Unique representation of a vector(the proof of the converse)

Question

I know that if $B$ is a basis for $\mathbf V$, there exist unique scalars $\alpha_1, \dots , \alpha_n$ with the desired property I wonder if the reverse proposition(vice versa proposition) is also true.

Can you show the proof that the reverse proposition(vice versa proposition) is also true?

Answer 1

If there exists unique scalars for each $v$ such that $v=\alpha_1v_1+\cdots+\alpha_nv_n$, then the only scalars for $v=0$ are $\alpha _1=\cdots =\alpha_n=0$. Thus the $v_1,\dots,v_n$ are linearly independent.

Answer 2

This is very easy and follows almost from the definition of basis. Suppose that $B=\lbrace x_1,\ldots,x_n\rbrace$ is the basis, and that a vector $x$ is written as $$x=\alpha_1x_1+\ldots +\alpha_n x_n=\beta_1 x_1 +\ldots \beta_n x_n$$this implies $$(\alpha_1-\beta_1)x_1 +\ldots +(\alpha_n - \beta_n)x_n=\overrightarrow{0}$$ However since $\lbrace x_1,\ldots,x_n\rbrace$ is a basis all its vectors are linearly independent, so $\alpha_i-\beta_i=0$ for all $i=1,\ldots,n$ follows.