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Dimension for Vector Space involving Planes

  1. Jun 10, 2010 #1
    I don't really understand what the dimension of a vector space for planes is. Is it 2 or 3 and why? What's the difference between the dimension of the plane as a surface (dimension of all surfaces is 2) and the dimension of plane in a vector space?

    Also, Wikipedia says that only planes passing through the origin is a subspace of R-3, so does this mean that such plane is equal to R-2? What of planes that do not satisfy this?
     
    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 11, 2010 #2
    The dimension of a finite vector space is the biggest number of vectors that it's possible to have in a set of linearly independent vectors of that space. To call a set of vectors linearly independent means that

    [tex]\sum_{i=1}^{p}s_i\textbf{v}_i=\textbf{0} \enspace \Rightarrow \enspace s_i=0[/tex]

    for all scalars si from 1 to p. (The right-arrow means "implies".)

    It happens that any maximal set of linearly independent vectors (i.e. a linearly independent set which can't be made bigger by including another vector and still be linearly independent) can be used as a basis for the vector space, meaning that there are scalars, si, such that any vector associated with the vector space can be expressed as a linear combination of basis vectors, ei, thus:

    [tex]\textbf{v}=\sum_{i=1}^{n}s_i\textbf{e}_i[/tex]

    You can also say that the dimension of a vector space is the span of any set of basis vectors for that space. The span, span(S), of a set, S, of vectors is the set of all vectors v that can expressed as a linear combination of vectors in S.

    The maximum number of linearly independent vectors you can find lying in a given plane is two, because for three or more, you can always find scalars such that

    [tex]s_1\textbf{v}_1+s_2\textbf{v}_2+s_3\textbf{v}_3+...=\textbf{0}[/tex]

    So if dimension is defined this way, and a plane is identified with the vector space whose vectors comprise R2, then the dimension of a plane is 2.

    I don't understand the grammar of your question, but I'm guessing it says that a set of vectors which describe a plane embedded in R3 is a 2-dimensional subspace of R3 only if it includes the origin, otherwise it's not a subspace of R3. That's because a subspace is itself a vector space, and every vector space must have a zero vector.

    Suppose you have a plane embedded in R3 which doesn't go through the origin. It has intrinsic properties--these are the properties that don't depend on the embedding--and it has extrinsic properties that do depend on the embedding. You can identify this plane with R2, and identify any point in it as the origin of R2. In other words, you can give it its own coordinate system. Now it's a vector space in its own right. But if you think of it in extrinsic terms as a subset of position vectors in R3, then it isn't a vector space because none of these position vectors are the zero vector.

    If I've misunderstood the question, could you try again to explain what is it that Wikipedia says about only planes passing through origin, or give a link to whichever Wikipedia article you're refering to?
     
  4. Jun 11, 2010 #3

    HallsofIvy

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    In mathematics you have to be very precise with your wording! "equal to R-2", no. Saying that "x= y", in mathematics, means that "x" and "y" are two ways of expressing exactly the same thing.

    The plane consisting of all points in R-3, (x, y, z), such that x=y, that is, (x, x, z), is a subspace of R-3. It is closed under sums- (u, u, z)+ (v, v, w)= (u+v, u+ v, z+ w) and the first two components are still the same: u+v= u+v. It is closed under scalar multiplication- a(x, x, z)= (ax, ax, z) and the first two components are still the same: ax= ax.

    That is a two dimensional subspace ( basis: {(1, 1, 0), (0, 0, 1)}) but is is not EQUAL to R-2. This subspace contains, for example, (1, 1, 0) while all members of R-2 are of the form (x, y) for real numbers x and y. They cannot be equal because their underlying sets do not contain the same thing.

    The word you are looking for is "isomorphic". This subspace is isomorphic to R-2 since the function, f, that takes (1, 1, 0) to (1, 0) and (0, 0, 1) to (0, 1) "preserves all operations". That is, this function maps (x, x, z) to (x, z), and, applying it to both sides of (a, a, b)+ (c, c, d)= (a+c, a+ c, b+ d) would give (a, b)+ (c, d)= (a+ c, b+d) while applying it to both sides of x(a, a, b)= (ax, ax, bx) would give x(a, b)= (ax, bx).

    It can be shown that any two (finite dimensional) spaces are isomorphic if and only if they have the same dimension.
     
  5. Jun 11, 2010 #4
    Does this only apply to vector spaces? I've seen space defined very broadly as a mathematical object consisting of a set (the underlying set) together with some further structure. Some such spaces don't have a dimension in the sense described here; are there spaces for which dimension isn't even defined?

    Presumably these isomorphic spaces would have to be spaces that share the same operations, i.e. the operations are defined for each of the spaces that are isomorphic to each other; they'd have to be, in some respect, the same sort of space. Suppose we had a vector space for which an inner product was defined, and a function that took its vectors to a vector space for which no inner product was defined, but which preserved addition and scaling. Would that be called an isomorphism, or would the fact that the inner product operation wasn't preserved disqualify it?
     
  6. Jun 12, 2010 #5

    Landau

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    No, it applies to vector spaces, just like the rest of this thread ;)
     
  7. Jun 13, 2010 #6

    HallsofIvy

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    I, on the other hand, have never seen "space" used that way. I would call a set with some further structure, involving operations defined on the set, an "algebraic structure", not a "space". Of course, "isomorphism" can be defined for all kinds of algebraic structures. One way of looking at it is that if A and B are isomorphic- that is, if there exist a function that maps each element of A into a unique element of B, then by "renaming" each element of B with its corresponding "name" from A, you get something that looks exactly like A.

    That is, the only difference between isomorphic structures is purely "cosmetic"- what the elements and operations are called.
     
  8. Jun 16, 2010 #7
    How would you define a "space" in general (if you have a general definition)? Examples I've met are: vector, affine, topological, sample and probability spaces. And inner product space, a kind of vector space.

    I've come across the term an "algebra" for a specific kind of algebraic structure consisting of a vector space with a "multiplication" (a bilinear vector product). I'm still puzzling over the meaning of terms like "field" and "algebra" in the names of certain structures like sigma algebra and field of sets, whose underlying set is a set of sets; the answers I've had here suggest that terms like "field" or "algebra" maybe don't have their general meaning if the underlying set is a set of sets [ https://www.physicsforums.com/showthread.php?t=409411 ], but "ring" does, which is a bit confusing...

    Let A be the vector space ((Rn, componentwise addition), R, scalar multiplication as usually refined). Let B be the inner product space consisting of the same vector space with the standard inner product. They're isomophic as vector spaces. From your answer, I think that as inner product spaces, they couldn't be isomorphic, since no function from A to B can preserve the inner product it doesn't have, and no function from B to A can preserve B's inner product, as none has been defined for A. Is that right? Or in the context of vector spaces, is it understood that isomorphism--unless otherwise specified--means a group isomorphism, regardless of what extra structure the two vector spaces might have?

    And, in the lingo, they're said to be the same "up to isomorphism".
     
  9. Jun 16, 2010 #8

    Landau

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    Yes. As vector spaces they are isomorphic (by defninition). They cannot be compared as inner product spaces, because the first is not an inner product space.
    No, of course not a GROUP isomorphism! A vector space is a group (under addition), but is has more structure, namely scalar multiplication. So a VECTOR SPACE isomorphism is just a linear isomorphism: a bijective linear map, which is a bijective map that preserves both addition AND scalar multiplication.

    Vector spaces are vector spaces: a vector space over a fiels K is a 3-tuple (V,+,*) such that (V,+) is an abelian group and scalar multiplication * is an associative map KxV->V, plus some axioms telling they interact nicely.

    Let's say we have two vector spaces [itex](V,+_v,*_v)[/itex] and [itex](W,+_w,*_w)[/itex]. An isomorphism between them would be a linear isomorphism, which preserves cardinality, addition, and scalar multiplication. If you forget about scalar multiplication, you can compare them as groups [itex](V,+_v)[/itex] and [itex](W,+_w)[/itex]. An isomorphism between them would be a group isomorphism, which preserves cardinality and addition. If you also forget about addition, you can compare them as sets V and W. An isomorphism between them would be a set isomorphism, also called a bijection, which preserves cardinality.

    An inner product space is not just a vector space, but a vector space equipped with an inner product: it has extra structure. Isomorphism are defined between the same 'structures', not between different 'structures'.
     
  10. Jun 16, 2010 #9
    Excellent! Thanks, Landau, for your very clear and detailed answers.
     
  11. Aug 18, 2010 #10
    Thanks for your response. I forgot about this thread until now :blushing:

    I was referring to this article:

    http://en.wikipedia.org/wiki/Plane_(geometry [Broken])

    where they talk about planes begin embedded in R3; do they mean that planes are actually subspace of R3 when they use the word embedded?

    But a plane is identified by a position vector and a normal vector, both of which are a part of R3 isn't it?
     
    Last edited by a moderator: May 4, 2017
  12. Aug 18, 2010 #11

    Mark44

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    Fixed your hyperlink...
    Only planes that pass through the origin in R3 are subspaces of R3. For example, The set {(x, y, z) | z = 2} is a plane embedded in R3. Since it doesn't pass through the origin, it is not a subspace of R3.

    The plane z = x + 2y in R3 does pass through the origin and can be shown to be a subspace of R3.
     
  13. Aug 18, 2010 #12
    So all planes that pass through the origin are part of a subspace of R3; what would such space look like? V = {(x, y, z) | ax + by + cz = 0 for some scalar a, b, c}? What would the dimension of this space be?
     
  14. Aug 19, 2010 #13
    Another thing I'm beginning to wonder is that, let's say a plane is a vector within its own vector space; a plane itself is an infinite number of points where as a vector from R3 is only one point, so how do we even compare the two? Only one vector in R3 is the 0 vector so why does every plane (if it is a vector in its own vector space) need to contain the 0 from R3 for the vector space to be a subspace of R3?
     
  15. Aug 19, 2010 #14
    The plane is a vector space. Points in the plane are vectors in this vector space. The plane itself is not a vector.

    A plane in R3 is a subspace of R3 if and only if it contains 0. Read the definition of subspace.

    U is a subspace of R3 if and only if
    1. 0 is in U.
    2. for all x, y in U, x + y is in U.
    3. for all x in U and k in R, kx is in U.
     
  16. Aug 19, 2010 #15

    Mark44

    Staff: Mentor

    No, they are not part of a subspace. Each plane (and line) through the origin is a subspace of R3. These planes have dimension two, and these lines have dimension one.

    Any plane looks pretty much like any other - the only difference would be the orientation of the plane. The dimension of the subspace (of R3) V is two.
     
  17. Aug 19, 2010 #16
    Thanks for clearing that up! I didn't know I approached the question entirely wrong.

    So V = {(x,y,z) | ax + by + cz = 0 for a, b, c as some scalar} -> so for every possible combination of scalars a, b, and c, would result in V being a specific subspace of R3 with dimension 2.

    So is there is no such thing as a vector space containing all possible planes (ax + by + cz = d for all possible a, b, c, d) or do you mean that such vector space has nothing to do with R3 and why?
     
    Last edited: Aug 19, 2010
  18. Aug 19, 2010 #17

    Mark44

    Staff: Mentor

    Yes.
    If d [itex]\neq[/itex] 0, the set {(x, y, z) | ax + by + cz = d} is a plane that doesn't go through the origin, so this set isn't a subspace of R3. I'm assuming that a, b, and c are arbitrary, but not all three can be zero, otherwise there is no solution, so you don't get even a single point, let alone a plane.

    Any plane in R3 that goes through the origin is a subspace of R3, and is embedded in R3, which means that every point in the plane is an element of R3. Such a plane is a vector space in its own right, but happens to be a two-dimensional vector space while R3 is three dimensional.

    Things don't quit with R3, as you'll probably discover. Even though it's difficult or impossible to visualize, there are Euclidian vector spaces of dimension 4 and higher. The concepts of Euclidian space (i.e., R3) carry over to R4, R5, and so on. In these higher dimension spaces, and equation such as ax + by + cz + dw = 0 represents a three-dimensional "hyperplane" in four-dimensional space. This hyperplane is a subspace of R4. Even though it's difficult to visualize, you can calculate the distance between points, the angle between vectors, and all the rest in these higher-dimension vector spaces.
     
  19. Aug 20, 2010 #18
    But is it possible for this set to be a vector space but just have nothing to do with R3 (ex. not a subspace of R3)? Functions, matrices etc. are all vectors within a specified vector space so is it possible that there is a vector space (that includes all planes) where a plane itself is a vector?
     
  20. Aug 20, 2010 #19

    Mark44

    Staff: Mentor

    No. For one thing, how can a set of points in R3 not have something to do with R3? For another thing, this set - {(x, y, z) | ax + by + cz = d, where d [itex]\neq[/itex]} doesn't go through the origin, so can't possibly be a vector space.

    A vector space is very precisely defined in terms of several axioms. If you don't know how a vector space is defined (and I suspect that you don't), here is a wiki article http://en.wikipedia.org/wiki/Vector_space.
    Not that I'm aware of. Loosely speaking, a vector space is a set of things (e.g., vectors, functions, matrices) together with an addition operation and a multiplication operation involving a member of the set and a scalar (a member of some field, typically the reals or the complex numbers). In addition there are a number of axioms that must be satisfied.

    For a plane in R3 that happens to be a subspace of R3, it is the set of all possible vectors in the plane that make up the subspace.
     
  21. Aug 20, 2010 #20
    By not having something to do with R3 I meant not a subspace of R3. I meant a set of planes, not a set of points, like:

    V = { ax + by + cz = d | a, b, c, d are scalars}. V is a set of all possible combinations of a, b, c, d, such that the equation satisfy; I think d = 0 is possible.

    An item in the set would be itself be a plane. I did learn about the axioms that define a vector space I think this set based on the axioms would still be a vector space but please correct me if I'm mistaken.

    Vector Addition:

    If vector addition is performed like the normal addition of 2 functions/equations.

    1. If two planes are defined as ax + by + cz = d and ex + fy + gz =h in V where all the coefficients are scalars, then their sum should also be included in V in the form of the plane (a+e)x + (b+f)y + (c+g)z = d+h as the equation is satisfied

    2. If addition is in this set is the same as functions, commutivity is obvious.

    3. This one I'm not sure, I would imagine the 0 plane be something like 0x + 0y + 0z = 0 (the trivial solution) but I'm not sure what this would look like geometrically (or if it's even possible) but mathematically it would satisfy this axiom since one is just adding some plane to 0.

    4. Inverse of a plane ax + by + cz = d would be -ax - by - cz = -d and the sum of these two would be the zero plane provided that is a valid 0 plane.

    Scalar Multiplication:

    Scalar multiplication would perform as for a normal function/equation.

    1. This would just be multiplying both sides of the equation by some scalar which should just shift the plane so the result should still be in V.

    let q and p be some scalar

    2a) (q+p)(ax + by + cz = d ) and p(ax + by + cz = d) plus q(ax + by + cz = d) should be the same based on the way the addition was defined

    2b) q(ax + by + cz = d plus ex + fy + gz =h) -> (qa + qe)x + (qb + qf)y+ (qc + qg)z =(qd + qh)

    This should be the same as q(ax + by + cz = d) plus q(ex + fy + gz =h)

    3. (pq)(ax + by + cz = d ) should be the same as (p)(q(ax + by + cz = d))

    4. 1 times both sides of the equation of a plane would not change it


    Did I make any mistakes? If so please tell me.
     
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