<p>Don’t worry, this isn’t for a homework grade or anything, I’m studiyng for a test and can’t figure this one out, if you could help, it’d be great</p>
<p>I can’t figure this out, I don’t think it should be that hard though</p>
<p>Show that if v1, v2, …vn is a linearly independent set of vectors, then any subset of thses vectors is also linearly independent. </p>
<p>any ideas?
Thanks
Ben</p>
<p>Say you have a dependent subset, then
a<em>i v</em>i + … + a<em>k v</em>k = 0 (*)
for some a’s not all zero.</p>
<p>Now look at
a<em>1 v</em>1 + … + a<em>n v</em>n = 0
and rearrange it into
(*) + (linear combination of remaining vectors) = 0.</p>
<p>We know that (*) can be set to zero without having all the coefficients 0, and we can then set the remaining linear combination to zero by setting all their coefficients to zero. Hence we have contradicted the independence of the original set.</p>
<p>Thanks a lot, I was looking for really how to show it and that’s perfect to show it in a contradiction, thanks a lot, I appreciate it</p>