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For a set of vectors to span R4 question

  1. Jul 9, 2010 #1
    1. The problem statement, all variables and given/known data
    The set of vectors u = {1,-2,2,1}, v = {1,3,1,1}, w = {3,4,4,3} cannot span R4. Complete this set to create a set of vectors that will span R4. Show that your set of vectors spans R4.

    3. The attempt at a solution
    Let [tex]y = {y_1,y_2,y_3,y_4}[/tex]. I write span{u,v,w,y} as the coefficient matrix:


    Using the first row to produce zeros in each row below yields:


    Using the second row to produce zeros in the row below yields:


    So to my understanding, it would seem given the set {u,v,w}, a fourth vector y cannot be chosen so that the set {u,v,w,y} spans R4 since not every row can contain a pivot position in this case. Please correct me if I am wrong.

    - Sam

    PS sorry for the messy work - I don't know how to write matrices in latex.
  2. jcsd
  3. Jul 9, 2010 #2


    Staff: Mentor

    The problem is that u, v, and w are linearly dependent. Have you learned about the Gram-Schmidt process for constructing a basis yet?
  4. Jul 9, 2010 #3
    No I haven't. To my understanding though, the Gram-Schmidt process creates a set (for this case) of vectors {u',v',w'} so that v' is orthogonal to u', w' is orthogonal to v' and I suppose u', v', and w' are linearly independent so I may choose a fourth vector y' to "complete" this set so that this completed set spans R4?
  5. Jul 9, 2010 #4


    Staff: Mentor

    You can use G-S to construct all four vectors of a basis for R4.
  6. Jul 9, 2010 #5


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    Staff Emeritus
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    You're right. You can't add just one more vector and span R4. You need to add two or more because, as Mark noted, the three you have are linearly dependent.
    You can use the bmatrix environment to produce matrices in LaTeX. Click on the one below to see what the code is.

    1 & 1 & 3 & y_1 \\
    -2 & 3 & 4 & y_2 \\
    2 & 1 & 4 & y_3 \\
    1 & 1 & 3 & y_4
  7. Jul 9, 2010 #6


    Staff: Mentor

    Don't include 0. Your u, v, and w are linearly dependent, which can be seen from the first 3 columns of your last matrix (but taken one step further).

    [tex]\begin{bmatrix}1 & 0 & 1 \\0 & 1 & 2 \\0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}[/tex]

    This says c1 + c3 = 0, and that c2 + 2c3= 0, where c1, cw, and c3 are the coefficients of u, v, and w. If you let c3 = 1, then c1 = -1, c2 = -2, and c3 = 1.

    In other words -u -2v + w = 0, so it can be seen that any one of these vectors can be solved for in terms of the other two.

    Without knowing G-S, I suppose the best way is to pick two of the three vectors, say u and v, and then try to find two more vectors out of the set of standard basis vectors for R4 - {<1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, <0, 0, 0, 1>} - so that you have four linearly independent vectors.

    For your spanning set you can throw in w.
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