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Kernel basis

  1. Jul 18, 2005 #1
    This is probably a simple question, but just to be sure:
    if the kernel of linear transformation is {0}, then the set is linearly dependent since 0-vector is LD, right? So dimension is 0, right? Then what's the basis of kernel? No basis?
    thanks in advance.
     
  2. jcsd
  3. Jul 18, 2005 #2

    AKG

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    What you said seems mostly right but instead of no basis, you would say that the basis is {}. However, it seems weird to me to say that the basis is an empty set, since the span of an empty set would have to be empty, but {0} is not empty.
     
  4. Jul 18, 2005 #3

    mathwonk

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    it is well established that the span of the empty set is the space {0}. e.g. to see it, you may take as a definition of span: "the smallest subspace containing the set".

    in other words, strange as it may seem, 0 is a linear combination of the vectors in the empty set. this is needed to make the summation symbol additive over all disjoint decompositions of the index set.

    i.e. the summation over an empty index set equals the zero vector.
     
  5. Jul 18, 2005 #4
    :eek:
    empty set has "real"' vectors?

    thanks for the explanations!
     
  6. Jul 18, 2005 #5

    mathwonk

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    you are welcome. this annoyed me greatly this past semester in trying to write up my linear algebra notes, as it causes havoc unless you come right out and explain it coprrectly. but most students do not notice so you often do not bother, out of fear it will cause even more havoc if explained!
     
  7. Jul 18, 2005 #6
    Well, I'd like to say that you explained definitely more than my instructor who just said: "we don't worry about finding a basis for this subspace", which basically meant "don't bother me, it's not on the test". So I posted it here, since it looked like a contradiction and I was just curious. :bugeye:
     
    Last edited: Jul 18, 2005
  8. Jul 19, 2005 #7

    mathwonk

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    the empty set does not have any vectors in it, but think about the meaning of the summation symbol.

    it acts on indexed families of vectors. and index set is any set at all, and an indexed family is a map from that set into the vector space, i.e. the indexed set {v1,v2,v3} is a map from the set {1,2,3} into V.

    and the summation symbol is a map from all indexed families to vectors, i.e. it maps the indexed set {v1,v2,v3} to v1+v2+v3.


    now the empty set is an index set, and a family indexed by it is a map from the empty set into V, and its graph is the empty set of ordered pairs.

    so the summation looks at the empty set and has to assign to it a vector in V. how to do it? i.e. the summation symbol is defined on all maps from sets into V, and the emptys et is a map from the emptys et minto V, so the summation symbol should be defiend on the empty map.


    now we ask ourselves one mroe question, how does the summation symbol behave under disjoint decompositions? i.e. what is the relation between the value of the summation symbol on {v1,v2,v3} and its value on {v1} and {v2,v3}? well it is additive. i.e. if an indexed family occurs as the disjoint union of two other indexed families than the value of the sumamtion symbol on the union should be the sum of the values on the two disjoing subsets.

    now we can write the index set {1,2,3} as the disjoint union of {1,2,3} and the empotys et, so the value of the sumamtion symbol on the union {1,2,3} should be its value on {1,2,3} and its value on the emptys et. hence the value on the empty set must be zero.
     
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