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Algebra- Vector ce and subspace

  1. Sep 30, 2007 #1
    Algebra- Vector space and subspace

    1. The problem statement, all variables and given/known data
    Here are some true or false statements given in my test.
    (a) R^2 is a subspace of R^3.
    (b) If {v1, v2, ..., vn} is a set of linearly dependent vectors, then it contains a zero vector.
    (c) If {v1, v2, ..., vn} is a spanning set, then {v1, v2, ..., vn} are linearly independent.

    2. Relevant equations

    3. The attempt at a solution
    (a) True, because R is a subspace of R^2 and R^2 is a subspace of R^3 and R^3 is a subspace of R^4, and so on.
    (b) False, because it may or may not contain a zero vector. I think that it is true for this statement: If {v1, v2, ..., vn} contains a zero vector, then it is linearly dependent. But the statement "If {v1, v2, ..., vn} is a set of linearly dependent vectors, then it contains a zero vector." is false.
    (c) False. Beacuse vectors in spanning sets can be expressed as linear combinations of each others, and hence it is consistent and they are linearly dependent.

    Any opinion on these questions? Thanks.
    Last edited: Sep 30, 2007
  2. jcsd
  3. Sep 30, 2007 #2
    a) you need to check the definition of subspace. Take a look at an element in R^2: (a,b) and an element in R^3: (x,y,z). R^2 is the set of all 2-tuples with real entries and R^3 is the set of all 3-tuples with real entires. I would say R^2 is not a subspace, but I'll leave it to you to justify why.

    b,c) you're on the right track, but the best way to show a T/F question is false is to provide a specific counter-example. For example if the statement is: "All odd numbers are divisible by 2" you might answer, "False, consider 3, 3 is odd and not divisible by 2" and you're done. Counterexamples can be hard to find, but they're usually easy to write down.
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