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Thread: Subspace in vectorspace

  1. #1
    Oct 2011

    Subspace in vectorspace

    First off, I would like to apologize if this is the wrong forum I am posting in. I looked at all the forums and I believe that the Calculus forum is the most appropriate for this question, but if I made a wrong decision then I'm sorry!

    The following equation system is a subspace in $\displaystyle \mathbb{R}^5$:

    $\displaystyle x_2+3x_3-x_4+2x_5=0$

    $\displaystyle 2x_1+3x_2+x_3+3x_4=0$

    $\displaystyle x_1+x_2-x_3+2x_4-x_5=0$

    How many dimensions is the subspace?
    What is a basis for this subspace?


    Okay so first off, I want to figure out how many dimensions the subspace is.
    I put the equation system into a totalmatrix:

    $\displaystyle \begin{bmatrix}0 & 1 & 3 & -1 & 2 & 0\\ 2 & 3 & 1 & 3 & 0 & 0\\ 1 & 1 & -1 & 2 & -1 & 0\end{bmatrix}$

    I then find its parametric solution:

    $\displaystyle \begin{bmatrix}x_1\\ x_2\\ -2x_1-3x_2-3x_4\\ x_4\\ 3x_1+4x_2+5x_4\end{bmatrix}$

    But here comes my problem...
    According to my textbook, the answer is that it has 3 dimensions!
    And in my parametric solution I only have 2 vectors (row 3 and 5) and two vectors only span over a 2 dimension!

    So I have no idea what it is I have done wrong. Why am I not getting 3 dimensions like my textbook?
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  2. #2
    Oct 2010

    Re: Subspace in vectorspace

    It looks like you have three dimensions. x1, x2, and x4. These three vectors form the basis of the solution of this system of equations, thus making the answer dim = 3. The other two are linear combinations of the three vectors you have. And also, I believe this question is more appropriately placed in the "Linear Algebra" section . Hope that helps!
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  3. #3
    MHF Contributor

    Mar 2011

    Re: Subspace in vectorspace

    if you choose x1,x2,and x4, then x3 and x5 are determined for you (they are not "free", being linear combinations of x1,x2 and x4).

    if x = (x1,x2,x3,x4,x5) is an arbitrary vector that is a solution, we can write:

    x = (x1,x2,-2x1-3x2-3x4,x4,3x1+4x2+5x4) = x1(1,0,-2,0,3) + x2(0,1,-3,0,4) + x4(0,0,-3,1,5),

    so {(1,0,-2,0,3),(0,1,-3,0,4),(0,0,-3,1,5)} is a basis for the solution space.

    "5" is just the number of coordinates your solutions have. "3" is the number of basis elements you have (how many variables you have to specify to get a solution).

    note that "2" (the number of linearly dependent vectors, in this case x3 and x5) is just what's left-over when we subtract 3 from 5.

    3 is the "nullity" of the matrix from the system, 2 is the "rank" of the system (you eventually wind up with 2 equations:

    -2x1-3x2-x3-3x4 = 0 <---> x3 = -2x1-3x2-3x4
    3x1+4x2+5x4-x5 = 0 <---> x5 = 3x1+4x2+5x4)

    nullity+rank (of a homogeneous set of linear equations) = number of variables to start with <---always true, and very important.

    if you row-reduce, the rank is also the number of non-zero rows you wind up with.
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