A Linear Combination of vectors is the vector of form
$$a_1v_1 + a_2v_2 + a_3v_3 + \dots + a_kv_k$$The set of all possible linear combinations of a set of vectors.
A set of vectors is linearly independent if no vector in the list is in the span of the others.
For example, the set of vectors $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ are linearly independent. Specifically, $(0, 0, 1)$ cannot be written as any linear combination of the other vectors. That is, given $a, b \in F$, $a(1, 0, 0) + b(0, 1, 0) = (a, b, 0)$.
A list of vectors $(v_1, v_2, \dots, v_n)$ is a basis for $V$ if its elements are linearly independent, and their span is (the whole vector space) $V$. The dimension of a vector space is the length of any basis.
A function $f : V \to W$ is a linear map if it preserves the operations of addition and scalar multiplication. That is, for all $v, w \in V, c \in F, f(v+w) = f(v) + f(w)$ and $f(cv) = cf(v)$.
Let $v$ be an arbitrary vector in $\mathbb{R}^3$ and $e_1 = (1, 0, 0), e_2 = (0, 1, 0), e_3 = (0, 0, 1)$ the basis vectors of $\mathbb{R}^3$. Then:
$$ (a_1,a_2,a_3) = (a_1, 0, 0) + (0, a_2, 0) + (0, 0, a_3) = a_1(1, 0, 0) + a_2(0, 1, 0) + a_3(0, 0, 1) = a_1e_1 + a_2e_2 + a_3e_3 $$Missing proof
Let $V, W$ be vector spaces and $f: V \to W$ a linear map. For all $v \in V$:
$$f(0 + v) = f(v) = f(0) + f(v) = 0 + f(v)$$Let $V, W$ be vector spaces and $f: V \to W$ a linear map. For all $v \in V$:
$$f(v + (-v)) = f(v) + f(-v) = f(0)$$