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Homework Help: Proving that P:V→V is linear

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose that V is the direct sum U[itex]\oplus[/itex]U' where U, U' are subspaces of V, which is a subspace of Fn. Define P:V→V as follows: if v[itex]\in[/itex]V then we know we can write v uniquely as v=u+u' for some u[itex]\in[/itex]U, u'[itex]\in[/itex]U'. Define P(v)=u. Show that:
    a) P is linear
    b) P2=P (a linear function with this property is called a projection).
    Let P'=I-P where I is the identity function
    c) PP'=0=P'P
    d) U=KerP', U'=KerP

    3. The attempt at a solution
    Since V is a direct sum of U and U', then U[itex]\bigcap[/itex]U'={0}
    To prove that P is linear, I need to prove that P(v+v')=P(v)+P(v') and P(cv)=cP(v)
    [itex]P(v+v') = u[/itex]
    [itex]P(v)+P(v') = u+u'[/itex]
    Which obviously doesn't work. I'm using the assumption that v+v' is still in V, which is clearly an incorrect assumption. I also tried this:
    [itex]P(v+v') = P((u_1+u_1')+(u_2+u_2')) = P(u_1'+u_2'+u_1+u_2)[/itex]
    [itex]P(v)+P(v') = P(u_1+u_1')+P(u_2+u_2')[/itex]
    For closed under multiplication, I didn't even know where to start.

    Sorry for not showing very much work, but I'm so stuck that there is no work to show...:confused: Thanks!
  2. jcsd
  3. Jan 31, 2012 #2


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    Your notation is awkward because you're using the ' to indicate two different ideas. Let's reserve ' to denote the part of the vector that lies in U'.

    Suppose [itex]v_1[/itex] and [itex]v_2[/itex] are in [itex]V[/itex]. The goal is to show that

    [tex]P(v_1 + v_2) = P(v_1) + P(v_2)[/tex].

    As V is the direct sum of U and U', there exist unique

    [tex]u_1, u_2 \in U[/tex]
    [tex]u_1', u_2' \in U'[/tex]

    such that

    [tex]v_1 = u_1 + u_1'[/tex]
    [tex]v_2 = u_2 + u_2'[/tex]

    Now start from [itex]P(v_1 + v_2)[/itex] and start plugging things in and simplifying based on the definition of P.
  4. Jan 31, 2012 #3
    hint: for x in U, p(x)=x, and for y in U', p(y)=0. now what happens to P(z+w) when z+w is in V? also you may need to know that U and U' only intersect at zero.
  5. Jan 31, 2012 #4
    Ok, so [itex]P(v_1+v_2) = P(u_1+u_1'+u_2+u_2') = P(u_1+u_2+u_1'+u_2')[/itex]

    Let [itex]u_3=u_1+u_2[/itex] which will still be in U. Let [itex]u_3'=u_1'+u_2'[/itex] which will be in U' because both U and U' are subspaces of V which are closed under addition.

    [itex]P(u_1+u_2+u_1'+u_2') = P(u_3+u_3') = P(v_3) = u = P(v)[/itex] where u is in U (because P(v) is defined to equal u)

    [itex]P(v_1)+P(v_2) = P(u_1+u_1')+P(u_2+u_2') = [/itex]

    Once again, I'm stuck. I know that I should use the fact that V is a direct sum of U and U', which means the intersection of U and U' is only at zero, v[itex]\in[/itex]V is uniquely the sum of u+u', if u+u'=0 then u=u'=0 but I don't see where to use this fact.
  6. Jan 31, 2012 #5
    How did you figure out that p(x)=x and p(y)=0 (if x is in U and y is in U')?

    Because U and U' only intersect at zero, the only time x=y is when x=y=0. So that taking x=y=0, we have P(x)=x=0 and P(y)=0 (which happens to equal y in this case). So if z+w is in V then P(z+w)=x=0=P(y). Or am I taking too far of a leap when I say x=y=0?
  7. Feb 1, 2012 #6
    I think I figured it out. Taking [itex]P(v_1+v_2) = P(u_1+u_1'+u_2+u_2') = P(u_1+u_2+u_1'+u_2')[/itex] and letting [itex]u_3=u_1+u_2[/itex] and [itex]u_3'=u_1'+u_2'[/itex], we get

    [itex]P(v_1+v_2) = P(u_1+u_2+u_1'+u_2') = P(u_3+u_3') = u_3[/itex]

    [itex]P(v_1)+P(v_2) = P(u_1+u_1')+P(u_2+u_2') = u_1+u_2=u_3[/itex]

    I didn't realize that [itex]P(v)=u[/itex] was referring to a specific v and u - I had thought it was an arbitrary v, u, and u' that they were referring to. Whoops!
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