Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Independence test

  1. Jan 29, 2005 #1
    Hello,
    I have a hw problem I need some help with.
    Given 3 matrices:
    A = [1,2,1,2]^t
    B = [2,3,-1,0]^t
    C = [1,0,1,0]^t
    I need to use test of independence to find out whether they are independent.
    So, the matrix I ended up with is this
    1 2 1 0
    0 1 2 0
    0 0 0 0
    0 0 0 0
    The answer says it's independent and I can see that from the original matrices. But I do not know how to proceed with the indep. test to prove that; for one thing, I have more equations than unknowns (I know they are all zeros, so I just disregarded the last row) and for another, if I make it a row-echelon form it looks like column 3 (not being a pivot matrix) is actually a lin. comb. of col.1 and 2 which means that the set is lin.dep.
    Where did I go wrong? :yuck:
    Thanks.
     
    Last edited: Jan 29, 2005
  2. jcsd
  3. Jan 30, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    How did you end with that matrix?

    Using the given vectors as rows of a matrix, you start with
    1 2 1 2
    2 3 -1 0
    1 0 1 0

    Now "row reduce":
    subtract twice the first row from the second row and subtract the first row from the third row to get
    1 2 1 2
    0 -1 -3 -4
    0 -2 0 -2

    Now subtract twice the second row from the third row:
    1 2 1 2
    0 -1 -3 -4
    0 0 -6 6

    Look independent to me!
     
  4. Jan 30, 2005 #3
    well, since they are transposes (I put '^t' there), following the dependence equation I got initially this:
    xA+yB+zD = 0 (vector, i.e. zero vector 4 by 1)
    which is
    1x + 2y + z = 0
    2x + 3y + 0 = 0
    1x - y + z = 0
    2x + 0 + 0 = 0
    and then I got the augmented matrix which reduced to what I posted above and I cannot 'make' it independent.
    Thank you for reply.
     
  5. Jan 31, 2005 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Take a look at your last equation: 2x= 0. Doesn't that say that x= 0? Now put that back into the second equation: 3y= 0. y must be 0! finally, either of the remaining equations and you get z= 0. Looks independent to me!
     
  6. Jan 31, 2005 #5
    That obvious...
    Thank you very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Independence test
  1. Linear Independence (Replies: 8)

  2. Algebraic independence (Replies: 2)

Loading...