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Homework Help: Linear Algebra (showing if a vector spans a vector space)

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data

    I was wondering if someone could explain the easiest way to determine if a set S spans V?

    some example questions would be: show that S = {v1, v2, v3, v4} spans R4 where
    v1 = [1 0 +1 0]
    v2 = [0 1 -1 2]
    v3 = [0 2 +2 1]
    v4 = [1 0 0 1]

    2. Relevant equations

    3. The attempt at a solution

    I know that you need to let x = [a, b, c, d] be any vector in R4 and form an eqn like
    a1v1 + a2v2 + a3v3 + a4v4 = x

    but now I'm lost... please help!
  2. jcsd
  3. Nov 23, 2011 #2


    Staff: Mentor

    This is a good start. You know v1, v2, v3, and v4, so you have essentially a system of four equations in four unknowns (a, b, c, and d). Write the system as an augmented matrix [A|x] and row reduce.
  4. Nov 23, 2011 #3
    Then what do I do?
  5. Nov 23, 2011 #4


    Staff: Mentor

    Start with that and tell me what you get.
  6. Nov 23, 2011 #5
    I get
    a1v1 + a2v2 + a3v3 + a4v4 = x

    a1 00 00 a4 = a
    00 a2 2a3 00 = b
    a1 -a2 2a3 00 = c
    00 2a2 00 a4 = d

    1 0 0 1 | a
    0 1 2 0 | b
    1 -1 2 0 | c
    0 2 0 1 | d

    => RRE form
    1 0 0 ....

    OOooohhh I see!! Now when I turn it into RREF, I'll get something like a1 = a+b-c etc.. correct?!

    => RREF
    1 0 0 1 | a
    0 1 2 0 | b
    0 0 1 (-1/4) | d-2b/-4
    0 0 0 0 | c-a+b+(d-2b)

    Does that look like I'm on the right track?
  7. Nov 23, 2011 #6

    The Electrician

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    Gold Member

    The easiest way is to calculate the determinant of the 4x4 matrix formed by v1, v2, v3 and v4. If the determinant isn't zero, then the vectors span R4.
  8. Nov 23, 2011 #7
    Right, but what happens if it's not a square matrix? I'm pretty sure my prof won't make it that easy on a test :(
  9. Nov 24, 2011 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I think you made a mistake somewhere (or I did). I didn't get all 0s on the last row.
  10. Nov 24, 2011 #9
    You didn't, I did it with Maple and I got all pivots except last column.

    Now @OP, what does this mean? Having no non-pivot columns? How does this relate to Span?
  11. Nov 24, 2011 #10


    Staff: Mentor

    And to continue with what vela and flyingpig said, think about what system of equations you're reduced augmented matrix represents. If <a, b, c, d> is any arbitrary vector in R4, is there a specific linear combination of the vi vectors that forms that vector <a, b, c, d>? If so, those vi vectors span R4.
  12. Nov 24, 2011 #11
    Umm I don't remember learning about pivots.

    and I did the thing again and I still got 0 0 0 0 for the last row :O Here are my steps:

    -r1 + r3 -> r3
    r2 + r3 -> r3

    r4-2r2 -> r4
    1/4 r3 -> r3

    r4 + 4r3 -> r4

    this gave me a final matrix of

    1 0 0 1 a
    0 1 2 0 b
    0 0 1 -1/4 [(c-a+b)/4]
    0 0 0 0 d-2b+4c

    Now what do I do? Or even better, could someone please explain to me the general steps of finding if a vector spans the v.s. so I can complete these problems?

    Thank you!

    EDIT: Mark I just saw your response, how would I know if there is a linear combination that forms <a, b, c, d>?
  13. Nov 24, 2011 #12

    The Electrician

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    Gold Member

    It's not clear to me whether you want to demonstrate a knowledge of basic methods of determining whether a set of vectors spans a space, or whether you just want to know how to use functions available in modern mathematical software such as Maple, Mathematica or Matlab to make the determination.

    If it's the latter, proceed as follows:

    If you have a bunch of N element vectors, let's say you have M of them, such that M≥N, let the vectors form the rows of an MxN matrix.

    Then you don't need to augment that matrix to determine if there are at least N linearly independent vectors in the set. Simply row reduce the MxN matrix. The number of 1's on the main diagonal of the result is the number of linearly independent vectors in the set. Only if that number is N does the set span the space R(N).

    The number of linearly independent rows (your vectors) is also equal to the rank of the MxN matrix of vectors, and you can use the "rank" function to determine this.

    You could also calculate the singular values of the MxN matrix; the rank is equal to the number of non-zero singular values (that's what the "rank" function does).

    See the attached image.

    Attached Files:

    • RANK.PNG
      File size:
      14.4 KB
  14. Nov 24, 2011 #13
    Sorry, I don't mean to be rude but I'm just kinda lost. This is my first course in linear algebra and that's why I'm pretty confused.

    What I was trying to do was to find a basis. The theorem states that vectors are said to form basis for V if
    a) if they span V
    b) they are linearly independent

    I know that to prove b) I need to put it in a matrix, reduce and if I get a matrix with trivial solution meaning everything equals to 0 then it's a trivial solution and it's linearly independant.

    But I'm having trouble with checking if they span or not... that's the whole point of this post :)

    The examples in my book don't show step by step solution so that's why I'm lost.

    e.g. Show that S = {v1, v2, v3, v4} where v1 = [1 0 0 1], v2 = [0 1 -1 2], v3 = [0 2 2 1], v4 = [1 0 0 1] is a basis for R4. (R4 is written as R subscript 4 meaning it's referring to the row spaces?...)

    to show that S is linearly independant, I formed the eqn a1v1 + a2v2 + a3v3 + a4v4 = 0

    turned it into a augmented matrix and got its RREF. This means that it's linearly independent.

    NOW, to show that S spans R4, I let x = [a, b, c, d] be a vector in R4

    what should I do next? thanks!
  15. Nov 24, 2011 #14


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    Science Advisor

    I don't mean to be rude but this is your 6th post in this thread and that is the first mention of finding a basis!

    That's not a theorem, that is the definition of "basis". You can add a "c" to this- the number of vectors in the set is the dimension of the space. And then there is a theorem that says any two implies the third.

    Well, no, you don't need to do that- that's one method. What you need to do is think about the definition: a set of vectors spans a space if and only if, any vector, y, in the space can be written as a linear combination of the vectors in the set. If the set is [itex]\{v_1, v_2, \cdot\cdot\cdot, v_n\}[/itex] then there exist scalars, [itex]a_1, a_2, \cdot\cdot\cdot, a_n[/itex] such that [itex]y= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n[/itex].

    Then you have a problem- either you did the problem wrong or you gave the vectors incorrectly here. Those vector are NOT independent because [itex]v_1[/itex] and [itex]v_2[/itex] are the same: [itex](1)v_1+ (-1)v_2= 0[/itex].

    If those vectors were actually the ones you were given, you do nothing now. Since there are 4 vectors and the are NOT indpendent, they cannot span R4. If you gave the fouth vector incorrectly and the ones you were given were independent, you still would not have to do anything more- four independent vectors must span R4.

    In either case, showing linear independence or spanning, you are, in effect, solving a system of equations. You are either trying to solve [itex]a_1v_1+ a_2v_2+ a_3v_3= 0[/itex] (for linear independence) or [itex]a_1v_1+ a_2v_2+ a_3v_3+ a_4v_4= y[/itex] for any y in the space (for spanning). You could do either by setting up the "augmented" matrix with the right side as the "fifth column". Of course, with all "0"s in the fifth column, no row operations will change those: [itex]a_1= a_2= a_3= a_4= 0[/itex] is an obvious solution. Showing that you can reduce the first four columns to the identity matrix is enough to show that is the only solution. Similarly, to show that the set spans the space you add the components of the vector y as the fifth column. You would row-reduce the matrix in exactly the same way, just doing the row operations on the fifth column on the y- components rather than "0"s. But the actual value of the coefficients is not important, only that you can find them. So the values you wind up with in the fifth column is not important, only that you can reduce the first four columns to the identity matrix- exactly what you did to show independence. They really are exactly the same thing.

    When someone said you could look at the determinant of the matrix, earlier, you asked what to do if the matrix is not square. In that case, there is not work to do- such a set of vectors cannot be a basis (though it may still span the space or be independent). The number of columns of the matrix, the number of components in each vector, is the dimension of the vector space. The number of rows is the number of vectors in the set. And those must be equal in order to have a basis. If there are fewer vectors than the dimension- if there are more columns than row- they might be independent but cannot span the space. If there are more vectors than the dimension- if there are more rows than columns, they might span the space but cannot be independent.
  16. Nov 24, 2011 #15
    Thank you all for being patient and helping me, I do understand it now! I really appreciate your help guys! :)

    and as a side note to anyone who comes across this, yes I did make a mistake in the vectors (late night mistakes...) v1 was actually [1 0 1 0]
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