Linear Algebra - Prove that E is not a basis for V.

In summary, the homework statement is that there is a definition of a basis, vector space, and all the axioms. The Attempt at a Solution states that part A is simple enough, but part B is violating the definition of a basis. The problem may be the infinite dimensions. The solution is to find a linear combination that uses a finite number of vectors.
  • #1
jdc15
40
0

Homework Statement



Since it's kind of hard to type out, I'll try to post a screenshot:

[PLAIN]http://img841.imageshack.us/img841/7357/questionq.jpg

Homework Equations



There's the definition of a basis, vector space, and all the axioms.

The Attempt at a Solution



I understand part A; it's simple enough, but I'm really stuck on part B. Had they been finite sets, as stated in our textbook E would be the standard basis for that finite set, i.e. {(1,0,0), (0,1,0), (0,0,1)} is the standard basis for R3. But somehow for infinite sequences this is not the case. Since I know by part A the set is linearly independent, it must be the second part of the definition of a basis that is violated, meaning it does NOT generate all infinite sequences. My initial guess is that sequences like {0,0,0,...0,0,1} can't be generated because j cannot reach infinity or something like that, but I don't know if it's correct to say that. Any ideas?

Thanks in advance.
 
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  • #2
Perhaps you should start with stating to yourself precisely and carefully what your fuzzy term "generate" means.
 
  • #3
Also {0,0,0,...0,0,1} isn't a sequence in the set, there is no infinity term.
 
  • #4
Alright so maybe {0,...,0,1} doesn't work. By generate I simply mean to span V. As in, when I say {0,0,0,...0,0,1} can't be generated I mean it is not in span(E).
 
  • #6
Well, what it means "span"? Can you provide the exact definition?
 
  • #7
jdc15 said:
Alright so maybe {0,...,0,1} doesn't work. By generate I simply mean to span V. As in, when I say {0,0,0,...0,0,1} can't be generated I mean it is not in span(E).
That, {0, 0, 0,..., 0, 0, 1} is not in your vector space because it has some unknown but finite number of components. {0, 0, 0, ..., 0, 0, 1, 0, ...} is[\b] in your vector space and is in the span of the given vectors.

Look at {1, 1, 1, ..., 1, 1, 1, ...}, the vector all of whose components are 1. Is that in the span?

(You have been asked to state the precise definition of "span". That is not just being picky. In mathematics you use the precise words of definitions in proofs. I can think of one word in the definition of "span" that is crucially important here!)
 
  • #8
Hint: this one important word should start with letter "f".
 
  • #9
I have been thinking about this question and I stumbled upon a similar one: the telescoping series. This has a basis of {1-x, x-x^2, x^2-x^3...}. So 1 = (1-x) + (x-x^2) ... . But 1 is not in the span, I'm guessing for the same reason as in this problem. So, answering your question, no, {1,1,... 1, 1} is not in the span of E. The definition of span (in my textbook, shortened,) is "the set consisting of all linear combinations of the vectors in S." Hmm.
 
  • #10
Ok I looked at the book's definition of linear combination and figured it out. In my book linear combinations need a finite number of vectors to make the sum. Therefore, since {1,1...1,1} requires is an infinite sequence and requires an infinite amount of vectors in E, E does not generate all infinite sequences. Thanks!
 

Related to Linear Algebra - Prove that E is not a basis for V.

1. What is a basis in linear algebra?

A basis in linear algebra is a set of linearly independent vectors that span a vector space. This means that any vector in the vector space can be written as a linear combination of the basis vectors.

2. How do you prove that a set of vectors is not a basis?

To prove that a set of vectors is not a basis, you need to show that the vectors are either linearly dependent or do not span the entire vector space. This can be done by finding a non-trivial linear combination of the vectors that equals the zero vector, or by showing that there are vectors in the vector space that cannot be written as a linear combination of the basis vectors.

3. What does it mean for a set of vectors to be linearly dependent?

A set of vectors is linearly dependent if at least one of the vectors in the set can be written as a linear combination of the other vectors in the set. In other words, there is a non-trivial solution to the equation c1v1 + c2v2 + ... + cnvn = 0, where c1, c2, ..., cn are scalars and v1, v2, ..., vn are the vectors in the set.

4. Can a set of vectors be both linearly dependent and span a vector space?

Yes, it is possible for a set of vectors to be linearly dependent and still span a vector space. This means that although there is a non-trivial linear combination of the vectors that equals the zero vector, the set of vectors can still be used to represent any vector in the vector space.

5. How does proving that E is not a basis for V affect the vector space V?

If E is not a basis for V, then the vector space V cannot be fully represented by the set of vectors in E. This means that there are vectors in V that cannot be written as a linear combination of the vectors in E. It also means that the vectors in E are not sufficient to describe all possible directions in V.

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