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What is the proof of this theorem in linear algebra ?

  1. Oct 18, 2012 #1
    hello, the theorem says :
    let V be a vector space over the field K , let { v1 , v2 , ... , vm } be a basis of V over K
    let {w1 , w2 , ... , wn} be elements of V and assume that n is bigger than m , then
    { w1 , w2 , ... , wn } are linearly dependent

    the proof is written here but I didn't understand it well , I hope that you give me a simple proof :)
  2. jcsd
  3. Oct 18, 2012 #2


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    First, do you understand that saying "{v1, v2, ..., vm} is a basis of V" means that every vector in V can be written as a unique linear combination of those vectors? And, I assume you know that saying that "{w1, w2, ..., wn} are linearly dependent" means that there exist members of K, a1, a2, ... , an, not all 0, such that a1w1+ a2w2+ ... + anwn= 0.

    To prove that, suppose a1w1+ a2w2+ ... + anwn= 0. Replace each wi with its expression as a linear combination of the basis vectors: wi= bi1v1+ bi2v2+ ...+ bimvm. The a1w1+ a2w2+ ...+ anwn= a1(b11v1+ b12v2+ ... b1nvn)+ a2(b21v1+ b22v2+ ...+ b2nvn)+ ...+ am(bn1v1+ bn2v2+ ...+ bmnvn)= (a1b11+ a2b12+ ...+ anbn1)v1+ (a1b12+ a2b22+ ... ambm2)v2+ ...+ (a1b1n+ a2b2n+ ... ambmn)vm= 0.

    Because the {v1, v2, ..., vn} are a basis, they are independent and so we must have a1b11+ a2b12+ ...+ anbn1= 0, a1b12+ a2b22+ ... ambm2= 0, ..., a1b1n+ a2b2n+ ...+ ambmn= 0. That is m linear equations in the n "unknowns" a1, a2, ..., am. Because n> m, we can always solve for m of the unknowns in terms of the other n- m. Taking any of the am+1, am+2, ..., an to be non- zero show that the m vectors are dependent.
  4. Oct 18, 2012 #3


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    there are many proofs.

    one concrete one is to put the vectors into a matrix and row reduce it until you see there are dependency solutions.

    another is to use riemann's "exchange" lemma, often falsely attributed to steinitz.

    i have given another in my math 4050 notes ion my webpage.

    most proofs proceed by induction. you might try making your own.
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