# Checking for linearity 2

1. Jul 19, 2012

### g.lemaitre

1. The problem statement, all variables and given/known data

here is theorem 4.5

3. The attempt at a solution

How can theorem 4.5 even relate to the question? The question deals with p's and q's and the theorem deals with u v w.

2. Jul 19, 2012

### Muphrid

The short version of what I said in the other thread is this:

The theorem deals with three arbitrary vectors that are elements of a given vector space. They do not need to be called u, v, and w.

3. Jul 19, 2012

### g.lemaitre

well how do i even find out if
p(x) = a0 + a1x + a2x2
q(x) = b0 + b1x + b2x2 and P2
meets the condition (u+v) *w = u*w + v*w?
What's the first question I ask myself? I don't even know what step 1 is, much less 2 and 3

4. Jul 19, 2012

### Muphrid

$P_2$ denotes the space of second-order polynomials. All you need to test this condition is to come up with a third "vector" (another second-order polynomial) that you can plug in as w.

This condition you're testing is the distributive property. It's the same thing as saying that (5+10)x6 = 5x6 + 10x6. I can plug in any three numbers there instead of 5, 10, and 6 and it works because the real line is a (very simple) vector space of sorts.

What you do to test this is simple. If I want to test the distributive property on real numbers, I go to what I just wrote and say, hm, is (5+10)x6 -> 15x6 really equal to 5x6 + 10x6?

At that point, you just evaluate both sides until you get something that's obviously (un)equal.

5. Jul 19, 2012

### g.lemaitre

ok, i got it.