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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.

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- Thread starter EvLer
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if the kernel of linear transformation is {

thanks in advance.

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AKG

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mathwonk

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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.

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mathwonk said:in other words, strange as it may seem, 0 is a linear combination of the vectors in the empty set.

empty set has "real"' vectors?

thanks for the explanations!

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mathwonk

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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.

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mathwonk

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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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