Is $A \cup B$ a subspace of $\Bbb R^3 $

Question

$u_1 = (0,4,10)$ $u_2 = (0,2,6)$ $u_3 = (4,-2,2)$

$A = span(u_1,u_2) $ , $B = span\langle u_3\rangle$

You have written three vectors and defined $A$ and $B$. That's all you've done. — Michael Albanese, Oct 31 '15 at 13:41
This is very similar to your other question http://math.stackexchange.com/questions/1506292/is-the-union-of-two-spans-a-subspace. What have you tried? Where are you stuck? If you show some effort you can be helped. If you don't help yourself, why are we supposed to do it? — mfl, Oct 31 '15 at 13:43
Well my primary problem is understanding the union of two spans. First I know that A and B are independent because you cannot create a linear combination of $u_3$ from $u_1$ and $u_2$. Therefore would I be correct in defining $A \cup B$ as $ {a_1u_1+a_2u_2+a_3u_3}$? Where $a_1,a_2,a_3$ are members of the field. — TheEnlightenedTomato, Oct 31 '15 at 13:47
@TheEnlightenedTomato, this is not a question because it lacks a question mark... — k170, Oct 31 '15 at 19:36

score 3 · Answer 1 · answered Oct 31 '15 at 13:48

3

The answer comes from the Avoidance lemma for vector spaces:

Over an infinite field, the union of a finite number of subspaces cannot be a subspace, unless they're all contained in one of them.

answered Oct 31 '15 at 13:48

Bernard

175,478

I have not studied the avoidance lemma :( How do you prove it simply using the definition of a span? – TheEnlightenedTomato Oct 31 '15 at 13:53
You actually just need the notion of subspace. The complete answer (and proof) can be found in my answer to this question. Please note in case of 2 subspaces, it comes from a more general (and elementary) fact from group theory: a group cannot be the union of two proper subgroups (but it can be of three!). – Bernard Oct 31 '15 at 14:01

hmakholm left over Monica · Answer 2 · 2015-10-31T20:26:08.497

The union here is the ordinary union operation from set theory: $A\cup B=\{x\mid x\in A \lor x \in B\}$. In particular, $\cup$ doesn't know it is working on subsets of a vector space; it works like it does on any other sets.

In order to show that $A\cup B$ is not a subspace, it will suffice to find $v_1$ and $v_2$ in $A\cup B$ such that $v_1+v_2$ is not in $A\cup B$.

Note that "not in $A\cup B$" is the same as "neither in $A$ nor in $B$".

Since $A$ and $B$ are subspaces, a good strategy would be to choose $v_1\in A$ and $v_2\in B$. For example you could let $v_1=u_1$ and $v_2=u_3$. Then you just need to compute $u_1+u_3$ and show that it is neither in $A$ nor in $B$.

mvw · Answer 3 · 2015-10-31T15:02:42.710

The subspaces of $\mathbb{R}^3$ are:

the set that contains the origin,
lines that contain the origin,
planes that contain the origin and
$\mathbb{R}^3$ itself.

In your case $A$ is a plane and $B$ is a line, both containing the origin. The only case that their union is a subspace is if $$ A \cup B = A $$ For this you can try to investigate if all vectors in $B$ are in $A$. Thus $B\subseteq A$.

That means trying to solve $$ a \, u_1 + b\, u_2 = c \, u_3 \quad (*) $$ where RHS vectors different from $0$ should have solutions too.

This is the same as asking if the $u_i$ are not linear independent. $$ a\,u_1+b\,u_2+(-c)\,u_3=0 $$

So you could try to solve $(*)$ for $(a,b,c)\ne 0$ or check if $\text{det}(u_i)=0$ to decide if your union is a subspace.

Is $A \cup B$ a subspace of $\Bbb R^3 $

3 Answers3