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Linear algebra linearly independent

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following vectors are linearly independent in P3 "not sure what P3 stands for maybe polynomial of third degree?"

    1,x2,x2-2

    let p1(x)=1
    p2(X)=x2
    p3=x2-2

    c1p1(x)+c2p2(x)+c3p3(x)=z

    where z=0x2+0x+0

    I then create a matrix using the above relation where I get

    (c2+c3)x2+c1-2c3

    The matrix i'm thinking if solving looks like this [0 1 1;0 0 0; 1 0 -2] I know the zero must be at the bottom so I switch things up and get [1 0 -2; 0 1 1;0 0 0]

    When I actually solve it I get c2=-c3
    which is a nontrivial solution, so I state that it is linearly dependent.

    Is my work correct, and or is there a faster way to come up with the conclusion that it is linearly dependent?

    Thanks




    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 11, 2013 #2

    Mark44

    Staff: Mentor

    It's probably "polynomials of degree less than 3."

    The notation isn't consistent from textbook to textbook, in my experience.
    This one is simple enough that you can tell that the set of functions is linearly dependent merely by observation. p3(x) is pretty obviously a linear combination of the other two functions.
     
  4. Feb 11, 2013 #3

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Most likely the set of all polynomials of degree 3 or less. The polynomials of degree 3 don't form a vector space. Nonetheless, shouldn't you find out for sure before attempting the problem?
     
  5. Feb 11, 2013 #4
    How so mark?

    Would it simply be because the first two could be represented as x2-2(1) which would give you px3?
     
  6. Feb 11, 2013 #5
    True.
     
  7. Feb 11, 2013 #6

    Mark44

    Staff: Mentor

    I understand what you're trying to say, but you aren't saying it very well. It is because p3(x) = (1) * p2(x) + (-2) * p1(x). This shows that p3 is a linear combination of p1 and p2.
     
  8. Feb 11, 2013 #7
    Oh! So just like in differential equations then? the same logic is being used in this problem?
     
  9. Feb 11, 2013 #8

    Mark44

    Staff: Mentor

    Yes. And the term "linear combination" means the same thing in both areas.
     
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