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Homework Help: Linear algebra problem.

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    W=[x,y,z]: 2x-y+3z=0 find W perp and give a basis for W perp

    3. The attempt at a solution
    None, I have no idea how to do this. No lecture on it, no textbook, and can't find anything on the net ;\

    If someone could point me in the right direction I'd really appreciate it.
  2. jcsd
  3. Dec 7, 2008 #2
    Okay.. I'm reading that the perp is the nullspace of the matrix.

    [2,-1,3], reduce row echelon form it to [1,-.5,1.5].

    which is just x-.5y+1.5z=0 now I know that isn't right.
  4. Dec 7, 2008 #3
    x_1 + -.5x_2 +1.5x_3 = 0
    and then x_1 = .5x_2 - 1.5x_3
    which means that [.5,1,0] and [-1.5,0,1] are the nullspaces which are also the complements?
  5. Dec 7, 2008 #4
    So I mean.. If that's right, then what is the complement and what is the basis? Argh I feel like nothing will give me a straight answer. Could someone explain what it is that's going on here? I'd truly appreciate it.
  6. Dec 7, 2008 #5


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    Homework Helper

    The formal definition of [tex]W^\perp[/tex] for a subspace W in a vector space V is that for a given definition of an inner product <..> on a vector space, So [tex]v \in W^\perp => v \cdot u = 0 | u \in W[/tex]
    If it's not defined the inner product is usually taken to be the Euclidean inner product:

    [tex]<\mathbf{u},\mathbf{v}> = \mathbf{u} \cdot \mathbf{v} = \sum_{i}^n u_i v_i [/tex].

    Think of it just the dot product in R^n, where here n=3. As for your question, first find a basis for W. Then use the definition of the inner product to find a basis for W_perp.
    Last edited: Dec 7, 2008
  7. Dec 7, 2008 #6
    So my solution is incorrect? And if I do that, and I find w perp's basis. I'm not finding w perp.
  8. Dec 8, 2008 #7


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    First find a basis for W. You have the eqn x_1 - 0.5x_2 + 1.5x_3 = 0. So that means you need to use some parameters to find the general solution of the equation. There should be two free variables, or parameters. Then with that you can find the basis for W_perp. Let a vector [tex]u = (x,y,z)^T \in W^\perp[/tex]. Then by definition, [tex]u \cdot v = 0 | v \in W [/tex]. So you need to find the dot product of u with the basis vectors of W, and in the process, solve for x,y,z. Expressing u in the general solution will give you a basis for W_perp.
  9. Dec 8, 2008 #8
    Yes and I got the two free variables, and then from there, which I think I got the general solution for. But I don't understand what you mean from there.
  10. Dec 8, 2008 #9
    Are what I have found incorrect? Are the nullspace of this solution not the basis for w perp? Because when I test it it comes out correct. But once again it's only the basis for w perp. NOT w perp itself.
  11. Dec 8, 2008 #10


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    Hi, I'm not sure what you understand at this point. W perp itself is completely described by its basis, just like any subspace. And the answer you have found is the basis for W, not W perp.
  12. Dec 8, 2008 #11
    I guess I just don't understand the latex code. The cryptic mathematical way of keeping things short and sweet. If you don't mind, could you explain how to get it in a much more detailed method? I'm rather slow with this stuff.
  13. Dec 8, 2008 #12
    Also a book I found claims that the null space IS the perp complement of the row space. And when I do a dot product of each of the null space vectors I found against the original they add up to 0.
  14. Dec 8, 2008 #13


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    Your book may say that but the problem, as you originally stated it, has nothing to do with either a "null space" nor a "row space". Both of those are properties of a linear transformation or matrix and you do not have either. You are given a subspace, W, of R3, given by W= {(x,y,z)| 2x-y+3z=0 }. If we take x= 1, z= 0, then 2- y= 0 so y= 2. (1, 2, 0) is in W. If we take x= 0, z= 1, then -y+ 3= 0 so y= 3. (0, 1, 3) is also in W and it should be easy to see that {(1, 2, 0), (0, 1, 3)} is a basis for W. The "orthogonal complement" of W consists of vectors that are perpendicular to all vectors in W and that is true if and only if a vector is perpendicular to both (1, 2, 0) and (0, 1, 3). Since two vectors in Rn are perpendicular if and only if their dot product is 0, (x, y, z) is in the orthogonal complement of W if and only if (1, 2, 0).(x, y, z)= x+ 2y= 0 and (0, 1, 3).(x, y, z)= y+ 3z= 0. What (x, y, z) satisfy both of those? Since those are two equations in 3 unknown values, you can solve for two of the, say x and z, in terms of the third, y, and write all vectors in the orthogonal complement of W in terms of y.
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