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Basis for given subspace - check answers?

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the basis for the subspaces of R3 and R4 below.

    2. Relevant equations

    A) All vectors of the form (a,b,c), where a=0
    B) All vectors of the form (a+c, a-b, b+c, -a+b)
    C) All vectors of the form (a,b,c), where a-b+5c=0

    3. The attempt at a solution

    I honestly had no idea what I was doing. I just scratched what I could make of it down. I'm going to do more research online but I'm terrified of the time limit coming up so I thought I'd post my work here meanwhile. I hope you all don't mind!
    Here's what I got (don't laugh please!):

    A) {(0,1,1),(1,1,0)}
    B) {(0,-1,0,1),(1,0,-1,0)}
    C) {(5,0,1),(1,0,5)}

    -Carmen :eek:
     
  2. jcsd
  3. Apr 29, 2010 #2

    lanedance

    User Avatar
    Homework Helper

    start with A)
    (a,b,c), where a=0 means

    (0,b,c) for any b or c...

    your first vector is in the set, but your 2nd has a = 1 so is not

    it may help to picture it geometrically in which case it is the yz plane
     
  4. Apr 29, 2010 #3
    A) Basis spans your subspace. So your answer indicates, for example that,

    [tex]
    \left[ \begin{array}{cccc}0 & 1 \\ 1 & 1 \\ 1 & 0 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ 2 \\ 5 \end{array} \right]
    [/tex]

    should have a solution, but under ur basis, this eqn has no solution. so ur basis cannot span this subspace

    Now if we use

    [tex]
    \left[ \begin{array}{cccc}0 & 0 \\ 1 & 0 \\ 0 & 1 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ a \\ b \end{array} \right]
    [/tex]

    this guy has a solution for all a, b. Now you have to show this is independent. Do this by showing


    [tex]
    \left[ \begin{array}{cccc}0 & 0 \\ 1 & 0 \\ 0 & 1 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ 0 \\ 0 \end{array} \right]
    [/tex]

    has only the trivial soln. which it does.
     
    Last edited: Apr 29, 2010
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