djeljenike

2021-02-15

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

l1koV

Expert

A zero vector in V can be written as, ${c}_{1}{u}_{1}+...+{c}_{p}{u}_{p}=0.$
Since the mapping $x|\to \left[x{\right]}_{\beta }$ is one-to-one,
the zero vector in ${R}^{n}$ is,
$\left[{c}_{1}{u}_{1}+...+{c}_{p}{u}_{p}{\right]}_{\beta }=\left[0{\right]}_{\beta }$
That is, $\left[{c}_{1}{u}_{1}{\right]}_{\beta }+...+\left[{c}_{p}{u}_{p}{\right]}_{\beta }=\left[0{\right]}_{\beta }$
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 $\left[{c}_{1}{u}_{1}{\right]}_{\beta }+...+\left[{c}_{p}{u}_{p}{\right]}_{\beta }=\left[0{\right]}_{\beta }$ implies that
$\left[{c}_{1}{u}_{1}{\right]}_{\beta },...,\left[{c}_{p}{u}_{p}{\right]}_{\beta }$ 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 $\left[{u}_{1}{\right]}_{\beta },...,\left[{u}_{p}{\right]}_{\beta }$ is linearly
independent in ${R}^{n}$

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