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I Same vector space for arbitrary independent vectors?

  1. May 28, 2016 #1

    kelvin490

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    If we use n linearly independent vectors x1,x2...xn to form a vector space V and use another set of n linearly independent vectors y1,y2...yn to form a vector space S, is it necessary that V and S are the same? Why?

    If we have a vector space Q that the dimension is n, can we say that any set of n linearly independent vectors k1,k2...kn can form a basis of Q? Why?

    Suppose only real numbers are involved.
     
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  3. May 28, 2016 #2

    micromass

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    I see no reason to conclude ##V## and ##S## are the same.

    Yes. This theorem should be proven in any standard linear algebra book.
     
  4. May 28, 2016 #3

    jbunniii

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    Consider the simplest case, where ##n=1##. Then ##x_1## and ##y_1## each span a one-dimensional subspace. If ##x_1## and ##y_1## are linearly independent, then these subspaces are distinct; indeed, they intersect trivially.
     
  5. May 28, 2016 #4

    kelvin490

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    Thanks. It's seems there is a theorem that a vector space Q that the dimension is n can have any set of n linearly independent vectors k1,k2...kn as its basis. How can we ensure that k1,k2...kn doesn't form a vector space other than Q but not Q?

    I have this question because two different set of n linearly independent vectors can form two different vector space with dimension n. It seems hard to ensure k1,k2...kn must form Q.
     
  6. May 28, 2016 #5

    jbunniii

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    Let ##R## denote the vector space spanned by ##k_1,k_2,\ldots,k_n##. Since each ##k_j## is in ##Q## and ##Q## is a vector space, every linear combination of the ##k_j##'s is in ##Q##, hence ##R \subseteq Q##.

    Now suppose that ##q## is an arbitrary element of ##Q##. Since ##k_1,k_2,\ldots,k_n## is a basis for ##Q##, we can write ##q = a_1 k_1 + a_2 k_2 + \cdots + a_n k_n## for some scalars ##a_1,a_2,\ldots a_n##. Therefore ##q## is contained in the subspace spanned by ##k_1,k_2,\ldots,k_n##, which is ##R##. This shows that ##Q \subseteq R##.

    Since we have shown both containments ##R \subseteq Q## and ##Q \subseteq R##, we conclude that ##Q = R##.
     
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