A subset u1,...,up in V is linearly independent if and only if the set of...

A zero vector in V can be written as, $c_1{u}_1+...+c_p{u}_p=0.$
Since the mapping $x|\to [x{]}_{\beta }$ is one-to-one,
the zero vector in $R^n$ is,
$[c_1{u}_1+...+c_p{u}_p{]}_{\beta }=[0{]}_{\beta }$
That is, $[c_1{u}_1{]}_{\beta }+...+[c_p{u}_p{]}_{\beta }=[0{]}_{\beta }$
Trivial solution of above equation implies that the equation
$c_1{u}_1+...+c_p{u}_p=0$ also has trivial solution as the mapping is one-to-one and onto.
The trivial solution of $[c_1{u}_1{]}_{\beta }+...+[c_p{u}_p{]}_{\beta }=[0{]}_{\beta }$ implies that
$[c_1{u}_1{]}_{\beta },...,[c_p{u}_p{]}_{\beta }$ are linearly independent.
Only a trivial solution of $c_1{u}_1+...+c_p{u}_p=0$ implies that vectors
$u_1,...,u_p$ are linearly independent.
Hence, a subset $u_1,...,u_p$ is linearly independent if and only
if the set of coordinate vectos $[u_1{]}_{\beta },...,[u_p{]}_{\beta }$ is linearly
independent in $R^n$