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Lin Alg - Rank of a Matrix

  1. Jul 20, 2010 #1
    Let S = {v1, v2, v3, v4, v5}

    v1 = <1,1,2,1>
    v2 = <1,0,-3,1>
    v3 = <0,1,1,2>
    v4 = <0,0,1,1>
    v5 = <1,0,0,1>

    Find a basis for the subspace V = span S of R^4.

    ----

    My attempt:

    I place the five vectors into a matrix, where each vector is a row of the matrix.
    I solve for row-echelon (not RREF). I get:

    Code (Text):

    1 1 2 1
    0 1 5 0
    0 0 1 0
    0 0 0 1
    0 0 0 0
     
    Therefore the basis of V = span S of R4 contains the vectors:

    <1,1,2,1>, <0,1,5,0>, <0,0,1,0>, <0,0,0,1>

    ---

    Question, besides the question "Is this correct?", I'd like to know the difference between solving for REF and RREF. I have two textbooks that I'm using to get through LinAlg and they differ on this section. One tells me to solve for RREF and one REF. Using the former method I get the identity matrix for this problem, and using the latter I get the basis above.
     
    Last edited: Jul 20, 2010
  2. jcsd
  3. Jul 20, 2010 #2

    hunt_mat

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    Not too sure what this is asking. span(S)=R^4 (ignore v1 and you can find yourself an orthonormal basis). Your working is correct, why did you leave out the fifth vector?

    Mat
     
  4. Jul 20, 2010 #3
    Good catch, left out a row of zeros.

    Are you not sure what I am asking, or the problem?
     
  5. Jul 20, 2010 #4

    hunt_mat

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    Both, S clearly spans R^4, so one of the vectors must be a linear combination of the others.
    Can you define what is meant by REF and RREF(row reduced echlon form?)

    Mat
     
  6. Jul 20, 2010 #5
    REF = row echelon form
    RREF = reduced row echelon form

    I am currently reading the section on the rank of a matrix and finding basis using row space/column space.

    In one book it has me setup a matrix and then solve for REF.
    The other book goes all the way and has me to RREF to the same matrix.

    Is there a preferred way to do these problems, I mean, obviously doing REF is less work than RREF...
     
  7. Jul 20, 2010 #6

    hunt_mat

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    I've done RREF but I haven't come across REF before (poor education on my part I suspect).
    Everyone has their own way of doing problem, there is no preferred way per se, just the persons preferred way.
     
  8. Jul 20, 2010 #7

    HallsofIvy

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    Here's a different, more fundamental way of doing this:

    If the five given vectors do not already form a basis for for their span, they must be dependent- there exist numbers a, b, c, d, e, not all 0, such that a<1, 1, 2, 1>+ b<1, 0, -3, 1>+ c<0, 1, 1, 2>+ d<0, 0, 1, 1>+ e<1, 0, 0, 1>= <0, 0, 0, 0>.

    That gives us the four equations a+ b+ e= 0, a+ c= 0, 2a- 3b+ c+ d= 0, and a+ b+ 2c+ d+ e= 0. Of course, equation 2 gives c= -a so the third equation is the same as a- 3b+ d= 0 and the last is -a+ b+ d+ e= 0. Since a+ b+ e= 0, b+ e= -a and the last equation becomes -2a+ d= 0 or d= 2a. Putting both c= -a and d= 2a into the 2a- 3b+ c+ d= 0 gives 3a- 3b= 0 or b= a. Finally, then a+ b+ e= 0 becomes 2a+ e= 0 so e= -2a. Our original equation, a<1, 1, 2, 1>+ b<1, 0, -3, 1>+ c<0, 1, 1, 2>+ d<0, 0, 1, 1>+ e<1, 0, 0, 1>= <0, 0, 0, 0>, becomes a<1, 1, 2, 1>+ a<1, 0, -3, 1>- a<0, 1, 1, 2>+ 2a<0, 0, 1, 1>- 2a<1, 0, 0, 1>= <0, 0, 0, 0>. We can divide through by a and then solve for any one of those vectors in terms of the other four- the five given vectors span all of [itex]R^4[/itex] and any four of the original five vectors will be a basis.
     
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