vector space without verification of the axioms

I'm trying to show that the functions $c_1 + c_2 \sin^2 x + c_3 \cos^2 x$ forms a vector space.

And I will need to find a basis of it, and its dimension.

Is there a way how to do this without verifying the 8 axioms for a vector space, and if we let the set $X = \{c_1 + c_2 \sin^2 x + c_3 \cos^2 x\}$ then we note that $1 = \sin^2 x + \cos^2 x$, and this is enough. So the dimension is $2$. Thanks.

Can you please provide clarification on how the argument of the subspace of the vector space follows? I think you did it already by inspection, but its not very complete to me, can you please write it down? Thanks

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If by the above you meant "the span of..." then it is trivial: the span of anything within a vector space, including the empty set, is always a vector space. – DonAntonio Aug 25 '12 at 23:05
I'm always amazed how undergraduate algebra can "cram four axioms into eight axioms". Sort of like baseball "crams two minutes of action into two hours". – rschwieb Aug 26 '12 at 0:53
a bounty? Really? If you'd like more help, why don't you show us how far you've gotten, and where you're getting stuck? – Dustan Levenstein Aug 30 '12 at 15:34
"Can you please write it down?" No. Put forth some effort. Show us what you've tried and where you got stuck. Stop being a help vampire. – user5137 Aug 30 '12 at 18:43

Yes, you can show that it's a subspace of some other vector space. Letting $$V=\{c_1 + c_2 \sin^2(x) + c_3 \cos^2(x)\,\vert\,c_i\in \mathbb{R}\},$$ and letting $W$ be the space of all functions from $\mathbb{R}$ to $\mathbb{R}$ (under the operations of point-wise addition and scalar multiplication), it is clear that $V\subseteq W$.

Now, all that you need to do is show that for all $\alpha,\beta\in V$ and all $a,b\in \mathbb{R}$, that we have $a\alpha + b \beta\in V$. That's simple enough that it practically writes itself.

(Note: I'm assuming that the underlying field is $\mathbb{R}$, as it usually is for an undergraduate-level linear algebra course.)

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"Undergraduate level linear algebra course"...in what career or university? As far as I know, in mathematics-physics and surroundings both the real and complex are basic ground fields, and many times also prime finite fields, at least, ar put into the game. Perhaps undergraduate linear algebra for sciences or engineering doesn't use but reals, and even this is moot... – DonAntonio Aug 25 '12 at 23:18
@Don, here at Macquarie you get two-thirds of the way through the Linear Algebra sequence before you see anything but real vector spaces, and even then it's only complex ones. You have to get to Galois Theory before you see vector spaces over the rationals, finite fields, etc. – Gerry Myerson Aug 25 '12 at 23:34
@GerryMyerson, my heart bleeds for Oz's students (just joking). Here (Israel) complex numbers are taught in High school (the extended usual program), and finite fields are introduced mostly in introductory (linear or not) algebra courses. That's why I thought real linear algebra would be just a tiny part of the program. – DonAntonio Aug 25 '12 at 23:37
@Don, complex numbers are one thing, complex vector spaces, another thing entirely (at least, here). – Gerry Myerson Aug 25 '12 at 23:50
I do understand that, @GerryMyerson, yet if complex numbers are well known there's no reason not to deal with complex vector spaces from the beginning, or almost. – DonAntonio Aug 26 '12 at 0:06

Every set $X$ of objects forms a vector space over chosen field.

It is the free vector space over $X$: $F(X)$. Its basis is the set $X$ and its dimension is the cardinality of $X$.

You must be careful what do you mean by saying that something forms something.

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The free vector space over $X$ always contains vectors that are not in $X$ itself -- at least the zero vector. So $X$ is not necessarily a vector space itself. (Or was that the point of the answer)? – Henning Makholm Aug 31 '12 at 19:02
@Henning Yes it was. – Godot Aug 31 '12 at 21:53
But "forms" is actually pretty standard terminology for "can (in a natural way) be given the structure of". Your meaning here I would call "generates" or "gives rise to". – Henning Makholm Aug 31 '12 at 22:44
Since @mary mentioned checking axioms of vector spaces I wanted to point out that the vector space is made out not only of the underlying set of vectors but also of operations on that set. Therefore to say that a set forms a vector space is at least ambigous, until you specify how the operations of that vector space work. Without specyfying the addition of vectors and multiplication by scalars there is only one natural way of forming a vector space out of a given set -by a free structure built over it. – Godot Sep 1 '12 at 0:04

To prove that the span of a set of vectors forms a subspace of a vector space one can use the subspace test theorem. Suppose $W = span\{ v_1,v_2, \dots v_k \}$ where "span" means the set of all linear combinations with coefficients from $\mathbb{R}$. Note $W \neq \emptyset$ as $0$ is a linear combination $0v_1+0v_2+ \cdots +0v_k=0$. Moreover, if $x,y \in W$ and $c \in \mathbb{R}$ then $x = x_1v_1+ \cdots x_kv_k$ and $y=y_1v_1+\cdots +y_kv_k$ for some real constants $x_i,y_j \in \mathbb{R}$. Consider then:

$$cx+y = c[x_1v_1+ \cdots x_kv_k]+y_1v_1+\cdots +y_kv_k = (cx_1+y_1)v_1+\cdots + (cx_k+y_k)v_k$$

Thus $cx+y \in W$. It follows that the nonempty $W$ is closed under scalar multiplication and vector addition and by the subspace test we find $W$ is a subspace. This means $W$ is a vector space with respect to the operations of the vector space $V$ which contains $W$.

Now, you can take the redundant set $\{ 1, \cos^2 \theta, \sin^2 \theta \}$ as a spanning set for your subspace $W$, however this would not be a basis.

To find a basis you need to select linearly independent vectors whose span is $W$. You already pointed out $\sin^2 \theta+\cos^2 \theta=1$ in your post. Think about this. You can see how to write one of the vectors in $\{ 1, \cos^2 \theta, \sin^2 \theta \}$ as a linear combination of the remaining vectors. You have at least three obvious choices for the basis here. Hope this helps.

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