Uniqueness of basis of a vector space

Question

If X is a vector space and:
$A \subset X$ s.t. $span(A)=X $ ( A basis of X ).
If $x \in span(A)$ : $x= \sum_{i=1}^{n} x_i k_i$ for $x_1,...,x_n \in X, k_1,...,k_n \in \mathbb{R}$.

How to prove that $x_i,k_i$ for $i=1,...,n$ are unique?

My attempt:
Suppose
$x= \sum_{i=1}^{n} x_i k_i$ for $x_1,...,x_n \in X, k_1,...,k_n \in \mathbb{R}$ and
$x= \sum_{i=1}^{n} y_i m_i$ for $y_1,...,y_n \in X, m_1,...,m_n \in \mathbb{R}$

Then : $0= \sum_{i=1}^{n} (x_ik_i-y_im_i)$ $\Rightarrow$ $\sum_{i=1}^{n} x_i k_i= \sum_{i=1}^{n} y_i m_i$ $\Rightarrow$ $x_i=y_i$ and $k_i=m_i$ for $i=1,...,n$.
Is this proof correct ?

Answer 1

In general, the basis of a vector space is not unique, and there are standard methods for transforming between bases.

For example, $\{(1, 0), (0, 1)\}$ is a basis for $\mathbb{R}^2$, but $\{(3, 5), (11, -6)\}$ is also a basis for $\mathbb{R}^2$.

However, for any specific basis, the representation of a vector in that basis is unique. In other words, if $\{b_1, \ldots, b_n\}$ is a basis for $V$, and for some $v \in V$ we have $v = a_1 b_1 + \ldots + a_n b_n = a_1' b_1 + \ldots + a_n' b_n$, then we do have $a_i = a_i'$ for all $i \in \{1, \ldots, n\}$, and your attempted proof is close to being a proof for this fact.