MurdocJensen said:
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 R
3, 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 (x, y, z)= (x_0, y_0, z_0)+ (1, 0, 5) where (x_0, y_0, z_0) satisfies x_0+ 2y_0+ z0= 0. That is because if x+ 2y+ z= 6, then (x_0+ 1)+ 2(y_0+ 0)+ (z_0+ 5)= (x_0+ 2y_0+ z_0)+ (1+ 0+ 5)= (x_0+2y_0+ z_0)+ 6= 6 and so x_0+ 2y_0+ z_0= 0.
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 R
3. 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, a_n(x) d^ny/dx^n+ a_{n-1}(x)d^{n-1}y/dx^{n-1}+ \cdot\cdot\cdot+ a_1(x)dy/dx+ a_0(x)y= 0 forms an n-dimensional vector space. That means that if we can find n "independent" solutions, y_1(x), y_2(x), ..., y_n(x), then
any solution can be written as a linear combination of them: if y(x) is any solution to that equation, then y(x)= C_1y_1(x)+ C_2y_2(x)+ \cdot\cdot\cdot+ C_ny_n(x).
From that, we can show that the set of all solutions to the non-homogeneous equation, a_n(x) d^ny/dx^n+ a_{n-1}(x)d^{n-1}y/dx^{n-1}+\cdot\cdot\cdot+ a_1(x)dy/dx+ a_0(x)y= f(x) form a linear manifold. Any solution to that equation can be written as a solution to the corresponding homogenesous equation, a_n(x) d^ny/dx^n+ a_{n-1}(x)d^{n-1}y/dx^{n-1}+\cdot\cdot\cdot+ a_1(x)dy/dx+ a_0(x)y= 0, plus a
single solution to the entire equation.
For example, the derivative of y= e^x is, of course, e^{x} so the second derivative is y= e^x again while the derivative of y= e^{-x} is -e^{-x} so the second derivative is again y= -(-e^{-x})= e^{-x}. That is, both e^x and e^{-x} satisfy the linear, homogeneous differential equation d^2y/dx^2- y= 0. 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 y(x)= Ce^x+ De^{-x} for constants C and D.
It is also easy to see that if y(x)= -1, the constant function, satisfies dy/dx= 0, so d^2y/dx^2= 0, and so d^2y/dx^2- y= 1. Putting that together with the above,
any solution to d^2y/dx^2- y= 1 must be of the form y(x)= Ce^x+ De^{-x}- 1 for some numbers C and D.
