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Test for subspace (is 1 condition sufficient?)

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

    U= { (x1, x2, x3, x4) | x1 x3 ≥ -5 }


    3. The attempt at a solution

    Let x = (1,2,3,4) and y = (1,2,3,4)

    x+ y = (2,4,6,8)

    x1x3 = 2x6 = 12

    12 >-5 so closure by addition is fulfilled.

    I've been hearing contradicting information-some state that any 1 test of the 3 is a sufficient condition.

    Should I perform the closure by scalar multiplication, it can be easily shown that U is not a subspace of R4. Is there a less tedious way to determine that a vector is not an subspace?
     
  2. jcsd
  3. Apr 1, 2014 #2

    Fredrik

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    It's not enough to show that U is closed under addition, or that U is closed under scalar multiplication, because the empty set is closed under both of these operations.

    The statement that U is closed under addition means that for all x,y in U, x+y is in U. You have only proved that there exist x,y in U such that x+y is in U.

    In general, when you prove a "for all x" statement, the proof should almost always begin with a "let x be arbitrary" statement. Then you prove that x has the desired property.
     
  4. Apr 1, 2014 #3

    HallsofIvy

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    In general if a statement says "all of these are true", then to show that the statement is false you need only show a single counter example.

    Here, if the condition is that x1 and x3 be greater than or equal to -5, you need only show that -10(-2, 1, -2, 1)= (-20, 10, -20, 10). (-2, 1, -2, 1) has x1= x3= -2> -5 but (-20,10, -20, 10) has x1= x3< -5.

    Another counter example, is (-3, 1, -3, 1)+ (-4, 1, -4, 1)= (-7, 2, -7, 2). You use a counter-example- am example to show that the conditions are NOT satisfied. Not, as you have here, an example in which the conditions are satisfied.
     
  5. Apr 1, 2014 #4


    Got it.

    In effect, if the question puts forth a universal proposition, all I do need to do to to demonstrate one counter example to falsify the suggested proposition.
     
    Last edited: Apr 1, 2014
  6. Apr 1, 2014 #5

    Mark44

    Staff: Mentor

    Just to be clear, you can disprove a general statement by finding a single counterexample, but no number of examples are sufficient to prove a general statement. The only exception is if you list every possible situation that the general statement represents and show that the statement is true for every one of them.
     
  7. Apr 2, 2014 #6
    Very logical.
     
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