2
$\begingroup$

Suppose we have 2 subspaces V,W. What is the base and the dimention of: V U W ?

Its clear to me that V U W is not always a subspace!

I was thinking about taking the base of U and the base of W and try to Gaussian elimination but that doesnt goes well becouse for example:

$V={(a,0,0)|a \text{ from } R}$. The base of V: $(1,0,0)$

$W={(0,b,0)|b \text{ from } R}$. The base of V: $(0,1,0)$

While doing Gaussian elimination to the matrix : $(1,0,0),(0,1,0)$ , we get that the base from the union of V with W is : ${(1,0,0),(0,1,0)}$ and this is not true becouse we create from this base the vector $(1,1,0)$ that doesnt exist in the uniton of V with W.

Thanks alot!

Stav

$\endgroup$
1
  • $\begingroup$ we cant accept the answer in the first 7 min so after that I forget. You can see in stackoverflow I do accept all answers and thank each person who helped. Thanks for the reminder. $\endgroup$
    – Stav Alfi
    Sep 3, 2016 at 15:40

1 Answer 1

1
$\begingroup$

I'm assuming that by "base", you mean "basis". What do you mean by a "basis of $V \cup W$"? As you noted, $V \cup W$ is not always a vector space; the definition of "basis" only applies to vector spaces. Usually people would talk about the basis of the span of $V \cup W$, in which case your method works fine.

Note that if $V \cup W$ did have a basis in the traditional sense (i.e., there existed linearly independent $v_1, \dots, v_n$ whose span was equal to $V \cup W$), then $V \cup W$ would certainly be a vector space, though this only happens when $V \subseteq W$ or $W \subseteq V$.

$\endgroup$
1
  • $\begingroup$ Okay thanks very much for the importent details. Now its clear to me! $\endgroup$
    – Stav Alfi
    Jul 4, 2016 at 19:30

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.