Determine whether there is linear independence or not Determine if the following set of vectors is linearly independent:
$$S =  \{e^x, e^{2x}\}$$ of continuous functions.
So my problem isn't in not knowing what linearly independence is, it is setting up a linear combination of this form:
$$a_1e^x + a_2e^{2x} = ????$$
What do I set it equal to to determine the coefficients? For example if it was in terms of vectors it would be something like:
$$a_1V_1 + a_2V_2 = (0,0,0,)$$ if I was in $R^3$ for instance.
 A: In this case the vectors are functions.  What is the zero vector?  It is the function that takes all real numbers to $0$.
What you want to prove is that, for the functions $e^x$ and $e^{2x}$, if we have $a$ and $b$ such that $a e^x + b e^{2x} = 0$ (the zero function), then $a = b = 0$.
One way to show this would be to first plug in $x = 0$ to conclude $a + b = 0$.  Then take the derivative and plug in $0$ to conclude $a + 2b = 0$.
A: Always the same thing, you set it equal to the zero vector.
In this case the "vectors" you are talking about are functions.  (Like $e^x$ for example.)  The "zero vector" is the function given by $f(x)=0$ for all $x$.
A: You would replace the question marks you above above with $0$. This zero isn't quite the same as an ordinary scalar zero (even though we write it the same way), rather it's the constant zero function. That is, it is the function $f \equiv 0$. 
A: your matrix will be in a 2x2 form where row 1 is ae^x + be^2x = 0 and row 2 = e^x + 2e^2x =0, you find the determinant by factoring out e^xe^(2x) multiplied by your simplified matrix with (1,1) and (1,2) as rows 1 and 2. Hence determinant is e^xe^(2x)* (2-1)= e^xe^(2x)(1) which is not zero, meaning you have a set of linearly independent vectors.. Sorry it is my first time using this and I dont know how to write properly yet. 
