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Elementary Linear Algebra Proof Basis/Dimension

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that if W is a subspace of a finite-dimensional space V, then (dimension of W) is less than or equal to (dimension of V).

    3. The attempt at a solution
    I've been trying to prove this by contradiction (assume dimension of W is greater than V and arrive at a contradiction) but am a little unclear on how to write the proof out. Here is my attempt so far:

    W is a subspace of V. Assume S is a basis for W. If span(S) = V then dim(W) = dim(V).

    Otherwise lets assume that dim(W) > dim(V).
    Let S = {V1,V2,...,Vn} be a basis for W. Let T = {V1,V2,...,Vk) be a basis for V where, because of our assumption, n>k (there are more vectors in the basis S than in T).

    Here is where i'm not sure where to proceed. I'm trying to see where i can use the fact that S is a subset of V and compare the number of vectors in S and T but i'm just not seeing the answer. Please help me understand the next step conceptually. Thank you for your time.
     
  2. jcsd
  3. Oct 22, 2011 #2

    verty

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    From a purely logical viewpoint, you seem to make a leap here: "otherwise let's assume ...". You should draw an implication: "otherwise, we have that ..." or "otherwise, there is ...".
     
  4. Oct 22, 2011 #3

    HallsofIvy

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    You are using a "proof by contradiction" so better wording would be "On the contrary, assume".... And, there is no reason to do dim(W)= dim(V) separately. Once you have shown a contradiction to "dim(W)> dim(V)", "less than or equal" follows.

    And I would recommend being more "precise". "Let m be the dimension of V, n the dimension of W, with n> m." That means that there exist as set of n independent vector in W. Do you see how that contradicts m begin the dimension of V?
     
  5. Oct 22, 2011 #4
    Thank you for helping me. It feels like this should be fairly simple but i'm having trouble writing out my ideas logically. Am I correct with my following reasoning?

    Let m be the dimension of V, n the dimension of W, and assume n > m. This means there exists a set of n independent vectors in W. Since W is a subspace of V, this implies that there is a set of independent vectors in V larger than m. But this is impossible since every set in V containing more than m vectors must be linearly dependent. So n cannot be greater than m. Therefore it can only be true that dim(W) is less than or equal to dim(V).
     
  6. Oct 23, 2011 #5

    HallsofIvy

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    Yes, that is perfectly good.
     
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