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Subspace test involving linear transformations

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine whether the subset W of the vector space V is a subspace of V.
    Let V = L(Q4) (the set of linear transformations from rational numbers with 4 coordinates to rational numbers with 4 coordinates).
    Let W = { T in V = L(Q4) | { (1,0,1,0) , (0,1,0,-1) } is contained in N(T) }
    where N(T) is the nullspace of T

    2. Relevant equations
    Subspace test involves checking if the zero vector is in W, whether W is closed under addition, and whether W is closed under scalar multiplication.

    3. The attempt at a solution
    1. Check if zero vector is in W
    From definition, N(T): {x in V | T(1,0,1,0) = 0 , T(0,1,0,-1) = 0 }
    So the zero vector is in W

    2. Check if closed under addition
    From linearity of T, T(x+y) = T(x) + T(y)
    where x = (a, b, c, d) in W
    y = (p, q, r, s) in W

    3. Check if closed under scalar multiplication
    From linearity of T, T(cx) = cT(x)
    where x = (a, b, c, d) in W

    So W is a subspace of V.

    Is it enough to use the linearity of T to check for closure under addition and multiplication? Is there something else I should include in my solution? I'm unsure if I'm supposed to do the subspace test on the actual vectors x, y, etc. or if I'm supposed to test the transformations, T(x), T(y), etc. For example, should closure under addition be asking whether x + y is in W or if T(x) + T(y) is in W?

    Thanks in advance!
     
  2. jcsd
  3. Dec 8, 2008 #2

    Mark44

    Staff: Mentor

    Your set W consists of all the transformations from Q^4 to Q^4 that send (1,0,1,0) and (0,1,0,-1) to (0, 0, 0, 0). This fact has ramifications on what the matrices for this set of transformations looks like.

    I don't see that your work uses the two given vectors.
    For the first part, you need to show that T(0, 0, 0, 0) = (0, 0, 0, 0)
     
  4. Dec 8, 2008 #3
    Okay, so checking if T(0,0,0,0) = 0 is asking if (0,0,0,0) is in N(T). Since (0,0,0,0) is not in N(T), then the zero vector is not in W. Is this right?
     
  5. Dec 8, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    W is a space of linear transformations. "Closed under addition" is not a question of 'x+ y' but of T+ S where T and S are in W. If T and S are in W is T+S in W?

    Again, the question is not about cx but about cT. If T is in W is cT in W?

     
    Last edited: Dec 8, 2008
  6. Dec 8, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Once again, the question is not about T(0,0,0,0)= 0. It is about the 0 transformation: T(x,y,z,w)= (0,0,0,0) for all x,y,z,w. If T= 0, the linear transformation that maps every (x,y,z,w) to (0,0,0,0), is it in W? That specifically, is asking whether T(1, 0, 1, 0)= (0,0,0,0) and whether T(0,1,0,-1)= (0,0,0,0) and the answer to that is trivial.
     
  7. Dec 8, 2008 #6
    Thanks for your reply.
    So it's obvious that the zero vector is in W.

    Check if T + S is in W (closure under addition):
    T(1,0,1,0) + S(0,1,0,-1) = 0 + 0 = 0
    Since T + S is in N(T), W is closed under addition.

    Check if cT is in W (closure under scalar multiplication):
    cT(1,0,1,0) = c * 0 = 0
    cT(0,1,0,-1) = c * 0 = 0
    Sin cT is in N(T), W is closed under scalar multiplication.

    Is this right or am I missing something?
     
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