Suppose the subspaces are S1,S2,S3.....Sk

Take a from S1, b from S2....k from Sk

then a+b+....k belongs to V and union of all subspaces.

By the definition of union, a+b+....k must belong to atleast one of the subspace which will make it equal to V.

Is this proof correct? Or am I missing something?

  • $\begingroup$ Your phrasing is quite unclear, please try to be more precise. Also, the statement itself clearly fails for finite dimensional vector spaces over finite fields, which are indeed unions of some proper subspaces. $\endgroup$ – A. Pongrácz Sep 1 '18 at 20:35
  • $\begingroup$ This is not clear. What does "its $k$ subspaces" mean? Vector spaces can have lots and lots of subspaces. $\endgroup$ – lulu Sep 1 '18 at 20:36
  • $\begingroup$ The statement is wrong over finite fields. $\endgroup$ – Arnaud Mortier Sep 1 '18 at 20:36
  • $\begingroup$ Of its some k subspaces. So, we have to prove that V is equal to any one of the k subspaces we have selected. $\endgroup$ – Avinash Bhawnani Sep 1 '18 at 20:37
  • 1
    $\begingroup$ See this question. $\endgroup$ – lulu Sep 1 '18 at 20:54

Actually, if a vector space over a field $F$ is the union of $k$ proper subspaces, then $|F|<k$. So if the field is infinite, one of the subspaces is equal to $V$. This result is known as the avoidance lemma for subspaces.

For a proof you can take a look at my answer to this closely related question

| cite | improve this answer | |
  • $\begingroup$ I didn't understand why at most 1 element? $\endgroup$ – Avinash Bhawnani Sep 1 '18 at 21:05
  • 1
    $\begingroup$ Oh! In the proof? It is because if $v+ku$ and $v+k'u\in W_i$ for some $k\ne k'$; as W_i is a subspace, we deduce that $(v+ku)-(v+k'u)=(k-k')u \in W_i$, and as $k-k'\ne 0$, this implies $u\in W_i$, contrary the the hypothesis. $\endgroup$ – Bernard Sep 1 '18 at 21:12
  • $\begingroup$ Thank you! For infinite elements, is my proof correct? (Given that written in Mathematical notations, don't know how to write here) $\endgroup$ – Avinash Bhawnani Sep 1 '18 at 21:18
  • 1
    $\begingroup$ No it is not correct, because you've forgotten that when you change the value of some of $a,b, \dots,k$, the sum $a+b+\dots+k$ lies in one of the subspaces, but not necessarily in the same subspace as for the previous values. For the formatting with Mathjax, it is very close to the formatting of maths formulæ with LaTeX, if you know it. $\endgroup$ – Bernard Sep 1 '18 at 21:28
  • $\begingroup$ I actually just tried to extend the proof of "if union of two subspaces is a subspace then one must contain the other" If a+b+c....+k belongs in V, then it must contain in one of its subspaces. Suppose S3. So, now a+b+....k belongs in S3. But we only took c from S3, so from the subspace property, a,b,d...,k must belong in S3. Which will make S3 =V? $\endgroup$ – Avinash Bhawnani Sep 1 '18 at 21:33

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.