Can the Empty Set Span the Zero Subspace? Insights on Spanning Sets

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In summary, the empty set spans the zero subspace because the intersection of all subspaces containing the empty set equals {0}. This is true because the sum of the vectors in the empty set is defined to be zero, following the convention that the empty product is 1. This convention is also used in the construction of primes in Euclid's proof of infinitude of primes.
  • #1
jwqwerty
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I have some questions about spanning sets

1. Why does empty set spans the zero subspace?

2. Why is this true: Since any vector u in A is dependent on A, A⊆<A>? (<A> is the set of all vecotrs in R^n that are dependent on A)
 
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  • #2
2. u = 1.u

1. the intersection of all subspaces containing the empty set equals {0}.

second answer: if you want the summation symbol to be additive over disjoint decomposition of the index set, i.e. if you want the sum of the vectors in SuT to equal the sum of the vectors in S plus the sum of the vectors in T, when S and T are disjoint, you have to agree that the sum of the vectors in the empty set is zero.

third answer: because i said so (i.e. it is defined that way).
 
  • #3
For non-empty sets S, "the smallest vector space that contains S", "the intersection of all vector spaces that contain S" and "the set of all linear combinations of members of S" are all the same. But if S=∅, the last one is ∅ and the first two are {0} (if your definition of vector space requires the set to be non-empty).

So it looks like your book uses one of the first two as the definition of span S, and requires vector spaces to be non-empty.
 
  • #4
I am arguing that the empty linear combination is zero. see above mumbo jumbo about empty index sets. for the same reason the empty product is 1. this is a pretty standard convention.
 
  • #5
Ah, I always forget to consider empty index sets. :smile: Yes, if we define ##\sum_{k=1}^0 a_k s_k=0##, then span S = ∅, even if the left-hand side is defined as the set of all linear combinations of members of S.
 
  • #6
this bothered me for a long time in connection with euclid's proof of infinitude of primes. i.e. to construct as many primes as desired start from any set of primes, multiply them together and add 1. then we claim that number has anew prime factor. but for this to work, it needs to equal 2 or more.

i thought this was a gap, and that one should exhibit at least one prime to begin the induction.

fortunately however, by this convention, even if we begin with the empty set of primes, their product is 1, and so the number constructed is 2.
 

1. What is a spanning set?

A spanning set is a set of vectors that can be used to create any vector within a given vector space. It is a fundamental concept in linear algebra and is used to determine the dimension of a vector space.

2. How do you determine if a set of vectors is a spanning set?

A set of vectors is a spanning set if every vector in the vector space can be written as a linear combination of the vectors in the set. This means that the vectors in the set must be linearly independent and must span the entire vector space.

3. Can a spanning set contain more vectors than the dimension of the vector space?

Yes, a spanning set can contain more vectors than the dimension of the vector space. However, if the vectors in the set are linearly dependent, then some of the vectors can be removed without affecting the span of the set.

4. How do you find the smallest spanning set for a vector space?

The smallest spanning set for a vector space can be found by taking the basis of the vector space. A basis is a set of linearly independent vectors that span the entire vector space. By definition, it is the smallest spanning set for a vector space.

5. Can a set of vectors that span a vector space still be linearly dependent?

Yes, a set of vectors that span a vector space can still be linearly dependent. This is because the spanning set only needs to span the vector space, it does not need to be linearly independent. However, if a spanning set is linearly dependent, then it can be reduced to a linearly independent set that still spans the vector space.

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