Show functions of this form are a vector space etc

[BustedBreaks]

Show that the functions $$(c_{1}+c_{2}sin^{2}x+c_{3}cos^2{x})$$ form a vector space. Find a basis of it. What is its dimension?

My answer is that it's a vector space because:

$$(c_{1}+c_{2}sin^{2}x+c_{3}cos^2{x})+(c'_{1}+c'_{2}sin^{2}x+c'_{3}cos^2{x}) =(c_{1}+c'_{1}+(c_{2}+c'_{2})sin^{2}x+(c_{3}+c'_{3})cos^2{x})$$ which is a function in the same form as the original function.

Basically all combinations of sums of multiples create functions of the same form as the original function. It's dimension is two because it's of one variable, x.

However, I not sure about the basis. I want to say it's just the original function, but I don't know why. I'm a little rusty when it comes to basis stuff.

 

[Mark44]

What you seem to be showing is that the set of functions {1, sin²x, cos²x} is a subspace of some function space. If you really need to show that this set of functions is a vector space (function space), you need to verify all 10 axioms.

I don't understand your reason for saying that the dimension of this subspace/function space is 2 -

It's dimension is two because it's of one variable, x.

How does it follow that the dimension is 2?

 

[BustedBreaks]

Well to be honest I have forgotten a lot of this stuff.

I'm looking at the function $$(c_{1}+c_{2}sin^{2}x+c_{3}cos^2{x})$$ with c1, c2, and c3, as distinct constants and that all functions of this form can be combined to create another function this form. This seems to be the wrong way to think about it?

As for the dimension. I just figured that its a function of just x so any choices of c1, c2, c3 constant, will give a graph in two dimensions

 

[Mark44]

Yeah, I think that's the wrong approach. You're not dealing with one function c₁ + c₂sin²x + c₃cos²x, you're dealing with three separate functions {1, sin²x, cos²x}.

Regarding the dimension, by your reasoning {sin²x} would also have dimension 2, which is not true.

 

[BustedBreaks]

Yeah, I think that's the wrong approach. You're not dealing with one function c₁ + c₂sin²x + c₃cos²x, you're dealing with three separate functions {1, sin²x, cos²x}.

Regarding the dimension, by your reasoning {sin²x} would also have dimension 2, which is not true.

I see what you mean by the difference in functions, however I feel like they would have written it the way you did, {1, sin2x, cos2x}, if that's what they meant? They way I see it, the function in the question represents all functions of that form which is why they have function plural.However, I'm not sure what you mean by sin^2(x) isn't in two dimensions? If you plot this it has a y direction and an x direction. This is probably the wrong way to think about as well.

 

[Mark44]

You should be thinking about these things as vectors, not as functions. The vectors <1,0,1>, <0,1,0>, and <1,1,1> form a subspace of R³. Every vector in this subspace can be represented as c₁<1,0,1> + c₂<0,1,0> + c₃<1,1,1> for some constants c₁, c₂, and c₃.
The fact that each of these vectors is a vector the 3-space has very little to do with anything. In the same way, the fact that the graph of y = sin²x is a graph in the plane also has very little to do with anything as far as this problem is concerned.

It could be proved that the vectors in my example here are a vector space, by verifying that all 10 axioms are satisfied. One could also find the dimension of this vector space, and find a basis for it. The problem I came up with is very similar to yours.