Understanding definition of Linear Independence Hello and happy holidays,
The definition of linear independence in my notes is:
(Linear independence of functions). The functions $y_i : \mathbb R \to \mathbb C^n, i \in \{1,...,k\}$ are said to be linearly independent if, given $c_i \in \mathbb C, i \in \{1,...,k\}$
$$\sum_{i=1}^{k} c_i y_i(t) = 0 \text{ for all } t \in \mathbb R \text{ implies } c_1 = c_2 = \cdots = c_k = 0.$$ 
Now I understand this. But in my notes in an example, they prove linear independence by proving the constants are equal to $0$ for only one value of $t$. Why can you do this? Do you just assume that the sum equals $0$ for all $t$ as in the definition?
 A: In the definition you are given that the $y_i$ are linearly independent if you have that 
$$
\sum_{i = 1}^k c_iy_i(t) = 0 \ \ \forall t\in\mathbb{R}  \quad\implies \quad c_i = 0 \ \  \forall i \in \{1, \ldots , k\}
$$
So when showing linear independence you will start by assuming the LHS of the implication. By this assumption you have that the sum above is zero for all $t$ and the goal is to show that from this assumption all the $c_i = 0$. You are free to use whatever specific value of $t$ that you want in order to prove this.
A: Linear independence of a bunch of objects always says that you can't scale the objects by any amounts and add them all together to get $0$, with the exception of where you're scaling by $0$.
Before we can apply this to functions, we have to define some things:


*

*Addition. Given functions $f, g, h$: $h = f+g$ iff $\forall x \cdot h(x) = f(x) + g(x)$.

*Scaling. Given functions $f, h$ and scalar $c$, $h = cf$ iff $\forall x \cdot h(x) = cf(x)$.

*$0$ as a function can be defined as the function such that for all $0(x) = 0$ for every value of $x$. That is, it is the function that's $0$ everywhere.


Then say we have functions $y_1, y_2, \ldots, y_k$. These are linearly independent iff for all scalars $c_1, c_2, \ldots, c_k$:
$$c_1f1 + c2_f2 + \ldots + c_kf_k = 0 \implies \forall i \cdot c_i = 0$$
All that says is, the only way to arrive at the function that's $0$ everywhere through a combination of any of these functions is to set all the coefficients to $0$.
