Proving that P:V→V is linear

PirateFan308

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... Thanks!

jbunniii

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

such that

[tex]v_1 = u_1 + u_1'[/tex]
and
[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.

xaos

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.

PirateFan308

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.

PirateFan308

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?

PirateFan308

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!