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brinethery

  • 4 years ago

Let v1, v2, ..., Vn be a list of nonzero vectors in a vector space V such that no vector in the list is a linear combination of its predecessors. Show that the vectors in the list form an independent set.

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  1. rulnick
    • 4 years ago
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    It's almost self-evident from the definition of independence, but here goes. A = no vector is a linear comb of its predecessors B = vectors are indep We will show A => B by showing not B => not A. Assume the vectors are not indep. Then there exists at least one vector that can be expressed as a linear combination of others. Call it v*. If all of the others in the linear combination are predecessors of v* then we are done, so assume not: assume at least one vector in the linear combination follows v*. Then this vector is a linear combination of the others together with v*, which contradicts A. Hence not B => not A, so A => B.

  2. brinethery
    • 4 years ago
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    I just read this twice, and I'll have to really read these definitions and proofs carefully. Thank you for your detailed answer, I appreciate it.

  3. rulnick
    • 4 years ago
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    No problem. It may help to think of this in the case of just three vectors. For example ...

  4. rulnick
    • 4 years ago
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    If you have v1, v2, and v3, and they are dependent, then a v1 + b v2 + c v3 = 0. Which means -a v1 + -c v3 = b v2 ( v2 is a lin comb, but not of its predecessors) but then -a v1 + -b v2 = c v3 (so v3 *is* a lin comb of its predecessors).

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