Importance of a vector space containing the origin.

[MurdocJensen]

Why does a vector space need to go through the origin? I am not sure I understand the geometric importance of that, or the importance of it as it pertains to the axioms of a vector space.

Is this to say I want any vector in the vector space to lie completely in the space, as oppose to say just its tip?

 

[Office_Shredder]

What do you mean by a vector space going through the origin?

The vector space axioms require that every vector space has a zero. if you're talking about subspaces of Rⁿ, then they have to have an additive zero and the only element that can satisfy that is the origin as long as you use the normal addition

 

[MurdocJensen]

For example: x+2y+z=6 and x+2y+z=0. These are parallel planes, but the latter intersects the origin, yes?

If I just get vectors by considering two points on either plane, technically I can get the same vectors for both planes as long as I'm considering two points on both planes that yield a similar vector, right? If that's the case, why should it matter that my plane go through the origin for it to be a vector space? I can still satisfy the zero vector condition in either case.

 

[Office_Shredder]

Ok then, what is the zero vector on the plane x+2y+z=6?

 

[Tac-Tics]

If I just get vectors by considering two points on either plane, technically I can get the same vectors for both planes as long as I'm considering two points on both planes that yield a similar vector, right? If that's the case, why should it matter that my plane go through the origin for it to be a vector space? I can still satisfy the zero vector condition in either case.

Before taking a stab at your question, I want to give an example in abstract algebra that will seem totally unrelated.

A group is a set with two operations called multiplication and inverse. Groups also have a special value called the unit.

Let's look at a few groups:

Mulitplication over the non-zero numbers
* multiplication -> multiplication.
* inverse -> the reciprocal (x^-1 = 1/x)
* unit -> 1 (because 1 * x = x)

Invertible real functions
* multiplication -> function application (f ∘ g, which is f(g(x)))
* inverse -> inverse function (f^-1 = f^-1(x))
* unit -> the identity function, f(x) = x

In this group, "multiplication" is function application. When we say "multiply x and y" what we mean is "give me the function that does y first, then does x afterwards". We also ascribe different meanings to the inverse and the unit.

There's also a group formed by addition over numbers
* multiplication -> addition
* inverse -> negation (x^-1 = -x)
* unit -> 0 (because 0 + x = x)

Note that the "multiplication" in this third group is actually addition(!). It's very very important not to confuse the "group multiplication" with "real number multiplication". When we are working with the group, multiplication means something totally different. Context is important ;)

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Now, let's bring it back to what your example says. You have two parallel planes. Only one of them intersects the origin. You want rationalize how only one passes through the origin, yet they are both vector spaces.

The answer is the "origin" depends on your vector space (just like "multiplication" above). In the first plane, it turns out the "origin" is, in fact, the vector (0, 0, 0). In the second plane, however, the "origin" will NOT be (0, 0, 0). It will be something else.

You might wonder what it would be. But to answer that, you have to be a little more careful with your definition. Note that "addition" and "scaling" (the two operations required for a vector space) also take on new meanings.

And you haven't properly defined them:

Suppose that I have a vector (x, y, z) such that x+2y+z=6. Then, the vector 2 * (x, y, z) = (2x, 2y, 2z) should also satisfy the equation (2x) + 2 (2y) + (2z) = 6... but clearly, it doesn't. Scaling doesn't work.

Take (x, y, z) as a vector. By closure over addition, (x, y, z) + (x, y, z) must also be a vector, but that equals 2 * (x, y, z), which we just showed above isn't a vector. Addition doesn't work.

You can redefine these terms to make them work, but as long as you see what's wrong with the way you've set it up, I think that's more important than working out the details.

 

[HallsofIvy]

For example: x+2y+z=6 and x+2y+z=0. These are parallel planes, but the latter intersects the origin, yes?

If I just get vectors by considering two points on either plane, technically I can get the same vectors for both planes as long as I'm considering two points on both planes that yield a similar vector, right? If that's the case, why should it matter that my plane go through the origin for it to be a vector space? I can still satisfy the zero vector condition in either case.

The problem with thinking of the plane x+ 2y+ z= 6 as a vector space is that, in a vector space, the two things we can do are add vectors and multiply vectors by numbers. And both of those will immediately give "non-sense answers" in the sense that they give answers outside the space.

If we take x= 1, y= 0, then 1+ 2(0)+ z= 6 gives z= 5 so (1, 0, 5) is in that plane. Similarly, if we take x= 0, y= 1, then 0+ 2(1)+ z= 6 gives z= 4 so (0, 1, 4) is also in that plane. The sum of those, as vectors in R³, is (1+ 0, 0+ 1, 5+ 4)= (1, 1, 9) and x+ 2y+ z= 1+ 2(1)+ 9= 12, not 6 so we are no longer in the same plane. Similarly, the scalar product of the number 3 with (1, 0, 5) is (3, 0, 15) and now x+ 2y+ z= 3+ 2(0)+ 15= 18, not 15.

A plane, not containing the origin, does have one nice property: if v is any vector (point) in that plane, then any vector, w, in that plane can be write in the form w= u+ v where u is in the plane, parallel to the given plane, that does contain the origin. Since (1, 0, 5) is in the plane x+2y+ z= 6, we can write any such point as [itex](x, y, z)= (x_0, y_0, z_0)+ (1, 0, 5)[/itex] where [itex](x_0, y_0, z_0)[/itex] satisfies [itex]x_0+ 2y_0+ z0= 0[/itex]. That is because if x+ 2y+ z= 6, then [itex](x_0+ 1)+ 2(y_0+ 0)+ (z_0+ 5)[/itex][itex]= (x_0+ 2y_0+ z_0)+ (1+ 0+ 5)[/itex][itex]= (x_0+2y_0+ z_0)+ 6= 6[/itex] and so [itex]x_0+ 2y_0+ z_0= 0[/itex].

We can extend that idea to the general concept of a "linear manifold" in a vector space as follows: A subset, U, of vector space V, is a linear manifold if and only if there exist a vector, v, such that the set "U- v", defined to be the set of all vectors of the form u- v for u in U, is a subspace of V. (Since a subspace must contain the 0 vector, there must be u in U such that u- v= 0. That is, u= v so v must be in U itself. Note, that, in fact, any v in U will work.)

While a "linear manifold" in a vector space is NOT a subspace in the sense that it does not form a vector space using the vector addition and scalar addition defined for the larger space, we can redefine them to make it a vector space. Suppose U is a linear manifold in vector space V and v is a vector in U. For any vectors, u and w, in U, we define "u+ w" by

"Subtract v from both u and w (so u- v and w- v are now in the subspace). Add (u- v) and (w- v) in the subspace. Because the subspace is closed under addition, (u- v)+ (w- v)= u+ w- 2v is in the subspace. Now add v back again, "lifting" that sum back to U: u+ w- 2v= u+ w- v."

Similarly, we can define scalar multiplication, au, by
"subtract v from u to get u- v in the subspace. Since the subspace is closed under scalar multiplication, 2(u- v)= 2u- 2v is in the subspace. Adding v again "lifts" that to the linear manifold U: 2u- v is in U.

For example, in the plane x+ 2y+ z= 6, we already know that u= (1, 0, 5) and w= (0, 1, 4) are in the plane. It is also easy to see that v= (1, 1, 3) is also in the plane. Subtracting v from any point in the plane, (x, y, z) gives (x- 1, y- 1, z- 3) which satifies x+ 2y+ z= 0 and forms a subspace of R³. Using the definition of "addition" above we subtract v= (1, 1, 3) from both (1, 0, 5) and (0, 1, 4), getting (0, -1, 2) and (-1, 0, 1) which satisfy x+ 2y+ z= 0. Their sum, (0, -1, 2)+ (-1, 0, 1)= (-1, -1, 3) also satisfies x+ 2y+ z= 0 and adding v back again, (1, 0, 5)"+" (0, 1, 4)= (0, 0, 6) which is in the plane x+ 2y+ z= 0.

Also, to multiply the number a times the vector (1, 0, 5) we subtract (1, 1, 3) to get (0, -1, 2), multiply by a: (0,-a, 2a), and add (1, 1, 3) again: a"*"(1, 0, 5)= (1, 1- a, 3+ 2a).

Of course, this new vector space formed by this new definition of addtion and scalar multiplication doeshave a 0 vector: it is v= (1, 1, 3).

This concept of "linear manifold" is of importance in linear differential equations. It is easy to show that the set of all nth order, linear, homogeneous differential equations, [itex]a_n(x) d^ny/dx^n+ a_{n-1}(x)d^{n-1}y/dx^{n-1}[/itex][itex]+ \cdot\cdot\cdot+ a_1(x)dy/dx+ a_0(x)y= 0[/itex] forms an n-dimensional vector space. That means that if we can find n "independent" solutions, [itex]y_1(x)[/itex], [itex]y_2(x)[/itex], ..., [itex]y_n(x)[/itex], then any solution can be written as a linear combination of them: if y(x) is any solution to that equation, then [itex]y(x)= C_1y_1(x)+ C_2y_2(x)+ \cdot\cdot\cdot+ C_ny_n(x)[/itex].

From that, we can show that the set of all solutions to the non-homogeneous equation, [itex]a_n(x) d^ny/dx^n+ a_{n-1}(x)d^{n-1}y/dx^{n-1}+[/itex][itex] \cdot\cdot\cdot+ a_1(x)dy/dx+ a_0(x)y= f(x)[/itex] form a linear manifold. Any solution to that equation can be written as a solution to the corresponding homogenesous equation, [itex]a_n(x) d^ny/dx^n+ a_{n-1}(x)d^{n-1}y/dx^{n-1}+[/itex][itex] \cdot\cdot\cdot+ a_1(x)dy/dx+ a_0(x)y= 0[/itex], plus a single solution to the entire equation.

For example, the derivative of [itex]y= e^x[/itex] is, of course, [itex]e^{x}[/itex] so the second derivative is [itex]y= e^x[/itex] again while the derivative of [itex]y= e^{-x}[/itex] is [itex]-e^{-x}[/itex] so the second derivative is again [itex]y= -(-e^{-x})= e^{-x}[/itex]. That is, both [itex]e^x[/itex] and [itex]e^{-x}[/itex] satisfy the linear, homogeneous differential equation [itex]d^2y/dx^2- y= 0[/itex]. Since they are clearly independent (two vectors are independent as long as one is not a multiple of the other), any solution to that equation must be of the form [itex]y(x)= Ce^x+ De^{-x}[/itex] for constants C and D.

It is also easy to see that if y(x)= -1, the constant function, satisfies dy/dx= 0, so [itex]d^2y/dx^2= 0[/itex], and so [itex]d^2y/dx^2- y= 1[/itex]. Putting that together with the above, any solution to [itex]d^2y/dx^2- y= 1[/itex] must be of the form [itex]y(x)= Ce^x+ De^{-x}- 1[/itex] for some numbers C and D.