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Math Help - Another spanning set question.

  1. #1
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    Another spanning set question.

    I keep getting inconsistent results with this one. I can't find a single vector that belongs to both sets. Is the question correctly formed?

    Question:
    Two subspaces S and T of R^3 are spanned by {(1,1,-2), (-1,1,0) } and { (1,1,0), (0,-11,1)} respectively. Find a non-zero vector X that belongs to S\cap T.

    This is what I did:

    We need X=a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1).

    So,
    a-b=c
    a+b=c-11d
    -2a=d

    Right so far?

    From here I can't get any consistent solution for a, b such that a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1).
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  2. #2
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    I don't know if I'm doing this right, but if I solve for a and b in terms of c and d, I get:

    -2a=d \Longrightarrow \boxed{a = -\frac{1}{2}d}

    a-b=c \Rightarrow \left(-\frac{1}{2}d\right)-b=c \Longrightarrow \boxed{b=-c-\frac{1}{2}d}

    But then if I solve the middle equation for b (to check), I get:

    a+b=c-11d \Rightarrow \left(-\frac{1}{2}d\right)+b=c-11d \Longrightarrow \boxed{b=c-\frac{21}{2}d}

    The last two b's are inconsistent! Have I done something wrong?
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  3. #3
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    Quote Originally Posted by scorpion007 View Post
    I keep getting inconsistent results with this one. I can't find a single vector that belongs to both sets. Is the question correctly formed?

    Question:
    Two subspaces S and T of R^3 are spanned by {(1,1,-2), (-1,1,0) } and { (1,1,0), (0,-11,1)} respectively. Find a non-zero vector X that belongs to S\cap T.

    This is what I did:

    We need X=a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1).

    So,
    a-b=c
    a+b=c-11d
    -2a=d

    Right so far?

    From here I can't get any consistent solution for a, b such that a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1).
    Let a = \lambda.

    Then d = -2 \lambda.

    So c = -10 \lambda and b = 11 \lambda.

    So one particular solution is a = 1, ~ b = 11, ~ c = -10, ~ d = -2.

    So one possible vector is (1, 1, -2) + 11 (-1, 1, 0) = -10 (1, 1, 0) - 2 (0, -11, 1) = (-10, 12, -2).
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    So c = -10 \lambda and b = 11 \lambda.
    Could you explain just this step please? It seems to be what's giving me trouble. I don't see how you derived this. The rest makes sense.
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  5. #5
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    Quote Originally Posted by scorpion007 View Post
    Could you explain just this step please? It seems to be what's giving me trouble. I don't see how you derived this. The rest makes sense.
    a - b = c .... (1)

    a + b = c - 11 d .... (2)

    (1) + (2): 2a = 2c - 11 d \Rightarrow 2 \lambda = 2c + 22 \lambda \Rightarrow -20 \lambda = 2c

    Substitute c = -10 \lambda and a = \lambda into equation (1) and solve for b.
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