GhostWhoWalks
Posts: 84
Joined: 12/13/2006
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Here's the definition of sub-space, courtesy of linear algebra; http://www.agolda.com/Linear_Algebra/Vector_Space.html
I've copied part of the explaination below.
(Er..if any of you subs get hot when someone whispers math
equations to you, let me know. I'm pretty sure that's a new kink, and I wanna get my name in the history books.  )
Definition 29: Let V be a Real vector space and let a set B of vectors in V be a basis of V. To keep things simple [in terms of the Set Theory] we assume that V is such that the set B is finite or countable infinite. Hence B={u1,u2,...} [this list might end or might not but in either case it is countable]. Then [see fact 22] every vector v of V can be represented as:
v=a1·u1+a2·u2+...
where the Real numbers a1,a2,... are unique and only finite amount of them is non-zero.
The Real numbers a1,a2,... are called the coordinates of the vector v with respect to the basis B
We also write: v=[a1,a2,...]B
Thus, with respect to some selected and then fixed basis B of V, the vector space V becomes either:
Rn if the dimension of V is n [so B has exactly n vectors in it];
R¥ if the dimension of V is À0 [so B has exactly À0 vectors in it];
Since there are some complications with infinite dimensional Real vector spaces which arise from Set-Theoretic properties of infinite sets we will restrict our attention to the finite dimensional Vector spaces only. In section 5 we will demonstrate one property of infinite dimensional Real vector spaces which makes them different from finite dimensional Real vector spaces and thus makes them a very hard object to work with.
Example 30: [See examples 31 & 33 in section 1 and example 04 in this section] Let U be the vector space of all ordered triples of Real numbers [x,y,z] such that x+y+z=0 (so z=-x-y).
Let B be the set {[1,0,-1], [0,1,-1]} of vectors in U (check that these vectors are in U !). Vectors of B are linearly independent (check it !).
Every vector [x,y,z] in U is given by:
[x,y,z]=[x,y,-x-y]=[x,0,-x]+[0,y,-y]=x·[1,0,-1]+y·[0,1,-1]
Hence B is a basis of U and
[x,y,z]=[x,y]B
Let C be the set {[1,1,-2], [2,3,-5]} of vectors in U (check that these vectors are in U !). Vectors of C are linearly independent (check it !).
Every vector [x,y,z] in U is given by:
[x,y,z]=[x,y,-x-y]=[3x-2y,3x-2y,-6x+4y]+[2y-2x,3y-3x,-5y+5x]=
=[(3x-2y),(3x-2y),-2(3x-2y)]+[2(y-x),3(y-x),-5(y-x)]=(3x-2y)·[1,1,-2]+(y-x)·[2,3,-5]
Hence C is also a basis of U and
[x,y,z]=[(3x-2y),(y-x)]C
In both cases once we have selected some basis of U and we calculate every vector with respect to this selected basis U becomes just R2 [with respect to the selected basis].
But selecting a different basis will produce a different representation of U as R2 [the same vector v of U has different coordinates with respect to different selected bases] !!!
In order to understand better how the coordinates of vectors in a Real vector space change with the change of the selected bases in that space consider the following 1-dimensional example which will make things clear:
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