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Show that the set W= {x member of R^n | Ax=Bx} is a subspace of R^n

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A be an n*n matrix and let B be a real number, Nothing which properties of matrix multiplication you use, show that the set W= {x member of R^n | Ax=Bx} is a subspace of R^n, where x in the equation Ax=bx is represented by a column vector instead of an n-tuple with commas.


    2. Relevant equations



    3. The attempt at a solution

    Can i have some hint plz ..i have no idea what to do ...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 20, 2012 #2

    Dick

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    What's the definition of 'subspace'?
     
  4. Nov 20, 2012 #3
    a subset of a vector space that is closed under addition and scalar multiplication
    Let H be a homogeneous system of linear equations in x1, ...,xn . Then the subset S of R^n which consists of all solutions of the system H is a subspace of R^n .
     
  5. Nov 20, 2012 #4

    Dick

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    So can you show W is closed under addition and scalar multiplication? That's what they want. Suppose x1 and x2 are in W. Is x1+x2 in W?
     
  6. Nov 20, 2012 #5
    Yes Ax1=Bx1 =>add Ax2 on both side=> Ax1+Ax2=Bx1+Bx2 => A(x1+x2)=B(x1+x2) is this enough to show close under addition?
     
  7. Nov 20, 2012 #6

    Dick

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    Sure it is. That means x1+x2 is in W, right? Now work on scalar multiplication.
     
  8. Nov 20, 2012 #7
    Ax1=Bx1 =>mutiply x2 on both side=> Ax1*x2=Bx1*x2 => A(x1*x2)=B(x1*x2) is this enough to show close under sclar multiplication ?
     
  9. Nov 20, 2012 #8

    Fredrik

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    Nitpick: It's a non-empty subset that that is closed under addition and scalar multiplication.

    An alternative (but equivalent) definition says that a subspace of a vector space V is a subset U that's also a vector space, with its addition and scalar multiplication operations defined as the restrictions of the original addition and scalar multiplication operations to U.

    So in addition to the two closure properties, you should also verify that the subset is non-empty. Usually, the easiest way to do this is to verify that the 0 vector of the original vector space is a member of the subset.
     
  10. Nov 20, 2012 #9
    Thank you
     
  11. Nov 20, 2012 #10

    Dick

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    Same answer, yes. So if x1 is in W then x1*x2 is. But I am a little worried if you know what both sides mean. A(x1)*x2 is A(x1*x2) because A is a matrix with x2 a scalar and x1 a vector. (B*x1)*x2=B*(x1*x2) because x1 is a vector and B and x2 are scalars. They are a little different. If you understand this, then no problem.
     
  12. Nov 20, 2012 #11
    Thank you for all the help.
     
  13. Nov 20, 2012 #12

    Fredrik

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    Since you have completed the problem, I can show you how I would do it:

    0 is in W since A0=0=B0.

    For all a,b in ℝ, and all x,y in W, we have A(ax+by)=aAx+bAy=aBx+bBy=B(ax+by), so ax+by is in W. (It's of course fine to break this up into two parts, one for scalar multiplication and one for addition).
     
  14. Nov 20, 2012 #13
    Thanks I should add A0=B0
     
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