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Linear Algebra: Polynomials subspaces

  1. Mar 20, 2010 #1
    U and W are subspaces of V = P3(R)
    Given the subspace U{a(t+1)^2 + b | a,b in R} and W={a+bt+(a+b)t^2+(a-b)t^3 |a,b in R}

    1) show that V = U direct sum with W
    2) Find a basis for U perp for some inner product

    Attempt at the solution:

    1) For the direct sum I need to show that it satisfy two conditions (a) V = U + W (which I did), (b) the intersection of U and W = {0} which I am not quite sure how to do.

    2) expanding U: a(t^2 + 2t +1) +b. as far as I understand U perp needs only to contain t^3 so that U + U perp span P3 (right?).

    Next I used the standard inner product for polynomials such that the integral from 0 to 1 of U.U perp = 0.

    so I have the following (at^2 + 2at +a +b)(-bt^3) integrate from 0 to 1, set it equal to zero and solve for b in terms of a. (here I chose U perp to be -bt^3 from some b in R).

    Is solution valid for finding a basis for U perp? Thanks
  2. jcsd
  3. Mar 21, 2010 #2


    Staff: Mentor

    For the b part of 1, you might try this.
    U = {a(t + 1)2 + b | a, b in R}
    W = {(c - d)t3 + (c + d)t2 + dt + c | c, d in R}
    (Notice that I changed to c and d in the 2nd subspace.)

    You can split these up like so:
    U = { a(t2 + 2t + 1) + b(1)}
    W = {c(t3 + t2 + 1) + d(-t3 + t2 + t)}

    If you think about the generators of each subspace as vectors (which are probably a bit easier to work with), U is all linear combinations of <0, 1, 2, 1> and <0, 0, 0, 1>, and W is all linear combinations of <1, 1, 0, 1> and <-1, 1, 1, 0>.

    If you can show that these vectors are linearly independent, that means that none of them is a linear combination of the others, and that means that the intersection of the two subspaces is {0}.
  4. Mar 23, 2010 #3
    Thanks Mark44 for your help
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