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Vector subspaces

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data

    (x,y,z) where 2x + 2y + z = 1

    Is this set a subspace of R^3?

    3. The attempt at a solution

    I am thinking it is not since it does not contain the origin since
    2(0)+2(0) + 0 = 1
    0 != 1
    (!= means not equal)
    Am I right? I am kind of having trouble with this part of linear algebra since the textbook is so vague about this.
     
  2. jcsd
  3. Mar 7, 2014 #2

    Mark44

    Staff: Mentor

    Yes, this is enough to show that this set is not a subspace of R3. BTW, the set is a plane in R3.
     
  4. Mar 7, 2014 #3
    Yea I know that, but how can I prove that something is a subspace of R^3? I know that proof against is easier, since you only need one tuple to disprove. But to do you need to do anything extra?
     
  5. Mar 7, 2014 #4

    Mark44

    Staff: Mentor

    To prove that a subset of a vector space (such as R3) is actually a subspace of that vector space, you need to show three things:
    1. The zero vector is an element of the set.
    2. If u and v are arbitrary elements of the set, then u + v is also an element of the set. (This shows that the set is closed under vector addition.)
    3. If u is an arbitrary element of the set, and k is a scalar, then ku is also an element of the set. (This shows that the set is closed under scalar multiplication.)
    That's it.

    It's a lot more work to show that a set together with two operations (vector addition and scalar multiplication) is actually a vector space - you have to verify that all 10 or so vector space axioms are satisfied. If the set already belongs to a vector space, then all you need to verify are the three I listed above.
     
  6. Mar 7, 2014 #5
    Thanks, I figured it out, and kind of on the same subject. If you are given u = (1,2,-1,1), v = (2,-1,1,0), and w = (-1,2,0,3) write x = (2,-3,-2,-8) as a linear combination of u,v, and w. I really don't know where to start, the text book gives examples of 3 3D vectors, not 3 4D vectors so how would I go ()u + ()v + ()w if you have 4 values for x?
     
  7. Mar 7, 2014 #6
    You can use good old row reduction to do this. Set up an augmented matrix whose first three columns are ##u, v, w##. Then make the last column ##x##. Then row reduce this.

    Note: You could also do it by inspection, although that is harder and may require a lot of playing around/ getting lucky.
     
    Last edited: Mar 7, 2014
  8. Mar 7, 2014 #7

    Mark44

    Staff: Mentor

    To elaborate on what kduna said, start with this vector equation:
    $$\begin{bmatrix} 2 \\ -3 \\ -2 \\ -8\end{bmatrix} = c_1\begin{bmatrix}1 \\ 2 \\ -1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix}2 \\ -1 \\ 1 \\0 \end{bmatrix} + c_3\begin{bmatrix}-1 \\ 2 \\ 0 \\ 3 \end{bmatrix}$$

    Your job is to find the three constants c1, c2, and c3 - the coordinates of x in terms of the vectors u, v, and w.
     
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