A zero vector in V can be written as, \(c_{1} u_{1} + ... + c_{p} u_{p} = 0.\)

Since the mapping \(x |\rightarrow [x]_\beta\) 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}

Since the mapping \(x |\rightarrow [x]_\beta\) 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}