TASK:
Show that in an n-dimensional vector space X, the representation of any x as a linear combination of given basis vectors $e_{1},..,e_{n}$ is unique.
Attempt:
Let X be a vector space with dimenstion $\dim{X}=n$.Let $e=\{e_{1},..,e_{n}\}$ be a cannonical basis.
We want to proof that $x \in X$ is a unique linear combination:
$$x=\alpha_{1}e_{1}+...+\alpha_{n} e_{n}$$
Assumption
Lets assume that for scalars $\beta_{1},..,\beta_{n},\alpha_{i} \neq \beta_{i}, i =\overline{1,n}$ we can represent
By definition, every vector space is a field, which means that for every vector $x \in X$ and every scalar $\alpha_{1},...,\alpha_{n}$ $$\alpha_i(x_1+x_2)=\alpha_i x_1 + \alpha_i x_2$$ $$(\alpha_i + \beta_i) x = \alpha_i x + \beta_i x$$
Question1:
if the two representations are the same, does that mean that $$\sum_{i=1}^{n} \alpha_{i} x_{i}=\sum_{i=1}^{n} \beta_{i} x_{i}$$
and if so, would that mean that: $$\sum_{i=1}^{n} (\alpha_{i} -\beta_{i}) x_{i} =0$$
But that would mean that they are linearly dependant?
Question 2:
maybe I have to make a construction that involves a vector of scalars $\gamma=(\gamma_1,..,\gamma_{n}),\gamma_{i}=\alpha_{i}-\beta_{i},\forall i=\overline{1,n}$
Question 3:
How can this statement be proven?
Your 'Question 1' is the key:
if a vector can be written as two different linear combinations of some vectors $e_i$, then - as you observed - it yields to a nontrivial zero combination of them, hence $e_i$ are linearly dependent.
Putting it in contrapositive: if $e_i$ is linearly independent, any vector can be written as at most one linear combination of them.