Remark 1 on Linear Mapping

Let $V$ be a finitely generated $K$-vector space and $b_1, \ldots, b_n$ be a basis of $V$. Then every $v \in V$ can be uniquely represented as

v = \sum_{i=1}^n \lambda_i b_i

with $\lambda_i = \lambda_i(v) \in K$.

Beweis

The existence of a representation follows from the fact that a basis is a generating set. Let

v = \sum_{i=1}^n \mu_i b_i,

with $\mu_i \in K$, be a second representation. It follows that

0 = \sum_{i=1}^n (\lambda_i - \mu_i) b_i.

Since the basis vectors $b_1, \ldots, b_n$ are linearly independent, it holds that

\lambda_i = \mu_i

for all $i = 1, \ldots, n$. Thus, the representation is unique. $\square$