Question

A subset {u_{1}, ..., u_{p}} in V is linearly independent if and only if the set of coordinate vectors {[u_{1}]_{beta}, ..., [u_{p}]_{beta}} is linearly independent in R^{n}

Alternate coordinate systems
A subset $${u_{1}, ..., u_{p}}$$ in V is linearly independent if and only if the set of coordinate vectors
$${[u_{1}]_{\beta}, ..., [u_{p}]_{\beta}}$$
is linearly independent in $$R^{n}$$

2021-02-16

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 $$R^{n}$$