let me know if u think i did this proof correctly...(adsbygoogle = window.adsbygoogle || []).push({});

use the definition of linear independence to prove that if [tex]{u_1,u_2,...,u_n}[/tex] is a linearly independent set, then [tex]{u_1,u_2,...,u_k}[/tex] is also a linearly independent set for any

[tex]k = 2,3,...,n-1[/tex]

well heres what i did...

if the set of vectors [tex]{u_1,u_2,...,u_n}[/tex] is linearly independent then none of the vectors in that set is linear combination of the others. therefore

[tex]c_1 u_1 + c_2 u_2 + ... + c_n u_n = \vect{0} [/tex] is satisfied only by the set of scalars [tex] {c_1,c_2,...,c_n}[/tex] that is [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex] then.. [tex] c_1 u_1 + c_2 u_2 +...+c_(n-1) u_(n-1) = -c_n u_n[/tex] should only be satisfied by the same set where [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex] because no vector is a linear combination of the others by definition. thus this can be expanded to...[tex]c_1 u_1 + c_1 u_2 + ... + c_k u_k = -c_n u_n -c_(n-1) u_(n-1) - ... -c_(k+1) u_(k+1)[/tex] the only solution is the set where [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex], thus the set [tex] {u_1,u_2, ...,u_k}[/tex] is linearly independent for any [tex]k=2,3,...,n-1.[/tex]

tell me what u think..

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# Yet another proof

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