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}

Question
Alternate coordinate systems
asked 2021-02-15
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 \(R^{n}\)

Answers (1)

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}
0

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