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    Linear Algebra : Span and Subspace

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    (The material is from Basis and Dimension. Please explain each step of your solution for # 8, 9. Thank you.)

    Proposition 3.7:

    Let V in Rn be a k-dimensional subspace. Then any k vectors that span V must be linearly independent and any k linearly independent vectors in V must span V.

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    (The material is from Basis and Dimension. Please explain each step of your solution for # 8, 9. Thank you.)

    Proposition 3.7:

    Let V in Rn be a k-dimensional subspace. Then any k vectors that span V must be linearly independent and any k linearly independent vectors in V must span V.

    Question 8.

    Proof. We prove it by contradiction. Suppose that l<k, and . Since is a basis of V, every vector of (in V) can be expressed as a linear combination of . This is impossible, as is a basis of V and so are linear independent. So, .

    Question 9.

    Proof.

    (1) We prove that any k vectors must form a basis of V, so they are independent.

    Proof. Let , and we know that V is k-dimensional. If are not linear independent, then we choose the maximum number of vectors of , say where l<k. Then will form a basis of V, contradicting that the dimension of V is k.

    (2) We prove that any linear independent k vectors in V must span V.

    Proof. Since are linear independent, if , then there is at least one vector, say w, in V such that can not be expressed as a linear combination of . In other words, these k+1 vectors , w are linear independent. This is a contradiction, as the subspace V is k-dimensional (which means that there are no such k+1 linear independent vectors)

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 4, 2022, 4:18 pm ad1c9bdddf>
    https://brainmass.com/math/linear-algebra/linear-algebra-span-subspace-175057

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