How do i find a basis for the span of a set of functions? 
Find a basis for Span($V)=\{sin(x),cos(x),e^x\}$, which is a subset of $C^0 (\Bbb R)$.

So if we consider $af(x) + bg(x) +ch(x)=0$
$\Rightarrow (a+b+c)(f(x)+g(x)+h(x))=0$
$\Rightarrow a=b=c=0$ so the set is Linearly Independent.
Also, the set clearly spans $C^0 (\Bbb R)$ as each function is continuous over real numbers. 
Hence, we can find a basis for V:
$$ \{(1,0,0),(0,1,0),(0,0,1)\}$$
I've just started with Linear Algebra, and I'm not convinced what I have done is sufficient or even correct. Any clarification on this? 
 A: Several hints/issues:


*

*$V$ is not a vector space as stated, it's just a subset.  It doesn't make sense to find a basis for a subset - basis is defined as a linearly independent spanning set of a vector space.  Perhaps you want to find a basis for $\text{span}(V)$?  (OK, the question was edited to fix this, however your notation is a bit off - you should say "find a basis for Span($V$) where $V=\{\sin(x),\cos(x),e^x\}$")

*Check your proof that $f,g,h$ are linearly independent.  You made a factoring mistake.

*Lastly, recall that a basis consists of vectors that are in the space.  The vectors you suggest are elements of $\Bbb{R}^3$, not $V$. 
Bigger hints are hidden below. 

 You just need to show that $\sin(x),\cos(x)$ and $e^x$ are linearly independent.  Then, those 3 functions form a basis for span($V$), by definition!  To show that $\sin(x),\cos(x)$ and $e^x$ are linearly independent, start as you did by assuming that $(a,b,c)\in\Bbb{R}^3$ are such that $a\sin(x)+b\cos(x)+ce^x=0$ for all $x\in\Bbb{R}$.  We want to show that $(a,b,c) = (0,0,0)$.  Here's the trick: evaluate this expression for 3 distinct $x$ values!  For instance, $a\sin(x)+b\cos(x)+ce^x=0$ means $a\sin(0) + b\cos(0) + ce^0 = 0$.  Pick 2 other values of $x$; now you'll have 3 equations for the three unknowns $(a,b,c)$.  You can write these equations as a matrix, and if this matrix is non-singular (compute the determinant, for instance), you know that $(a,b,c) = (0,0,0)$ is the only solution.

A: $af(x) + bg(x) + ch(x) \ne (a+b+c)(f(x)+g(x) + h(x))$
Nonetheless, $\sin x, \cos x, e^{x}$ are linearly independent.
So, make them your basis!  $e_1 = \cos x, e_2 = \sin x, e_3 = e^x$
We haven't proven that these "vectors" are linearly independent.  However, if they were dependent and,  $a e_1 + b e_2 + c e_3 = 0$ for some $a,b,c \ne 0$
It must be true for all values of $x.$  Pick some convenient values of $x.$ 
For example $x = 0, 2\pi, \pi/2, -\pi/2$
And you will get an set of equations with no solution other than the trivial.
