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

1. Jan 31, 2012

### PirateFan308

1. The problem statement, all variables and given/known data
Suppose that V is the direct sum U$\oplus$U' where U, U' are subspaces of V, which is a subspace of Fn. Define P:V→V as follows: if v$\in$V then we know we can write v uniquely as v=u+u' for some u$\in$U, u'$\in$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$\bigcap$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)
$P(v+v') = u$
$P(v)+P(v') = u+u'$
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:
$P(v+v') = P((u_1+u_1')+(u_2+u_2')) = P(u_1'+u_2'+u_1+u_2)$
$P(v)+P(v') = P(u_1+u_1')+P(u_2+u_2')$
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!

2. Jan 31, 2012

### 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 $v_1$ and $v_2$ are in $V$. The goal is to show that

$$P(v_1 + v_2) = P(v_1) + P(v_2)$$.

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

$$u_1, u_2 \in U$$
and
$$u_1', u_2' \in U'$$

such that

$$v_1 = u_1 + u_1'$$
and
$$v_2 = u_2 + u_2'$$

Now start from $P(v_1 + v_2)$ and start plugging things in and simplifying based on the definition of P.

3. Jan 31, 2012

### 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.

4. Jan 31, 2012

### PirateFan308

Ok, so $P(v_1+v_2) = P(u_1+u_1'+u_2+u_2') = P(u_1+u_2+u_1'+u_2')$

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

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

$P(v_1)+P(v_2) = P(u_1+u_1')+P(u_2+u_2') =$

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$\in$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.

5. Jan 31, 2012

### 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?

6. Feb 1, 2012

### PirateFan308

I think I figured it out. Taking $P(v_1+v_2) = P(u_1+u_1'+u_2+u_2') = P(u_1+u_2+u_1'+u_2')$ and letting $u_3=u_1+u_2$ and $u_3'=u_1'+u_2'$, we get

$P(v_1+v_2) = P(u_1+u_2+u_1'+u_2') = P(u_3+u_3') = u_3$

$P(v_1)+P(v_2) = P(u_1+u_1')+P(u_2+u_2') = u_1+u_2=u_3$

I didn't realize that $P(v)=u$ 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!