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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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    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?2012-08-30

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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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    Replace "undergraduate" with "freshman", and its pretty accurate, me thinks :-)2012-09-01
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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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    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.2012-09-01
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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.