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Homework Help: Decide if specified elements are linearly independent, span V, and form a basis

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    "In each of the given cases, decide whether the specified elements of the given vector space V (i) are linearly independent, (ii) span V, and (iii) form a basis. Show all reasoning.

    V is the space of all infinite sequences (a0, a1, a2, ...) of real numbers v1 = (1,0,1,0,1,0,...), v2 = (0,1,0,1,0,1,...).

    2. Relevant equations
    I know the basic properties of how to row reduce to find linear independence and span, and the fact that an element that has both of these forms a basis.

    3. The attempt at a solution
    The problem is that I don't know how to deal with these infinite sequences. Should I just write them in a matrix and do as I usually do? If I do this, I don't know how to handle the "..." part of either vector. So basically I have no idea how to start the question :confused: anyone shed some light?
  2. jcsd
  3. Oct 1, 2011 #2


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    You don't need any big ol' matrix. Just answer the questions:

    1. Are v1 and v2 linearly independent? Use the definition to figure it out.

    2. Do they span the vector space. Try an example. Try something. Play around with it.

    3. If you know the answers to 1 and 2, you will know the answer to 3.
  4. Oct 2, 2011 #3
    1. That is the problem, I don't know how I can use the definition to just find the answer. Unless it's extremely obvious and I've been over thinking the question: they are linearly independent because the rows will both have a pivot, meaning a unique solution?

    2. I still really have no way to go about this. Not sure what I'm supposed to "play around" with. I think I'm probably overlooking this part as well...
  5. Oct 2, 2011 #4
    I think you may be getting confused about what the "definition" of linear independence is. Without going too technical, a set [itex] \{ v_1, ... , v_n \} [/itex] is linearly independent if the whenever [itex] a_1 v_1 + \cdots + a_n v_n = 0 [/itex] then all [itex] a_i = 0 [/itex]. Thus in order to see if the elements are linearly independent, write
    [tex] a_1 v_1 + a_2 v_2 = 0 [/tex]
    Is there any way this is possible without [itex] a_1 = a_2 = 0?[/itex]

    If v1 and v2 span the space, then every element of the space can be written as some linear combination. To play around, try choosing actual numbers, say 3 and 4, and look at what [itex] 3 v_1 + 4 v_2 [/itex] looks like. Choose another set of numbers, and see what that looks like. Do you see a pattern? Must all elements follow that pattern?
  6. Oct 2, 2011 #5
    I do understand the definition, I must have worded it weird. By unique solution, I meant that all the coefficients [itex] a_1 , a_2 , ... , a_n = 0 [/itex], aka, the trivial solution. So, it is not possible that they are linearly independent without all the coefficients equal to 0. Is this correct?

    As for playing around using example numbers like 3 and 4, I'm not sure I can do this. Our prof usually wants us to prove all this stuff using general cases only, i.e. not picking any specific numbers for any part of it. Is there any other way of finding if they span?
  7. Oct 2, 2011 #6
    Okay, so choose fixed "numbers" [itex] a_1,a_2 [/itex]. What is [itex] a_1 v_1 + a_2 v_2 [/itex]? Write it out!
  8. Oct 2, 2011 #7
    I end up with [itex] (a_1 , a_2 , a_1 , a_2 , a_1 , a_2 , ...) [/itex]. So, that answer is a linear combination of [itex] v_1 [/itex] and [itex] v_2 [/itex] because any multiple of [itex] v_1 [/itex] and [itex] v_2 [/itex] will give me the above result, therefore they span V? Is this the right answer, or at least close?...
  9. Oct 2, 2011 #8
    So there is a pattern right? In particular, any linear combination has alternating a1 and a2 components. Do ALL POSSIBLE sequences follow this pattern?
  10. Oct 2, 2011 #9
    Ahh.. all sequences EXCEPT [itex] a_1 = a_2 [/itex]. Then they are not alternating. So the vectors are not a linear combination for all real numbers, therefore they do not span V.
    Last edited: Oct 2, 2011
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