Prove this is a Real Vector Space

1. Feb 9, 2009

fk378

1. The problem statement, all variables and given/known data
Let V be the real functions y=f(x) satisfying d^2(y)/(dx^2) + 9y=0.

a. Prove that V is a 2-dimensional real vector space.
b. In V define (y,z) = integral (from 0 to pi) yz dx. Find an orthonormal basis in V.

3. The attempt at a solution
part A:
I integrated and got that f(x)= (-3/2)y^3 + Cy, C is a real number.
It seems like I need to use dot product here. I don't know how, though.

part B:
Completely lost.

2. Feb 9, 2009

Staff: Mentor

First off, f(x) should be a function of x.
Second, you made a mistake in solving your DE: the solutions to y'' + 9y = 0 are not what you show.

What do you know about the characteristic equation of a linear DE?

3. Feb 9, 2009

fk378

Oh, I never took a DE class. All I did was integrate it twice, and I didn't even realize it wasn't a function of x. What should I do?

4. Feb 10, 2009

Staff: Mentor

OK. You don't need to know how to solve the equation in order to verify that the set of all solutions to it is a vector space.
You know that you have to verify a set of axioms involving addition and scalar multiplication, right?

5. Feb 10, 2009

fk378

Yes..
Under addition we must show it is well-defined and
1. Associativity
2. commutativity
3. zero element
4. inverses
5. closed

Under multiplication we need to show well-definedness and
1. L and R distribution
2. associativity
3. id (unit)
4. zero element
5. closed

do I have it covered here?

6. Feb 10, 2009

Staff: Mentor

Yes, those are the axioms that you have to verify.

Let f, g, and h be elements of V. (What does that imply in regard to your DE?. IOW, what exactly does it mean for a function to be an element of V?)
Let a and b be real numbers.

To pick a couple, #4 in the addition group and #4 in the scalar multiplication group:

If g is in V, is there another element h in V so that g + h = 0?
If g and h are in V, is g + h in V?

7. Feb 10, 2009

HallsofIvy

The point was that you wrote f(x)= (-3/2)y^3 + Cy. "y" is not "x"! If you had had d^2y/dx^2= a function of x, then you could integrate twice but you cannot integrate an expression in "y" with respect to x when you don't know y as a function of x.

Presumably you already know that addition of functions is associative and commutative, that f(x)= 0 for all x is the additive identity, etc and don't need to prove that here. All you really need to show is that if f(x) and g(x) are functions satisfying d^2f/dx^2+ 9f= 0 and d^2g/dx^2+ 9g= 0 then any linear combination of them, af(x)+ bg(x) for a and b any numbers, also satisfies that.

8. Feb 10, 2009

fk378

Well how do I use the second derivative in helping me prove that V satisfies all the axioms?

9. Feb 11, 2009

Staff: Mentor

By using the theorems of differentiability. If f and g are differentiable functions, then so is f + g, and (f+g)' = f' + g'. Also, if a is a constant, then af is differentiable, and (af)' = af'. These ideas can be extended to the next higher derivative.