im pretty sure that if a vector is in the span of a set of vectors, then it can be written as a linear combination of the vectors in the set, whch you can find out by setting up a system of equations. But what if you have an arbitrary object and want to find out if it is in the span of a set?
such as this:
$S = \{\cos^2x, \sin^2x, \tan^2x\}$
a) $1$
b) $\sec^2x$
c) $\cos2x$
d) $0$
which of these is in the span of this set? I want to say that $1$ can't be written as a linear combination of the three functions, but don't know how to find out concretely, I also want to say $0$ can be, because you can have $k1,k2,k3 = 0$, but am I allowed to have all scalars be $0$? and I'm not sure how to concretely figure out if the other functions or a value in general is in the span of this set. what is the most guaranteed way to find out whether something spans a set or not?
You do it the same way as with Euclidean vectors -- you check if you can represent a given function (/vector) by a linear combination of your basis:
That is, you try to solve:
$a)\ 1=a\cos^2(x)+b\sin^2(x)+c\tan^2(x)$
$b)\ \sec^2(x)=a\cos^2(x)+b\sin^2(x)+c\tan^2(x)$
$c)\ \cos(2x)=a\cos^2(x)+b\sin^2(x)+c\tan^2(x)$
$d)\ 0=a\cos^2(x)+b\sin^2(x)+c\tan^2(x)$
NOTE: the $a$'s, $b$'s, and $c$'s need not be the same for each problem.
If there is at least one solution $(a,b,c)$ to any of these, then that function is in the space spanned by $\cos^2(x)$, $\sin^2(x)$, and $\tan^2(x)$.
You should also note that some sort of $0$ (scalar, function, n-tuple, etc) will ALWAYS be in your space precisely because you can always have $a=b=c=0$.
So let's do $(a)$: Remember that we have the Pythagorean identity: $\sin^2(x) + \cos^2(x) =1$. So clearly one solution is $1=(1)\cos^2(x)+(1)\sin^2(x)+(0)\tan^2(x)$, AKA $(a,b,c) = (1,1,0)$. Because we were able to find this solution, it means that $1 \in \text{span}(\cos^2(x),\sin^2(x),\tan^2(x))$.