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What is the dimension of this special stuff

  1. Aug 11, 2007 #1
    1. The problem statement, all variables and given/known data

    I am learning linear algebra (basic) and using Lang's book. In talking about vector spaces (finite dimensional only), it is easy to understand what is the dimension of the vector space (i.e. the number of elements in any basis). But I met the following case (not a vector space), in which I am puzzled.

    2. Relevant equations

    a) I know the dimension of the set of the solutions for AX=0 (usual meaning of the symbols in many text books,X belongs to R^n). Because such a set forms a vector space (actually a subspace), the dimension is easily understood.

    b) For a non-homogeneous system of linear equations AX=B. And suppose solution(s) exist.Then what is the dimension of the set of the solutions for AX=B? The set of solutions does not form a subspace (is only a subset of R^n), so the defintion of such a dimension must be defined newly.



    3. The attempt at a solution

    Lang took dimension in (b) as that corresponds in (a). But he did not DEFINE it ,only CALLED it.


    Thanks
     
    Last edited: Aug 11, 2007
  2. jcsd
  3. Aug 12, 2007 #2

    radou

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    b) linear variety.
     
  4. Aug 12, 2007 #3
    Thanks. But I don't know linear variety.

    I can guess one case, e,g. A first straight line in R^3, that line can be passing the origin, Shifting the first line makes the second line. Now both are lines (can be extended to planes etc), so they have the same dimension? Is this what linear variety about?

    Does the above have some generalities. For example, a set in R^3, which is not a line or plane. e.g. the set is a set consisting some lines plus some points(points do not lie on the lines). (1,2,3) and a line (x=y=z) can make such a set. Then can I question what is the dimension for such a set ?
     
  5. Aug 12, 2007 #4

    radou

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    Well, you can't question the dimension of any set, first of all. You can question the dimension of a vector space. Now, is such a set a vector space (i.e. its underlying set)?
     
    Last edited: Aug 12, 2007
  6. Aug 12, 2007 #5
    But my case b) is also NOT A VECTOR SPACE, it is only a subset in R^n. But the subset is only a little special. Since you can define the dimension of this little special stuff (not a vector space), it is natural to question whether such a definition has some generality.

    Note:The special thing about case b) is that the solution space is a set below

    one special solution (of AX=B) + Ker (La)

    ,where Ker (La) is solution space of AX=0

    Thanks
     
  7. Aug 12, 2007 #6

    HallsofIvy

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    I would say "linear manifold" rather than "linear variety" but I think Radou and I mean the same thing.

    Think R2: any subspace is a straight line through the origin. It must pass through the origin because a subspace must contain the 0 vector. Also, of course, the sum of any two vectors in that line is again in the line.
    Example: any point on the line y= 3x is of the form (x, 3x) of course. Adding two such things gives (x1+ x2,3x1+3x2)= (x1+ x2,3(x1+x2)), again a point on the line.

    A straight line that does not pass through the origin is a "linear manifold". It is not a subspace because two vectors in it do not add to give vector in the line: Example: y= 3x+ 2 is straight line not passing through the plane. Any point on it is of the form (x, 3x+ 2). Adding to such "vectors" gives (x1+ x2,3x1+ 2+3x2+2)= (x1+ x2,3(x1+x2)+ 4) which is NOT on the same line.

    But such a linear manifold is parallel to a subspace: the line y= 3x+ 2 is parallel to y= 3x. We can find a "vector" in y= 3x+2, say, (0, 2), and write any vector in y= 3x+ 2 as that plus a vector in the subspace y= 3x: obviously, (x, 3x+ 2)= (x, 3x)+ (0, 2).

    This is extermely important in the theory of linear differential equations as well as for matrix equations. If you can find a "general solution" for Ax= 0 (i.e. define the subspace of such solutions), then find a single solution, say u, such that Au= b, then their sum is the "general solution" to Ax= b. That's why the dimensions are the same.
     
  8. Aug 12, 2007 #7
    Hallsofivy,

    I appreciate your linear manifold stuff, which is parallel to a subspace. Then the dimension of the linear manifold is DEFINED to be the dimension of the subspace (parral to the manifold) (I am not sure whether this is a definition or not)? Or is there another general definition of the dimension for the linear manifold so that the claim (for the linear manifold and the subspace have the same dimension) is a theorem?
     
  9. Aug 12, 2007 #8
    Last edited: Aug 12, 2007
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