonal to $spanleft{ v_{1},v_{2}right} $, etc.
Since determinant is multilinear, anti-symmetric, then
begin{eqnarray*}
detleft(v_{1},v_{2},v_{3},ldots,v_{n}right) & = & detleft(v_{1},c_{12}v_{1}+v_{2}^{perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp},ldotsright)\
& = & detleft(v_{1},v_{2}^{perp},v_{3}^{perp},ldots,v_{n}^{perp}right)\
& = & mbox{signed volume}left(v_{1},ldots,v_{n}right)
end{eqnarray*}
$endgroup$
add a comment |
$begingroup$
In 2d, you calculate the area of a parallelogram spanned by two vectors using the cross product. In 3d, you calculate the volume of a parallelepiped using the triple scalar product. Both of these can be written in terms of a determinant, but it's probably not clear to you what the proper generalization is to higher dimensions.
That generalization is called the wedge product. Given $n$ vectors $v_1, v_2, ldots, v_n$, the wedge product $v_1 wedge v_2 wedge ldots wedge v_n$ is called an $n$-vector, and it has as its magnitude the $n$-volume of that $n$-parallelepiped.
What is the relationship between the wedge product and the determinant? Quite simple, actually. There is a natural generalization of linear maps to work on $k$-vectors. Given a linear map $underline T$ (which can be represented as a matrix), the action of that map on a $k$-vector is defined as
$$underline T(v_1 wedge v_2 wedge ldots wedge v_k) equiv underline T(v_1) wedge underline T(v_2) wedge ldots wedge underline T(v_k)$$
When talking about $n$-vectors in an $n$-dimensional space, it's important to realize that the "vector space" of these $n$-vectors is in fact one-dimensional. That is, if you think about volume, there is only one such unit volume in a given space, and all other volumes are just scalar multiples of it. Hence, when we talk about the action of a linear map on an $n$-vector, we can see that
$$underline T(v_1 wedge v_2 wedge ldots wedge v_n) = alpha [v_1 wedge v_2 wedge ldots wedge v_n]$$
for some scalar $alpha$. In fact, this is a coordinate system independent definition of the determinant!
When you build a matrix out of $n$ vectors $f_1, f_2, ldots, f_n$ as the matrix's columns, what you're really doing is the following: you're saying that, if you have a basis $e_1, e_2, ldots, e_n$, then you're defining a map $underline T$ such that $underline T(e_1) = f_1$, $underline T(e_2) = f_2$, and so on. So when you input $e_1 wedge e_2 wedge ldots wedge e_n$, you get
$$underline T(e_1 wedge e_2 wedge ldots wedge e_n) = (det underline T) e_1 wedge e_2 wedge ldots wedge e_n= f_1 wedge f_2 wedge ldots wedge f_n$$
This is how you can use a matrix determinant to calculate volumes: it's just an easy way of constructing something that automatically computes the wedge product.
Edit: how one can see that the wedge product accurately gives the volume of a parallelepiped. Any vector can be broken down into perpendicular and parallel parts with respect to another vector, to a plane, and so on (or to any $k$-vector). As such, if I have two vectors $a$ and $b$, then the wedge product $a wedge b = a wedge b_perp$, where $b_perp$ is effectively the height of the parallelogram. Similarly, if I construct a parallelepiped with a vector $c$, then the wedge product $a wedge b wedge c = (a wedge b_perp) wedge c_perp$, where $c_perp$ lies entirely normal to $a wedge b_perp$. So we can recursively do this for any $k$-vector, looking at orthogonal vectors instead, which is much simpler to see the volumes from.
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Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question.
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– ahala
Jun 23 '13 at 15:39
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I've added a section to this effect.
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– Muphrid
Jun 23 '13 at 16:27
add a comment |
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The determinant of a matrix A is the unique function that satisfies:
$det(A)=0$ when two columns are equal- the determinant is linear in the columns
- if A is the identity $det(A)=1$.
You can easily convince yourself that the oriented volume $operatorname{vol}(v_1,v_2,ldots,v_n)$ between $v_1, v_2,ldots, v_n$ vectors is a function that satisfies exactly the same properties if we place the vectors as the columns of a matrix $A=(v_1,ldots,v_n)$. Hence $operatorname{vol}(v_1,v_2,ldots,v_n)=det(A)$.
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2
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one has to be convinced why such function is unique to reach the conclusion.
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– ahala
Dec 12 '16 at 14:37
1
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True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix.
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– jacopoviti
Dec 13 '16 at 14:59
add a comment |
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Determinant involves a cross-product of the first two vectors and a dot of the result with the third. The result of a cross product is a vector whose magnitude is the area of its null space. Said simply, any plane in 3D is the null space of its normal.The size of the plane is defined by the length of the normal. The volume is found by projecting this normal onto the third vector.
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This question is not about 3D case, it is about nD cases.
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– ahala
Dec 12 '16 at 14:39
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Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD.
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– Shadi
Dec 26 '16 at 3:14
add a comment |
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You can also invoke the change-of-variable theorem in higher dimensions. A $n$-dimensional parelellopiped $mathcal P=mathcal P(a_1,dots,a_n)$ in $mathbb R^n$ (where the $a_i$ are independent vectors in $mathbb R^n$) is the set of all $x$ such that:
$$
x=c_1a_1+dots+c_ka_n,
$$
with $0leq c_ileq 1$. We can define the linear transformation $h(x)=Acdot x$, where $A$ is the $ntimes n$ matrix with $a_i$ as its columns. This gives us $mathcal P=h([0,1]^n)$. The volume of $h([0,1]^n)$ is equal to $h((0,1)^n))$ (those sets are equal modulo a set of measure zero), so we can apply the change-of-variable theorem:
$$
v(mathcal P)=int_{h((0,1)^n)}1=int_{(0,1)^n}vertdet Dhvert=vertdet Avert.
$$
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1
$begingroup$
Isn't the change of variables theorem based on what we are asked to prove?
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– Theorem
Aug 16 '18 at 14:56
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
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$begingroup$
If the column vectors are linearly dependent, both the determinant and the volume are zero.
So assume linear independence.
The determinant remains unchanged when adding multiples of one column to another. This corresponds to a skew translation of the parallelepiped, which does not affect its volume.
By a finite sequence of such operations, you can transform your matrix to diagonal form, where the relation between determinant (=product of diagonal entries) and volume of a "rectangle" (=product of side lengths) is apparent.
$endgroup$
1
$begingroup$
thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant?
$endgroup$
– ahala
Jun 23 '13 at 15:34
1
$begingroup$
Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional
$endgroup$
– Hagen von Eitzen
May 2 '14 at 11:58
$begingroup$
I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear.
$endgroup$
– Mitch
Oct 19 '18 at 19:52
add a comment |
$begingroup$
If the column vectors are linearly dependent, both the determinant and the volume are zero.
So assume linear independence.
The determinant remains unchanged when adding multiples of one column to another. This corresponds to a skew translation of the parallelepiped, which does not affect its volume.
By a finite sequence of such operations, you can transform your matrix to diagonal form, where the relation between determinant (=product of diagonal entries) and volume of a "rectangle" (=product of side lengths) is apparent.
$endgroup$
1
$begingroup$
thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant?
$endgroup$
– ahala
Jun 23 '13 at 15:34
1
$begingroup$
Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional
$endgroup$
– Hagen von Eitzen
May 2 '14 at 11:58
$begingroup$
I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear.
$endgroup$
– Mitch
Oct 19 '18 at 19:52
add a comment |
$begingroup$
If the column vectors are linearly dependent, both the determinant and the volume are zero.
So assume linear independence.
The determinant remains unchanged when adding multiples of one column to another. This corresponds to a skew translation of the parallelepiped, which does not affect its volume.
By a finite sequence of such operations, you can transform your matrix to diagonal form, where the relation between determinant (=product of diagonal entries) and volume of a "rectangle" (=product of side lengths) is apparent.
$endgroup$
If the column vectors are linearly dependent, both the determinant and the volume are zero.
So assume linear independence.
The determinant remains unchanged when adding multiples of one column to another. This corresponds to a skew translation of the parallelepiped, which does not affect its volume.
By a finite sequence of such operations, you can transform your matrix to diagonal form, where the relation between determinant (=product of diagonal entries) and volume of a "rectangle" (=product of side lengths) is apparent.
edited Jun 23 '13 at 16:58
Pedro Tamaroff♦
96.4k10152296
96.4k10152296
answered Jun 23 '13 at 13:35
Hagen von EitzenHagen von Eitzen
277k21269496
277k21269496
1
$begingroup$
thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant?
$endgroup$
– ahala
Jun 23 '13 at 15:34
1
$begingroup$
Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional
$endgroup$
– Hagen von Eitzen
May 2 '14 at 11:58
$begingroup$
I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear.
$endgroup$
– Mitch
Oct 19 '18 at 19:52
add a comment |
1
$begingroup$
thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant?
$endgroup$
– ahala
Jun 23 '13 at 15:34
1
$begingroup$
Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional
$endgroup$
– Hagen von Eitzen
May 2 '14 at 11:58
$begingroup$
I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear.
$endgroup$
– Mitch
Oct 19 '18 at 19:52
1
1
$begingroup$
thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant?
$endgroup$
– ahala
Jun 23 '13 at 15:34
$begingroup$
thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant?
$endgroup$
– ahala
Jun 23 '13 at 15:34
1
1
$begingroup$
Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional
$endgroup$
– Hagen von Eitzen
May 2 '14 at 11:58
$begingroup$
Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional
$endgroup$
– Hagen von Eitzen
May 2 '14 at 11:58
$begingroup$
I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear.
$endgroup$
– Mitch
Oct 19 '18 at 19:52
$begingroup$
I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear.
$endgroup$
– Mitch
Oct 19 '18 at 19:52
add a comment |
$begingroup$
Here is the same argument as Muphrid's, perhaps written in an elementary way.
Apply Gram-Schmidt orthogonalization to ${v_{1},ldots,v_{n}}$, so that
begin{eqnarray*}
v_{1} & = & v_{1}\
v_{2} & = & c_{12}v_{1}+v_{2}^{perp}\
v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp}\
& vdots
end{eqnarray*}
where $v_{2}^{perp}$ is orthogonal to $v_{1}$; and $v_{3}^{perp}$
is orthogonal to $spanleft{ v_{1},v_{2}right} $, etc.
Since determinant is multilinear, anti-symmetric, then
begin{eqnarray*}
detleft(v_{1},v_{2},v_{3},ldots,v_{n}right) & = & detleft(v_{1},c_{12}v_{1}+v_{2}^{perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp},ldotsright)\
& = & detleft(v_{1},v_{2}^{perp},v_{3}^{perp},ldots,v_{n}^{perp}right)\
& = & mbox{signed volume}left(v_{1},ldots,v_{n}right)
end{eqnarray*}
$endgroup$
add a comment |
$begingroup$
Here is the same argument as Muphrid's, perhaps written in an elementary way.
Apply Gram-Schmidt orthogonalization to ${v_{1},ldots,v_{n}}$, so that
begin{eqnarray*}
v_{1} & = & v_{1}\
v_{2} & = & c_{12}v_{1}+v_{2}^{perp}\
v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp}\
& vdots
end{eqnarray*}
where $v_{2}^{perp}$ is orthogonal to $v_{1}$; and $v_{3}^{perp}$
is orthogonal to $spanleft{ v_{1},v_{2}right} $, etc.
Since determinant is multilinear, anti-symmetric, then
begin{eqnarray*}
detleft(v_{1},v_{2},v_{3},ldots,v_{n}right) & = & detleft(v_{1},c_{12}v_{1}+v_{2}^{perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp},ldotsright)\
& = & detleft(v_{1},v_{2}^{perp},v_{3}^{perp},ldots,v_{n}^{perp}right)\
& = & mbox{signed volume}left(v_{1},ldots,v_{n}right)
end{eqnarray*}
$endgroup$
add a comment |
$begingroup$
Here is the same argument as Muphrid's, perhaps written in an elementary way.
Apply Gram-Schmidt orthogonalization to ${v_{1},ldots,v_{n}}$, so that
begin{eqnarray*}
v_{1} & = & v_{1}\
v_{2} & = & c_{12}v_{1}+v_{2}^{perp}\
v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp}\
& vdots
end{eqnarray*}
where $v_{2}^{perp}$ is orthogonal to $v_{1}$; and $v_{3}^{perp}$
is orthogonal to $spanleft{ v_{1},v_{2}right} $, etc.
Since determinant is multilinear, anti-symmetric, then
begin{eqnarray*}
detleft(v_{1},v_{2},v_{3},ldots,v_{n}right) & = & detleft(v_{1},c_{12}v_{1}+v_{2}^{perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp},ldotsright)\
& = & detleft(v_{1},v_{2}^{perp},v_{3}^{perp},ldots,v_{n}^{perp}right)\
& = & mbox{signed volume}left(v_{1},ldots,v_{n}right)
end{eqnarray*}
$endgroup$
Here is the same argument as Muphrid's, perhaps written in an elementary way.
Apply Gram-Schmidt orthogonalization to ${v_{1},ldots,v_{n}}$, so that
begin{eqnarray*}
v_{1} & = & v_{1}\
v_{2} & = & c_{12}v_{1}+v_{2}^{perp}\
v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp}\
& vdots
end{eqnarray*}
where $v_{2}^{perp}$ is orthogonal to $v_{1}$; and $v_{3}^{perp}$
is orthogonal to $spanleft{ v_{1},v_{2}right} $, etc.
Since determinant is multilinear, anti-symmetric, then
begin{eqnarray*}
detleft(v_{1},v_{2},v_{3},ldots,v_{n}right) & = & detleft(v_{1},c_{12}v_{1}+v_{2}^{perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{perp},ldotsright)\
& = & detleft(v_{1},v_{2}^{perp},v_{3}^{perp},ldots,v_{n}^{perp}right)\
& = & mbox{signed volume}left(v_{1},ldots,v_{n}right)
end{eqnarray*}
answered Nov 1 '13 at 2:50
JamesJames
1,10768
1,10768
add a comment |
add a comment |
$begingroup$
In 2d, you calculate the area of a parallelogram spanned by two vectors using the cross product. In 3d, you calculate the volume of a parallelepiped using the triple scalar product. Both of these can be written in terms of a determinant, but it's probably not clear to you what the proper generalization is to higher dimensions.
That generalization is called the wedge product. Given $n$ vectors $v_1, v_2, ldots, v_n$, the wedge product $v_1 wedge v_2 wedge ldots wedge v_n$ is called an $n$-vector, and it has as its magnitude the $n$-volume of that $n$-parallelepiped.
What is the relationship between the wedge product and the determinant? Quite simple, actually. There is a natural generalization of linear maps to work on $k$-vectors. Given a linear map $underline T$ (which can be represented as a matrix), the action of that map on a $k$-vector is defined as
$$underline T(v_1 wedge v_2 wedge ldots wedge v_k) equiv underline T(v_1) wedge underline T(v_2) wedge ldots wedge underline T(v_k)$$
When talking about $n$-vectors in an $n$-dimensional space, it's important to realize that the "vector space" of these $n$-vectors is in fact one-dimensional. That is, if you think about volume, there is only one such unit volume in a given space, and all other volumes are just scalar multiples of it. Hence, when we talk about the action of a linear map on an $n$-vector, we can see that
$$underline T(v_1 wedge v_2 wedge ldots wedge v_n) = alpha [v_1 wedge v_2 wedge ldots wedge v_n]$$
for some scalar $alpha$. In fact, this is a coordinate system independent definition of the determinant!
When you build a matrix out of $n$ vectors $f_1, f_2, ldots, f_n$ as the matrix's columns, what you're really doing is the following: you're saying that, if you have a basis $e_1, e_2, ldots, e_n$, then you're defining a map $underline T$ such that $underline T(e_1) = f_1$, $underline T(e_2) = f_2$, and so on. So when you input $e_1 wedge e_2 wedge ldots wedge e_n$, you get
$$underline T(e_1 wedge e_2 wedge ldots wedge e_n) = (det underline T) e_1 wedge e_2 wedge ldots wedge e_n= f_1 wedge f_2 wedge ldots wedge f_n$$
This is how you can use a matrix determinant to calculate volumes: it's just an easy way of constructing something that automatically computes the wedge product.
Edit: how one can see that the wedge product accurately gives the volume of a parallelepiped. Any vector can be broken down into perpendicular and parallel parts with respect to another vector, to a plane, and so on (or to any $k$-vector). As such, if I have two vectors $a$ and $b$, then the wedge product $a wedge b = a wedge b_perp$, where $b_perp$ is effectively the height of the parallelogram. Similarly, if I construct a parallelepiped with a vector $c$, then the wedge product $a wedge b wedge c = (a wedge b_perp) wedge c_perp$, where $c_perp$ lies entirely normal to $a wedge b_perp$. So we can recursively do this for any $k$-vector, looking at orthogonal vectors instead, which is much simpler to see the volumes from.
$endgroup$
$begingroup$
Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question.
$endgroup$
– ahala
Jun 23 '13 at 15:39
$begingroup$
I've added a section to this effect.
$endgroup$
– Muphrid
Jun 23 '13 at 16:27
add a comment |
$begingroup$
In 2d, you calculate the area of a parallelogram spanned by two vectors using the cross product. In 3d, you calculate the volume of a parallelepiped using the triple scalar product. Both of these can be written in terms of a determinant, but it's probably not clear to you what the proper generalization is to higher dimensions.
That generalization is called the wedge product. Given $n$ vectors $v_1, v_2, ldots, v_n$, the wedge product $v_1 wedge v_2 wedge ldots wedge v_n$ is called an $n$-vector, and it has as its magnitude the $n$-volume of that $n$-parallelepiped.
What is the relationship between the wedge product and the determinant? Quite simple, actually. There is a natural generalization of linear maps to work on $k$-vectors. Given a linear map $underline T$ (which can be represented as a matrix), the action of that map on a $k$-vector is defined as
$$underline T(v_1 wedge v_2 wedge ldots wedge v_k) equiv underline T(v_1) wedge underline T(v_2) wedge ldots wedge underline T(v_k)$$
When talking about $n$-vectors in an $n$-dimensional space, it's important to realize that the "vector space" of these $n$-vectors is in fact one-dimensional. That is, if you think about volume, there is only one such unit volume in a given space, and all other volumes are just scalar multiples of it. Hence, when we talk about the action of a linear map on an $n$-vector, we can see that
$$underline T(v_1 wedge v_2 wedge ldots wedge v_n) = alpha [v_1 wedge v_2 wedge ldots wedge v_n]$$
for some scalar $alpha$. In fact, this is a coordinate system independent definition of the determinant!
When you build a matrix out of $n$ vectors $f_1, f_2, ldots, f_n$ as the matrix's columns, what you're really doing is the following: you're saying that, if you have a basis $e_1, e_2, ldots, e_n$, then you're defining a map $underline T$ such that $underline T(e_1) = f_1$, $underline T(e_2) = f_2$, and so on. So when you input $e_1 wedge e_2 wedge ldots wedge e_n$, you get
$$underline T(e_1 wedge e_2 wedge ldots wedge e_n) = (det underline T) e_1 wedge e_2 wedge ldots wedge e_n= f_1 wedge f_2 wedge ldots wedge f_n$$
This is how you can use a matrix determinant to calculate volumes: it's just an easy way of constructing something that automatically computes the wedge product.
Edit: how one can see that the wedge product accurately gives the volume of a parallelepiped. Any vector can be broken down into perpendicular and parallel parts with respect to another vector, to a plane, and so on (or to any $k$-vector). As such, if I have two vectors $a$ and $b$, then the wedge product $a wedge b = a wedge b_perp$, where $b_perp$ is effectively the height of the parallelogram. Similarly, if I construct a parallelepiped with a vector $c$, then the wedge product $a wedge b wedge c = (a wedge b_perp) wedge c_perp$, where $c_perp$ lies entirely normal to $a wedge b_perp$. So we can recursively do this for any $k$-vector, looking at orthogonal vectors instead, which is much simpler to see the volumes from.
$endgroup$
$begingroup$
Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question.
$endgroup$
– ahala
Jun 23 '13 at 15:39
$begingroup$
I've added a section to this effect.
$endgroup$
– Muphrid
Jun 23 '13 at 16:27
add a comment |
$begingroup$
In 2d, you calculate the area of a parallelogram spanned by two vectors using the cross product. In 3d, you calculate the volume of a parallelepiped using the triple scalar product. Both of these can be written in terms of a determinant, but it's probably not clear to you what the proper generalization is to higher dimensions.
That generalization is called the wedge product. Given $n$ vectors $v_1, v_2, ldots, v_n$, the wedge product $v_1 wedge v_2 wedge ldots wedge v_n$ is called an $n$-vector, and it has as its magnitude the $n$-volume of that $n$-parallelepiped.
What is the relationship between the wedge product and the determinant? Quite simple, actually. There is a natural generalization of linear maps to work on $k$-vectors. Given a linear map $underline T$ (which can be represented as a matrix), the action of that map on a $k$-vector is defined as
$$underline T(v_1 wedge v_2 wedge ldots wedge v_k) equiv underline T(v_1) wedge underline T(v_2) wedge ldots wedge underline T(v_k)$$
When talking about $n$-vectors in an $n$-dimensional space, it's important to realize that the "vector space" of these $n$-vectors is in fact one-dimensional. That is, if you think about volume, there is only one such unit volume in a given space, and all other volumes are just scalar multiples of it. Hence, when we talk about the action of a linear map on an $n$-vector, we can see that
$$underline T(v_1 wedge v_2 wedge ldots wedge v_n) = alpha [v_1 wedge v_2 wedge ldots wedge v_n]$$
for some scalar $alpha$. In fact, this is a coordinate system independent definition of the determinant!
When you build a matrix out of $n$ vectors $f_1, f_2, ldots, f_n$ as the matrix's columns, what you're really doing is the following: you're saying that, if you have a basis $e_1, e_2, ldots, e_n$, then you're defining a map $underline T$ such that $underline T(e_1) = f_1$, $underline T(e_2) = f_2$, and so on. So when you input $e_1 wedge e_2 wedge ldots wedge e_n$, you get
$$underline T(e_1 wedge e_2 wedge ldots wedge e_n) = (det underline T) e_1 wedge e_2 wedge ldots wedge e_n= f_1 wedge f_2 wedge ldots wedge f_n$$
This is how you can use a matrix determinant to calculate volumes: it's just an easy way of constructing something that automatically computes the wedge product.
Edit: how one can see that the wedge product accurately gives the volume of a parallelepiped. Any vector can be broken down into perpendicular and parallel parts with respect to another vector, to a plane, and so on (or to any $k$-vector). As such, if I have two vectors $a$ and $b$, then the wedge product $a wedge b = a wedge b_perp$, where $b_perp$ is effectively the height of the parallelogram. Similarly, if I construct a parallelepiped with a vector $c$, then the wedge product $a wedge b wedge c = (a wedge b_perp) wedge c_perp$, where $c_perp$ lies entirely normal to $a wedge b_perp$. So we can recursively do this for any $k$-vector, looking at orthogonal vectors instead, which is much simpler to see the volumes from.
$endgroup$
In 2d, you calculate the area of a parallelogram spanned by two vectors using the cross product. In 3d, you calculate the volume of a parallelepiped using the triple scalar product. Both of these can be written in terms of a determinant, but it's probably not clear to you what the proper generalization is to higher dimensions.
That generalization is called the wedge product. Given $n$ vectors $v_1, v_2, ldots, v_n$, the wedge product $v_1 wedge v_2 wedge ldots wedge v_n$ is called an $n$-vector, and it has as its magnitude the $n$-volume of that $n$-parallelepiped.
What is the relationship between the wedge product and the determinant? Quite simple, actually. There is a natural generalization of linear maps to work on $k$-vectors. Given a linear map $underline T$ (which can be represented as a matrix), the action of that map on a $k$-vector is defined as
$$underline T(v_1 wedge v_2 wedge ldots wedge v_k) equiv underline T(v_1) wedge underline T(v_2) wedge ldots wedge underline T(v_k)$$
When talking about $n$-vectors in an $n$-dimensional space, it's important to realize that the "vector space" of these $n$-vectors is in fact one-dimensional. That is, if you think about volume, there is only one such unit volume in a given space, and all other volumes are just scalar multiples of it. Hence, when we talk about the action of a linear map on an $n$-vector, we can see that
$$underline T(v_1 wedge v_2 wedge ldots wedge v_n) = alpha [v_1 wedge v_2 wedge ldots wedge v_n]$$
for some scalar $alpha$. In fact, this is a coordinate system independent definition of the determinant!
When you build a matrix out of $n$ vectors $f_1, f_2, ldots, f_n$ as the matrix's columns, what you're really doing is the following: you're saying that, if you have a basis $e_1, e_2, ldots, e_n$, then you're defining a map $underline T$ such that $underline T(e_1) = f_1$, $underline T(e_2) = f_2$, and so on. So when you input $e_1 wedge e_2 wedge ldots wedge e_n$, you get
$$underline T(e_1 wedge e_2 wedge ldots wedge e_n) = (det underline T) e_1 wedge e_2 wedge ldots wedge e_n= f_1 wedge f_2 wedge ldots wedge f_n$$
This is how you can use a matrix determinant to calculate volumes: it's just an easy way of constructing something that automatically computes the wedge product.
Edit: how one can see that the wedge product accurately gives the volume of a parallelepiped. Any vector can be broken down into perpendicular and parallel parts with respect to another vector, to a plane, and so on (or to any $k$-vector). As such, if I have two vectors $a$ and $b$, then the wedge product $a wedge b = a wedge b_perp$, where $b_perp$ is effectively the height of the parallelogram. Similarly, if I construct a parallelepiped with a vector $c$, then the wedge product $a wedge b wedge c = (a wedge b_perp) wedge c_perp$, where $c_perp$ lies entirely normal to $a wedge b_perp$. So we can recursively do this for any $k$-vector, looking at orthogonal vectors instead, which is much simpler to see the volumes from.
edited Jun 23 '13 at 16:27
answered Jun 23 '13 at 14:51
MuphridMuphrid
15.5k11541
15.5k11541
$begingroup$
Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question.
$endgroup$
– ahala
Jun 23 '13 at 15:39
$begingroup$
I've added a section to this effect.
$endgroup$
– Muphrid
Jun 23 '13 at 16:27
add a comment |
$begingroup$
Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question.
$endgroup$
– ahala
Jun 23 '13 at 15:39
$begingroup$
I've added a section to this effect.
$endgroup$
– Muphrid
Jun 23 '13 at 16:27
$begingroup$
Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question.
$endgroup$
– ahala
Jun 23 '13 at 15:39
$begingroup$
Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question.
$endgroup$
– ahala
Jun 23 '13 at 15:39
$begingroup$
I've added a section to this effect.
$endgroup$
– Muphrid
Jun 23 '13 at 16:27
$begingroup$
I've added a section to this effect.
$endgroup$
– Muphrid
Jun 23 '13 at 16:27
add a comment |
$begingroup$
The determinant of a matrix A is the unique function that satisfies:
$det(A)=0$ when two columns are equal- the determinant is linear in the columns
- if A is the identity $det(A)=1$.
You can easily convince yourself that the oriented volume $operatorname{vol}(v_1,v_2,ldots,v_n)$ between $v_1, v_2,ldots, v_n$ vectors is a function that satisfies exactly the same properties if we place the vectors as the columns of a matrix $A=(v_1,ldots,v_n)$. Hence $operatorname{vol}(v_1,v_2,ldots,v_n)=det(A)$.
$endgroup$
2
$begingroup$
one has to be convinced why such function is unique to reach the conclusion.
$endgroup$
– ahala
Dec 12 '16 at 14:37
1
$begingroup$
True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix.
$endgroup$
– jacopoviti
Dec 13 '16 at 14:59
add a comment |
$begingroup$
The determinant of a matrix A is the unique function that satisfies:
$det(A)=0$ when two columns are equal- the determinant is linear in the columns
- if A is the identity $det(A)=1$.
You can easily convince yourself that the oriented volume $operatorname{vol}(v_1,v_2,ldots,v_n)$ between $v_1, v_2,ldots, v_n$ vectors is a function that satisfies exactly the same properties if we place the vectors as the columns of a matrix $A=(v_1,ldots,v_n)$. Hence $operatorname{vol}(v_1,v_2,ldots,v_n)=det(A)$.
$endgroup$
2
$begingroup$
one has to be convinced why such function is unique to reach the conclusion.
$endgroup$
– ahala
Dec 12 '16 at 14:37
1
$begingroup$
True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix.
$endgroup$
– jacopoviti
Dec 13 '16 at 14:59
add a comment |
$begingroup$
The determinant of a matrix A is the unique function that satisfies:
$det(A)=0$ when two columns are equal- the determinant is linear in the columns
- if A is the identity $det(A)=1$.
You can easily convince yourself that the oriented volume $operatorname{vol}(v_1,v_2,ldots,v_n)$ between $v_1, v_2,ldots, v_n$ vectors is a function that satisfies exactly the same properties if we place the vectors as the columns of a matrix $A=(v_1,ldots,v_n)$. Hence $operatorname{vol}(v_1,v_2,ldots,v_n)=det(A)$.
$endgroup$
The determinant of a matrix A is the unique function that satisfies:
$det(A)=0$ when two columns are equal- the determinant is linear in the columns
- if A is the identity $det(A)=1$.
You can easily convince yourself that the oriented volume $operatorname{vol}(v_1,v_2,ldots,v_n)$ between $v_1, v_2,ldots, v_n$ vectors is a function that satisfies exactly the same properties if we place the vectors as the columns of a matrix $A=(v_1,ldots,v_n)$. Hence $operatorname{vol}(v_1,v_2,ldots,v_n)=det(A)$.
edited Nov 30 '18 at 17:19
José Carlos Santos
153k22123225
153k22123225
answered Oct 27 '16 at 17:48
jacopovitijacopoviti
684
684
2
$begingroup$
one has to be convinced why such function is unique to reach the conclusion.
$endgroup$
– ahala
Dec 12 '16 at 14:37
1
$begingroup$
True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix.
$endgroup$
– jacopoviti
Dec 13 '16 at 14:59
add a comment |
2
$begingroup$
one has to be convinced why such function is unique to reach the conclusion.
$endgroup$
– ahala
Dec 12 '16 at 14:37
1
$begingroup$
True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix.
$endgroup$
– jacopoviti
Dec 13 '16 at 14:59
2
2
$begingroup$
one has to be convinced why such function is unique to reach the conclusion.
$endgroup$
– ahala
Dec 12 '16 at 14:37
$begingroup$
one has to be convinced why such function is unique to reach the conclusion.
$endgroup$
– ahala
Dec 12 '16 at 14:37
1
1
$begingroup$
True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix.
$endgroup$
– jacopoviti
Dec 13 '16 at 14:59
$begingroup$
True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix.
$endgroup$
– jacopoviti
Dec 13 '16 at 14:59
add a comment |
$begingroup$
Determinant involves a cross-product of the first two vectors and a dot of the result with the third. The result of a cross product is a vector whose magnitude is the area of its null space. Said simply, any plane in 3D is the null space of its normal.The size of the plane is defined by the length of the normal. The volume is found by projecting this normal onto the third vector.
$endgroup$
$begingroup$
This question is not about 3D case, it is about nD cases.
$endgroup$
– ahala
Dec 12 '16 at 14:39
$begingroup$
Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD.
$endgroup$
– Shadi
Dec 26 '16 at 3:14
add a comment |
$begingroup$
Determinant involves a cross-product of the first two vectors and a dot of the result with the third. The result of a cross product is a vector whose magnitude is the area of its null space. Said simply, any plane in 3D is the null space of its normal.The size of the plane is defined by the length of the normal. The volume is found by projecting this normal onto the third vector.
$endgroup$
$begingroup$
This question is not about 3D case, it is about nD cases.
$endgroup$
– ahala
Dec 12 '16 at 14:39
$begingroup$
Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD.
$endgroup$
– Shadi
Dec 26 '16 at 3:14
add a comment |
$begingroup$
Determinant involves a cross-product of the first two vectors and a dot of the result with the third. The result of a cross product is a vector whose magnitude is the area of its null space. Said simply, any plane in 3D is the null space of its normal.The size of the plane is defined by the length of the normal. The volume is found by projecting this normal onto the third vector.
$endgroup$
Determinant involves a cross-product of the first two vectors and a dot of the result with the third. The result of a cross product is a vector whose magnitude is the area of its null space. Said simply, any plane in 3D is the null space of its normal.The size of the plane is defined by the length of the normal. The volume is found by projecting this normal onto the third vector.
answered Dec 11 '16 at 22:27
ShadiShadi
111
111
$begingroup$
This question is not about 3D case, it is about nD cases.
$endgroup$
– ahala
Dec 12 '16 at 14:39
$begingroup$
Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD.
$endgroup$
– Shadi
Dec 26 '16 at 3:14
add a comment |
$begingroup$
This question is not about 3D case, it is about nD cases.
$endgroup$
– ahala
Dec 12 '16 at 14:39
$begingroup$
Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD.
$endgroup$
– Shadi
Dec 26 '16 at 3:14
$begingroup$
This question is not about 3D case, it is about nD cases.
$endgroup$
– ahala
Dec 12 '16 at 14:39
$begingroup$
This question is not about 3D case, it is about nD cases.
$endgroup$
– ahala
Dec 12 '16 at 14:39
$begingroup$
Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD.
$endgroup$
– Shadi
Dec 26 '16 at 3:14
$begingroup$
Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD.
$endgroup$
– Shadi
Dec 26 '16 at 3:14
add a comment |
$begingroup$
You can also invoke the change-of-variable theorem in higher dimensions. A $n$-dimensional parelellopiped $mathcal P=mathcal P(a_1,dots,a_n)$ in $mathbb R^n$ (where the $a_i$ are independent vectors in $mathbb R^n$) is the set of all $x$ such that:
$$
x=c_1a_1+dots+c_ka_n,
$$
with $0leq c_ileq 1$. We can define the linear transformation $h(x)=Acdot x$, where $A$ is the $ntimes n$ matrix with $a_i$ as its columns. This gives us $mathcal P=h([0,1]^n)$. The volume of $h([0,1]^n)$ is equal to $h((0,1)^n))$ (those sets are equal modulo a set of measure zero), so we can apply the change-of-variable theorem:
$$
v(mathcal P)=int_{h((0,1)^n)}1=int_{(0,1)^n}vertdet Dhvert=vertdet Avert.
$$
$endgroup$
1
$begingroup$
Isn't the change of variables theorem based on what we are asked to prove?
$endgroup$
– Theorem
Aug 16 '18 at 14:56
add a comment |
$begingroup$
You can also invoke the change-of-variable theorem in higher dimensions. A $n$-dimensional parelellopiped $mathcal P=mathcal P(a_1,dots,a_n)$ in $mathbb R^n$ (where the $a_i$ are independent vectors in $mathbb R^n$) is the set of all $x$ such that:
$$
x=c_1a_1+dots+c_ka_n,
$$
with $0leq c_ileq 1$. We can define the linear transformation $h(x)=Acdot x$, where $A$ is the $ntimes n$ matrix with $a_i$ as its columns. This gives us $mathcal P=h([0,1]^n)$. The volume of $h([0,1]^n)$ is equal to $h((0,1)^n))$ (those sets are equal modulo a set of measure zero), so we can apply the change-of-variable theorem:
$$
v(mathcal P)=int_{h((0,1)^n)}1=int_{(0,1)^n}vertdet Dhvert=vertdet Avert.
$$
$endgroup$
1
$begingroup$
Isn't the change of variables theorem based on what we are asked to prove?
$endgroup$
– Theorem
Aug 16 '18 at 14:56
add a comment |
$begingroup$
You can also invoke the change-of-variable theorem in higher dimensions. A $n$-dimensional parelellopiped $mathcal P=mathcal P(a_1,dots,a_n)$ in $mathbb R^n$ (where the $a_i$ are independent vectors in $mathbb R^n$) is the set of all $x$ such that:
$$
x=c_1a_1+dots+c_ka_n,
$$
with $0leq c_ileq 1$. We can define the linear transformation $h(x)=Acdot x$, where $A$ is the $ntimes n$ matrix with $a_i$ as its columns. This gives us $mathcal P=h([0,1]^n)$. The volume of $h([0,1]^n)$ is equal to $h((0,1)^n))$ (those sets are equal modulo a set of measure zero), so we can apply the change-of-variable theorem:
$$
v(mathcal P)=int_{h((0,1)^n)}1=int_{(0,1)^n}vertdet Dhvert=vertdet Avert.
$$
$endgroup$
You can also invoke the change-of-variable theorem in higher dimensions. A $n$-dimensional parelellopiped $mathcal P=mathcal P(a_1,dots,a_n)$ in $mathbb R^n$ (where the $a_i$ are independent vectors in $mathbb R^n$) is the set of all $x$ such that:
$$
x=c_1a_1+dots+c_ka_n,
$$
with $0leq c_ileq 1$. We can define the linear transformation $h(x)=Acdot x$, where $A$ is the $ntimes n$ matrix with $a_i$ as its columns. This gives us $mathcal P=h([0,1]^n)$. The volume of $h([0,1]^n)$ is equal to $h((0,1)^n))$ (those sets are equal modulo a set of measure zero), so we can apply the change-of-variable theorem:
$$
v(mathcal P)=int_{h((0,1)^n)}1=int_{(0,1)^n}vertdet Dhvert=vertdet Avert.
$$
edited Apr 28 '18 at 13:32
answered Apr 28 '18 at 12:49
Sha VukliaSha Vuklia
1,3651717
1,3651717
1
$begingroup$
Isn't the change of variables theorem based on what we are asked to prove?
$endgroup$
– Theorem
Aug 16 '18 at 14:56
add a comment |
1
$begingroup$
Isn't the change of variables theorem based on what we are asked to prove?
$endgroup$
– Theorem
Aug 16 '18 at 14:56
1
1
$begingroup$
Isn't the change of variables theorem based on what we are asked to prove?
$endgroup$
– Theorem
Aug 16 '18 at 14:56
$begingroup$
Isn't the change of variables theorem based on what we are asked to prove?
$endgroup$
– Theorem
Aug 16 '18 at 14:56
add a comment |
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