0
$\begingroup$

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

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

then

$\alpha_1=\beta_1,...,\alpha_n=\beta_n$

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 $i=1,...,n$?

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.