Let $\left\{v_1,v_2,...,v_n\right\}$ linearly independent set and $\left\{\alpha _1,\alpha _2,...,\alpha _n\right\}$ scalars . Let $u=\alpha _1v_1+\alpha _2v_2+...+\alpha _nv_n$ Show that $\left\{\alpha _1,\alpha _2,...,\alpha _n\right\}$ Are determined singly
My attempted :
Let $u=\alpha _1v_1+\alpha _2v_2+...+\alpha _nv_n$ and $u=\beta _1v_1+\beta _2v_2+...+\beta _nv_n$ such that $\alpha_i \ne \beta_i,1 \le i \le n$
so we get $u-u=0=(\alpha _1v_1+\alpha _2v_2+...+\alpha _nv_n)-(\beta _1v_1+\beta _2v_2+...+\beta _nv_n) \rightarrow 0= (\alpha _1-\beta_1)v_1 + ... + (\alpha_n-\beta_n)v_n$
we know that $\left\{v_1,v_2,...,v_n\right\}$ linearly independent
then $\alpha _1-\beta_1=0 , \alpha _2-\beta_2=0 , ...,\alpha _n-\beta_n=0 \rightarrow \alpha _1=\beta_1 , \alpha _2=\beta_2 , ...,\alpha _n=\beta_n$
is this prove it ? if not how can i prove it ?
thanks.