Prop. Let $V$ be a vectos space over $K$ and let $v_1,...,v_n$ be linearly independt. Then, if

$\alpha_1 v_1+...+\alpha_n v_n=\beta_1 v_1+...+\beta_n v_n$



Proof. Suppose $\alpha_1 v_1+...+\alpha_n v_n=\beta_1 v_1+...+\beta_n v_n$. Then $(\alpha_1-\beta_1)v_1+...+(\alpha_n-\beta_n)v_n=0$

So, by assumption $\alpha_i - \beta_i=0$ for $i=1,...,n.$

My question is: Which assumption $\alpha_i - \beta_i=0$ for $i=1,...,n$?

  • 1
    $\begingroup$ The assumption you are using is that $v_1,\dots,v_n$ are linearly independent. This means that if $c_i \in K$ are scalars such that $c_1v_1 + \cdots + c_nv_n = 0$, then $c_i = 0$ for $i=1,\dots,n$. $\endgroup$ – suchan Feb 27 '17 at 0:59
  • $\begingroup$ @suchan yes. thanks. $\endgroup$ – James Ensor Feb 27 '17 at 1:00

Definition of linear independence: $v_{1},...,v_{n}$ are said to be linearly dependent if and only if $\lambda_{1}v_{1} +...+\lambda_{n}v_{n} = 0 \Rightarrow \lambda_{1}=\lambda_{2} =...=\lambda_{n}=0$

This is what you have assumed, and then your proof follows.


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