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We are given that $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis. My question is does it necessarily follow that $\{u_1,...,u_n\}$ or $\{v_1,...,v_k\}$ are linearly independent? I suspect the answer is yes, but I think I'm missing something in my argument.

So: if $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis, then this list of vectors is linearly independent. That is,

$a_1 u_1 + \cdots + a_n u_n + b_1 v_1 + \cdots + b_k v_k = 0 \implies a_1 = \cdots = a_n = b_1 = \cdots = b_k = 0$.

Then I want to say the following: Since all the $b_i \,'s$ are $0$, we have $a_1 u_1 + \cdots + a_n u_n = 0 \implies a_1 = \cdots = a_n = 0$, meaning that {$u_1,...,u_n$} is linearly independent. With a similar argument, we can also conclude that {$v_1,...,v_k$} is linearly independent.

Am I missing something though? I feel like my argument is kind of making a bit of a stretch. Thanks for your help!!

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    $\begingroup$ You're right! in general, any subset of a linearly independent set is again linearly independent, as your proof shows (you're not using more than linearly independency). $\endgroup$
    – Daniel
    Commented Oct 1, 2016 at 20:25
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    $\begingroup$ Yes, any subset of a linearly independent set is lin. indep. itself. $\endgroup$
    – DonAntonio
    Commented Oct 1, 2016 at 20:25
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    $\begingroup$ It's fine for me. Actually, whether it's a basis is not the relevant hypothesis. What you proved indeed is a little more general: any sublist of a list of linearly independent vectors is linearly independent. $\endgroup$
    – Bernard
    Commented Oct 1, 2016 at 20:26
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    $\begingroup$ Thank you so much guys!!! :D $\endgroup$
    – Javier
    Commented Oct 1, 2016 at 20:36

1 Answer 1

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Your argument is fine, but perhaps this is a case where it can be nicely put as a proof by contradiction. Suppose some sublist were not linearly independent, then we can find some $\{a_i\}$, not all zero, such that $\sum_{i=1}^n a_iu_i=0$. We can then use this to show that the bigger list is not linearly independent.

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    $\begingroup$ +1; I might add that this argument goes through as a proof by contrapositive without appealing to contradiction at all. :) $\endgroup$ Commented Oct 2, 2016 at 7:18
  • $\begingroup$ Good point @AlexWertheim! I never really thought about the difference much. For those interested, here's a good summary: math.stackexchange.com/a/705291/374223 $\endgroup$ Commented Oct 2, 2016 at 22:21

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