Dimension for Vector Space involving Planes

[Red_CCF]

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?

 

[Rasalhague]

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

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

for all scalars s_i 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, s_i, such that any vector associated with the vector space can be expressed as a linear combination of basis vectors, e_i, thus:

\textbf{v}=\sum_{i=1}^{n}s_i\textbf{e}_i

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

s_1\textbf{v}_1+s_2\textbf{v}_2+s_3\textbf{v}_3+...=\textbf{0}

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.

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?

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 referring to?

 

[HallsofIvy]

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?

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.

 

[Rasalhague]

It can be shown that any two (finite dimensional) spaces are isomorphic if and only if they have the same dimension.

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?

 

[Landau]

No, it applies to vector spaces, just like the rest of this thread ;)

 

[HallsofIvy]

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.

 

[Rasalhague]

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".

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...

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.

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?

That is, the only difference between isomorphic structures is purely "cosmetic"- what the elements and operations are called.

And, in the lingo, they're said to be the same "up to isomorphism".

 

[Landau]

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?

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.

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?

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 (V,+_v,*_v) and (W,+_w,*_w). 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 (V,+_v) and (W,+_w). 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'.

 

[Rasalhague]

Excellent! Thanks, Landau, for your very clear and detailed answers.

 

[Red_CCF]

Thanks for your response. I forgot about this thread until now

I was referring to this article:

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

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?

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.

But a plane is identified by a position vector and a normal vector, both of which are a part of R³ isn't it?

 

[Mark44]

Fixed your hyperlink...

Thanks for your response. I forgot about this thread until now

I was referring to this article:

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

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?

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.

But a plane is identified by a position vector and a normal vector, both of which are a part of R³ isn't it?

 

[Red_CCF]

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.

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?

 

[Red_CCF]

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?

 

[adriank]

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.

 

[Mark44]

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?

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.

 

[Red_CCF]

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.

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?

 

[Mark44]

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.

Yes.

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?

If d \neq 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.

 

[Red_CCF]

Y
If d \neq 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.

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?

 

[Mark44]

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)?

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 \neq} 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.

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?

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.

 

[Red_CCF]

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 \neq} doesn't go through the origin, so can't possibly be a vector space.

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

 

[Mark44]

What you're calling the zero "plane", 0x + 0y + 0z = 0, is not actually a plane. Every point in R³ is a solution; hence, this "plane" is all of R³ itself.

For your additive inverse, the equation ax + by + cz = d is equivalent to -ax - by - cz = -d, so you don't get anything different.

Also, for scalar multiplication, k*(ax + by + cz) = k*d is equivalent to ax + by + cz = d, so you don't get anything different there, either.

What I think you are missing is that a plane is not an equation. The equation is merely a restriction on all of the points in whatever space we're talking about, so that we can identify the points that are on the plane. It doesn't make a whole lot of sense geometrically to me to talk about adding two planes together. And multiplying a plane by a constant makes about as much sense as subracting "red" from a circle.

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.

Each plane consists of a set of points, though, and each point is an element of R³, so you can't get away from the fact that these planes are embedded in R³.

 

[adriank]

You have defined a formal 4-dimensional vector space of "equations of the form ax + by + cz = d", but that wouldn't really be useful. In particular: if k is a number, then the equations ax + by + cz = d and kax + kby + kcz = kd represent the same plane (they have the same set of solutions). Also, 0x + 0y + 0z = d is not a plane, since for d = 0 it gives the entire space, and for nonzero d there are no solutions.

You can make a space of all planes, but it won't be a vector space in any sensible way. The set of planes in R³ containing the origin (that is, the set of 2-dimensional vector subspaces of R³) is an example of what is called a Grassmannian; this particular one is denoted Gr₂(3) or Gr₂(R³). But it's not a vector space. Likewise, there is something called an affine Grassmannian, for the space Graff₂(3) of all planes (2-dimensional affine subspaces of R³), not necessarily passing through the origin, but again it's not a vector space.

(Whoops, missed Mark44's post, which says some of the same things I did in the first paragraph.)[/size]

 

[Red_CCF]

Ah okay I see the (many) flaws in my logic. Thank you, Mark44, for being patient and explaining it to me and for responding so quickly!

So basically it's illogical to classify all planes into its own vector space plus some of the axioms are not satisfied (ex. the 0 plane doesn't actually exist).

Just have a general question regarding the axioms; for example what if the operation does not actually change a member of a set (like the scalar multiplication I did above), would the particular set/space be satisfying that particular axiom (ex. did the set I define actually satisfy the particular axiom even though the equations did not change)?

Lastly, how come two planes cannot be added, I thought it would just result in a new plane?

 

[Mark44]

Just have a general question regarding the axioms; for example what if the operation does not actually change a member of a set (like the scalar multiplication I did above), would the particular set/space be satisfying that particular axiom (ex. did the set I define actually satisfy the particular axiom even though the equations did not change)?

Maybe, but if at least one of the axioms isn't satisfied, then the set isn't a vector space, so I'm not sure what the point is of continuing.

Lastly, how come two planes cannot be added, I thought it would just result in a new plane?

If you look at planes strictly from a geometric perspective, what does it mean to add two planes? On the other hand, if you define addition to mean adding the left sides of the equation together and the right sides together, as you have done, then what do you get?

Here's a simple example. Consider p₁ = {(x, y, z)| x = 2} and p₂ = {(x, y, z)| x = 3}. Using your definition of addition, we get 2x = 5, which is equivalent to x = 5/2. In this case, adding two planes gives the somewhat surprising result of a plane that lies between the two we started with. This seems a lot like adding 1/2 and 1/3 using so-called "baseball addition" and getting 2/5. (In baseball addition, a/b + c/d = (a + c)/(b + d).) The surprising thing about this kind of addition is that it's possible to add one positive number to another positive number to get a sum that is between the two numbers.

Here's another example, with p₃ = {(x, y, z)| x = 1} and p₄ = {(x, y, z)| y = 2}. If we add p₃ and p₄, we get the plane whose equation is x + y = 3. The only thing interesting about this, I think, is that the line of intersection of p₃ and p₄ is also contained in the plane whose equation is x + 2y = 3.

 

[adriank]

Continuing: That addition isn't well-defined on planes. For example, continuing Mark44's example above, you could also write p₁ = {(x, y, z) | 2x = 4}. But then if you try to add this to p₂, you get 3x = 7, which is a different plane from 2x = 5.

 

[Red_CCF]

Yeah I see how it doesn't make any sense now. Thanks!