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Math Help - Basis of Vector Space

  1. #1
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    Basis of Vector Space

    Hi, I have some questions about basis.

    1-)Let's say w1,w2,......,wL is in the Span(U), (U={u1,u2,.....,uK} , u1,u2,.....,uK are the basis of U)
    If L>K then w1,w2,......,wL is linearly independent.

    I understand the statement intuitively very good, but can someone give me a formal proof? I don't feel fully convinced.

    2-) If we have a Vector Space U, and its basis consists of , lets say n vectors, all other bases that I will find for the vector space U will consist of n vectors? I can't find more linearly independent vectors that spreads the whole vector space that is also the base?

    Thanks you
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  2. #2
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    Re: Basis of Vector Space

    Quote Originally Posted by davidciprut View Post
    Hi, I have some questions about basis.

    1-)Let's say w1,w2,......,wL is in the Span(U), (U={u1,u2,.....,uK} , u1,u2,.....,uK are the basis of U)
    If L>K then w1,w2,......,wL is linearly independent.

    I understand the statement intuitively very good, but can someone give me a formal proof? I don't feel fully convinced.
    Perhaps you do not understand it! What you wrote is NOT true- unless you simply mistyped it- If L> K then w1,w2,......,wL is linearly dependent, not independent.

    To see that use the definition of "dependent-independent". The vectors w1,w2,......,wL are linearly dependent if and only if there exist numbers, a1, a2, ..., aL, NOT all 0, such that a1w1+ a2w2+ ...+ a[sub]L[sub]wL= 0.

    Since u1, u2, ..., uK form a basis for U, they span the space: we can write each of the "w" vectors as a linear combination of the "u" vectors:
    w1= b21u1+ b22u2+ ...+ b2LuL
    w2= b21u1+ b22u2+ ...+ b2LuL
    ...
    wK= bK1u1+ bK2u2+ ...+ bKLu[sub][L/sub]

    And we can replace all those "w"s in the original linear combination to get
    a1w1+ a2w2+ ...+ a[sub]L[sub]wL= a1(b21u1+ b22u2+ ...+ b2LuL)+ a2(b21u1+ b22u2+ ...+ b2LuL)+ ... + aL(bK1u1+ bK2u2+ ...+ bKLu[sub][L/sub])
    = (a1b11+ a2b21+ ...+ aLbL1)u1+ (a1b12+ a2b22+ ...+ aLbL2)u2+ ... + (a1b1K+ a2b2K+ ...+ aLbLK)uK)uK= 0.

    Since the u vectors are a basis, they are independent so each of the coefficients must be 0:
    a1b11+ a2b21+ ...+ aLbL1= 0
    a1b12+ a2b22+ ...+ aLbL2= 0
    ...
    a1b1K+ a2b2K+ ...+ aLbLK= 0

    Now, since we have "K" equations in "L" unknowns, and L> K, we could, for example, solve the equations for specific values of K unknowns, in terms of the other L- K unknowns. Any particular, we could choose one or more of the L- K unknowns to be non-zero proving that the vectors are dependent.



    2-) If we have a Vector Space U, and its basis consists of , lets say n vectors, all other bases that I will find for the vector space U will consist of n vectors? I can't find more linearly independent vectors that spreads the whole vector space that is also the base?

    Thanks you
    If there exist a basis for a vector space, consisting of n vectors, then, by definition of basis, they span the space and are independent.

    To show that a set with m> n vectors cannot be independent, write a linear combination equal to 0. Replace each of the m vectors with its expression as a linear combination of the n basis vectors and combine coefficients of the same basis vector. That gives a linear combination of the basis vectors equal to 0 so each coefficient of each basis vector is equal to 0. As above, that gives n equations in more than n unknowns which must have a non-trivial solution.

    To show that a set with m< n vectors cannot span the space, assume that one does, then do effectively the same thing but writing a linear combination of the basis vectors in terms of the m vectors that span the space. That shows that the original set of n vectors cannot be independent, contradicting the definition of a basis.
    Thanks from davidciprut
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    Re: Basis of Vector Space

    Quote Originally Posted by HallsofIvy View Post
    Perhaps you do not understand it! What you wrote is NOT true- unless you simply mistyped it- If L> K then w1,w2,......,wL is linearly dependent, not independent. .
    Ups, my fault, I didn't notice , I meant dependent!
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    Re: Basis of Vector Space

    Quote Originally Posted by davidciprut View Post
    Ups, my fault, I didn't notice , I meant dependent!
    If U is spanned by K independent vectors, then we we say \text{dim}(U)=K.

    Therefore, if L>K then any set of L vectors must be dependent.
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