Linearly independent vectors coefficients Suppose that the set $\{\vec u_1, \vec u_2, \ldots \vec u_n\}$ is linearly independent and that $\vec w$ is a linear combination of these vectors. Is it true that the coefficients of the linear combination are unique? How so?
I do not understand how to prove that the coefficients are unique.
 A: It's true that if the $u_i$ are linearly independent then every vector can be written uniquely as a linear combination of the $u_i$. Suppose
$$a_1u_1 + a_2u_2 + \cdots + a_nu_n = b_1u_1 + b_2u_2 + \cdots + b_nu_n.$$
Then
$$(a_1 - b_1)u_1 + (a_2 - b_2)u_2 + \cdots + (a_n - b_n)u_n = 0.$$
Since the $u_i$ are linearly independent we conclude $a_i - b_i = 0$.
A: $w=\displaystyle\sum_{k=1}^n\alpha_kv_k$ where $v_1,...,v_n$ is a set of linearly independent vectors and $\alpha_1,...,\alpha_n$ are its correcponding coefficients.
We will prove this statement using proof by contradiction.
We start by assuming that the choice of $\alpha_1,...,\alpha_n$ is not unique. i.e.
$w=\displaystyle\sum_{k=1}^n\alpha_kv_k=\sum_{k=1}^n\beta_kv_k$ where at least one $\alpha_k\neq\beta_k$
Here,
$\displaystyle\sum_{k=1}^n\alpha_kv_k=\sum_{k=1}^n\beta_kv_k\\
\displaystyle\sum_{k=1}^{n-1}\alpha_kv_k+\alpha_nv_n=\displaystyle\sum_{k=1}^{n-1}\beta_kv_k+\beta_nv_n\\
v_n=\displaystyle\frac{\displaystyle\sum_{k=1}^{n-1}\beta_kv_k-\displaystyle\sum_{k=1}^{n-1}\alpha_kv_k}{\alpha_n-\beta_n}$
However, this contradicts our assumption that the vectors $v_1,...,v_n$ are linearly independent.
So, our assumption that the choice of $\alpha_1,...,\alpha_n$ is not unique is false.
So, the choice of $\alpha_1,...,\alpha_n$ is unique.
