8
$\begingroup$

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!!

$\endgroup$
4
  • 4
    $\begingroup$ 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
    Oct 1 '16 at 20:25
  • 2
    $\begingroup$ Yes, any subset of a linearly independent set is lin. indep. itself. $\endgroup$
    – DonAntonio
    Oct 1 '16 at 20:25
  • 3
    $\begingroup$ 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
    Oct 1 '16 at 20:26
  • 1
    $\begingroup$ Thank you so much guys!!! :D $\endgroup$
    – Javier
    Oct 1 '16 at 20:36
1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ +1; I might add that this argument goes through as a proof by contrapositive without appealing to contradiction at all. :) $\endgroup$ Oct 2 '16 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 $\endgroup$ Oct 2 '16 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.