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Thread: [SOLVED] Affine subset

  1. #1
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    [SOLVED] Affine subset

    I've just been handed a project for my Linear Algebra course, so far I can't even begin to grasp the concept of affine subsets. Anyway I was wondering if anyone could help with the first question in which I am meant to illustrate the concept:

    Show that:

    M = {x=(x1,...x4) exists in R^4 : x1 - 2*x2 + x3 = 3 and x1 + 5*x3 - 2*x4=1}

    is an affine subset of R^4.

    Sorry about the notation here. x1 means x subscript 1.
    If anyone can help I would be really grateful.
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  2. #2
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    Quote Originally Posted by Galbraith005 View Post
    I've just been handed a project for my Linear Algebra course, so far I can't even begin to grasp the concept of affine subsets. Anyway I was wondering if anyone could help with the first question in which I am meant to illustrate the concept:

    Show that:

    M = {x=(x1,...x4) exists in R^4 : x1 - 2*x2 + x3 = 3 and x1 + 5*x3 - 2*x4=1}

    is an affine subset of R^4.
    Geometrically, an affine subset is something that looks like a subspace except that it doesn't contain the origin. Algebraically, if S is a subspace and v is a vector, then the set v+S (in other words, the set of all vectors of the form v plus something in S) is an affine set, and every affine set is of this form.

    In the above example, start by finding a vector in M. The easiest one to find is probably v=\bigl(0,\tfrac32,0,-\tfrac12\bigr). Now let S = \{x\in\mathbb{R}^4:x_1 - 2x_2 + x_3 =  x_1 + 5x_3 - 2x_4= 0\}. Show that S is a subspace of \mathbb{R}^4, and M = v + S. Therefore M is an affine set.

    An affine set is simply a subspace that has been translated away from the origin.
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  3. #3
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    Thanks...

    How can you set the two eqn's equal to each other when one equals 1 and the other equals 3?
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    Quote Originally Posted by Galbraith005 View Post
    How can you set the two eqn's equal to each other when one equals 1 and the other equals 3?
    We're talking about two different sets of equations here. There's the affine set M, which is the set of all vectors x = (x_1,x_2,x_3,x_4) such that x_1 - 2x_2 + x_3 = 3 and x_1 + 5x_3 - 2x_4= 1. Then there is the subspace S, which is the set of all vectors x = (x_1,x_2,x_3,x_4) such that x_1 - 2x_2 + x_3 = 0 and x_1 + 5x_3 - 2x_4= 0.

    It might have made it clearer if I had used a different letter, say y instead of x, for elements of the subspace S. Suppose I had said that S = \{y = (y_1,y_2,y_3,y_4)\in\mathbb{R}^4:y_1 - 2y_2 + y_3 = y_1 + 5y_3 - 2y_4= 0\}. Then the point is that if  y\in S then v+y\in M, where v=\bigl(0,\tfrac32,0,-\tfrac12\bigr). Conversely, if x\in M, then x = v+y for some y\in S.

    Does that make it any clearer?
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    Thank you.

    Thanks, you've been a great help. I think I've got it now.
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  6. #6
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    Thanks for the help
    Last edited by Kazzy2906; Feb 27th 2010 at 03:28 PM.
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  7. #7
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    We're talking about two different sets of equations here. There's the affine set M, which is the set of all vectors such that and . Then there is the subspace S, which is the set of all vectors such that and .

    It might have made it clearer if I had used a different letter, say y instead of x, for elements of the subspace S. Suppose I had said that . Then the point is that if then , where . Conversely, if , then for some .
    Let  y=(1,1,1,3) \in S) then using your definition of  v we have  v+y=(1, \frac{5}{2}, 1 , \frac{5}{2}) \not\in M .

    P.S. Guys try to do this project on your own. There is no point in begging for answers on math forums.
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  8. #8
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    Quote Originally Posted by patpat View Post
    Let  y=(1,1,1,3) \in S then using your definition of  v we have  v+y=(1, \frac{5}{2}, 1 , \frac{5}{2}) \not\in M .
    Careless typo on my part. The vector v should have been (0,-\tfrac32,0,-\tfrac12). Then v+y = (1,-\tfrac12,1,\tfrac52), which is in M of course.
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