Get the solution to "A subset u1,...,up in V is linearly independent..."

djeljenike

Answered question

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}]_{β}, . . ., [u_{p}]_{β}

is linearly independent in

R^{n}

Answer & Explanation

l1koV

Skilled2021-02-16Added 100 answers

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

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