Prove that the vector space is 3D So we consider a vector space V which is defined by $f(x)=ax+be^x+csin(x)$ where a, b and c are real. I was able to prove that this was in fact a vector space in another problem that I had. Now, the problem is asking me to prove that this vector space is a space of 3 dimensions. How do you do that ? Never done this. Any help would be appreciated.
 A: This $V$ is a subspace of the real vector space $C(\mathbb{R})$, i.e. continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Now note that $V=span\{x,e^{x},\sin(x)\}$ as any member in $V$ is a unique linear combination of these vectors. These three vectors are also linearly independent so they form a basis for $V$. Implied is that $V$ has dimension $3$.
To prove that $x,e^{x},\sin(x)$ are linearly independent, consider $ax+be^{x}+c\sin(x)\equiv 0$ for some real $a,b,c$. Now this holds for all $x$, so plugging in $x=0$ gives $b=0$. Apply $b=0$ to the equation and plug in $x=\pi$ to $ax+c\sin(x)=0$ to get $a\pi=0$, i.e. $a=0$. Now finally $x=\frac{\pi}{2}$ gives $c=0$. So $a=b=c=0$, which concludes the argument.
A: You need to show that your three functions are linearly independent.
That is $ax + be^x + c\sin x = 0 \iff a,b,c = 0$
To do that, choose some values for $x$.  This simplifies your problem to a system of linear equations, and it should be easy enough to show that the only solution is $a,b,c =0$
