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Prove that the union of three subspaces of V is a subspace iff one of the subspaces contains the other two.

I can do this problem when I am working in only two subspaces of $V$ but I don't know how to do it with three.

What I tried is: If one of the subspaces contains the other two, Then their union is obviously a subspace because the subspace that contains them is a subspace. (Is this sufficient??).

If the union of three subspaces is a subspace..... How do I prove that one of the subspaces must contain the other two from here?

*When proving this for two I said that there is an element in one of the subspaces that is not the other and proved by contradiction that one of the subspaces must be contained in the other. How would I do this for three?

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    $\begingroup$ "(Is this sufficient??)." Yes, I'd say so. The phrasing might need some work, but no more arguments are required. Something along the lines of "If one subspace $V_0$ contains the other two, then their union is just $V_0$, which is a subspace by definition." $\endgroup$ – Arthur Jul 7 '14 at 3:47
  • $\begingroup$ I don't think this statement is true (because I can't find a proof), but still can't construct a counterexample yet. $\endgroup$ – Gina Jul 7 '14 at 5:34
  • $\begingroup$ Found the counterexample! I guess you need to strengthen the hypothesis (say, some restriction on the underlying field) to obtain the result. $\endgroup$ – Gina Jul 7 '14 at 5:44
  • $\begingroup$ @Gina is right. This depends on the cardinality of the underlying field. See for example this earlier question. $\endgroup$ – Jyrki Lahtonen Jul 7 '14 at 6:38
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The statement is false. Consider the following counterexample:

Consider the vector space $V=(\mathbb{Z}/2\mathbb{Z})^{2}$ where $F=\mathbb{Z}/2\mathbb{Z}$. Let $V_{1}$ be spanned by $(1,0)$. Let $V_{2}$ be spanned by $(0,1)$. Let $V_{3}$ be spanned by $(1,1)$. Then we have $V=V_{1}\cup V_{2}\cup V_{3}$, but none of the $V_{1},V_{2},V_{3}$ are subspace of another.

You can usually count on field of characteristic $2$ to give you counterexample. There are many similar counterexample, too. In finite dimension, I think all counterexamples can be constructed this way. My intuition tells me that there are infinite dimensional counterexamples of other form, but have not checked clearly.

EDIT. Here is a proof of the statement with the restriction $F\not=\mathbb{Z}/2\mathbb{Z}$:

Without loss of generality, we can assume the whole space $V$ is in fact $V_{1}+V_{2}+V_{3}$. Easily seen that in fact we must also have $V=V_{1}\cup V_{2}\cup V_{3}$.

There exist $a,b\in F$ such that $a,b\not=0$ and $a-b=1$ (take $a$ to be anything except $0,1$, and take $b=a-1$).

Assume $V_{1}$ and $V_{2}$ neither contains another (otherwise this reduce to the 2-subspace case). For any $u\in V_{1}\setminus(V_{1}\cap V_{2})$ we take an arbitrary $w\in V_{2}\setminus(V_{1}\cap V_{2})$ (it exists due to the fact that neither $V_{1}$ nor $V_{2}$ contains another). Then $au+w$ is in neither $V_{1}$ nor $V_{2}$ (if in $V_{1}$ then since $au\in V_{1}$ we must have $w\in V_{1}$ so $w\in V_{1}\cap V_{2}$ contradiction; same for the other case but now using the fact that $a\not=0$), so $u+aw\in V_{3}$. Same argument apply to show $bu+w\in V_{3}$. Hence $u=(bu+w)-(au+w)\in V_{3}$. Hence $V_{1}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Same argument apply to show $V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Now for any $v\in V_{1}\cap V_{2}$ we pick a $w\in V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Then $w+v\notin V_{1}\cap V_{2}$ (otherwise $w\in V_{1}\cap V_{2}$). But $w+v\in V_{2}$. Hence $w+v\in V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Thus $v=(w+v)-w$ so $v\in V_{3}$. Hence $V_{1}\cap V_{2}\subset V_{3}$. Therefore $V_{1},V_{2}\subset V_{3}$.

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    $\begingroup$ Why can we assume that the whole space $V$ is in fact $V_{1}+V_{2}+V_{3}$? $\endgroup$ – ubadub Mar 4 at 1:51
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Gina gave an excellent answer,in fact,we can have : If $V$ is a vector space over the field $F$ and there is a collection of finite number of subspaces of $V$, $\{U_1,U_2,U_3,\cdots ,U_n\}$,and $n$,the number of the elements of the collection above,is not more than the cardinality of $F$,when $F$ is finite,or $F$ is just infinite,then the union of all the subspaces $U_1,U_2,U_3,\cdots ,U_n$ is a subspace of $V$ if and only if one of the subspaces $U_1,U_2,U_3,\cdots ,U_n$ contains all other subspaces.The proof is similar to the way one proves that "a vector space over an infinite field cannot be a finite union of proper subspaces of its own", and using the technique "prove by contradiction".(Using the pigeonhole principle to deduce absurdity: Imagine that the elements of $F$ "fly" into the subspaces $U_1,U_2,U_3,\cdots ,U_n$)

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  • $\begingroup$ Readers who are interested in this may also be interested in the proof of the 'prime avoidance lemma'. $\endgroup$ – W.Leywon Feb 18 '17 at 14:58
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Suppose $W=V_1\cup V_2\cup V_3$ is a subspace. Since $W$ contains $V_1,V_2,V_3$ and is closed under linear combinations, we have $V_1+V_2+V_3\subset W$ as well. The reverse inclusion also holds, since each $V_1,V_2,V_3$ lie in the sum. Thus $$W=V_1+V_2+V_3.$$

Observe the inclusions $$ W=V_1\cup V_2\cup V_3\subset (V_1+V_2)\cup V_3\subset V_1+V_2+V_3=W. $$ Thus $(V_1+V_2)\cup V_3=W$, so by the result for two subspaces, either $V_3\subset V_1+V_2$ or $V_1+V_2\subset V_3$. In the former case, $W=V_1+V_2$ so repeating the two-subspace argument yields $W=V_1$ or $W=V_2$. In the latter case, $W=V_3$. The result follows.

Note: The same proof extends by induction to any finite union of subspaces.

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    $\begingroup$ "$W=V_1+V_2$ so repeating the two-subspace argument yields $W=V_1$ or $W=V_2$". This does not follow. The "two-subspace argument" only apply when $W=V_1\bigcup V_2$ and this not the case here. I don't think this error can be fixed at all. $\endgroup$ – Gina Jul 7 '14 at 5:20

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