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Let $V$ be a vector space over a field $F$. Suppose $V_1,V_2$ are subspaces of $V$ that lies strictly between $\{ 0_V \}$ and $V$. Show that there is a basis $\mathcal B$ for $V$ such that $\mathcal B\cap ( V_1 \cup V_2)=\emptyset $.

My work so far: My guess is that $V \setminus (V_1 \cup V_2)$ will make a linearly independent set and therefore can be extended to a basis for $V$. I showed that this set is non empty. So we can consider $S:=V\setminus (V_1 \cup V_2)$ How do I show that $S$ is linearly independent? Or is that even the right way to proceed?.

Any hints, suggestions? Thanks for your time.

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$V\setminus (V_1 \cup V_2)$ is not a linearly independent set. Have you tried any examples to get a feel for what you're trying to do? Maybe do the exercise for $\mathbb R^2$ and let $V_1$ be the subspace generated by $(1,0)$ and $V_2$ the subspace generated by $(0,1)$. –  JSchlather Feb 14 '13 at 3:39
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You should in fact focus on the opposite aspect: show that (with additional hypothesis on $F$ which my answers shows is necessary) $V \setminus (V_1 \cup V_2)$ is a spanning set; once this is established, you can select a basis from it. –  Marc van Leeuwen Feb 14 '13 at 5:21
    
Does the characteristic of the field $F$ matter here?. What if I assume that $char (F) \neq 0$, does that matter at all here? –  user54755 Feb 14 '13 at 5:22
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This result does not hold in general. If $F=\Bbb F_2$ and $\dim V=2$ (whence $\dim V_i=1$ for $i=1,2$) and $V_1\neq V_2$, then $V\setminus(V_1\cup V_2)$ is a singleton, so you cannot choose a basis from its element.

The example is somewhat marginal, and the result ought to hold for any field with at least three elements, but this counterexample shows one will need to use the fact of having three distinct elements in your fields in a proof of that restricted statement. For $F=\Bbb F_2$ you can get counterexamples in any dimension by taking $V_1,V_2$ to be distinct hyperplanes (subspace of dimension $\dim V-1$); reduction modulo $V_1\cap V_2$ proves this (it reduces the question to the initial counterexample, since a basis must project to a spanning set of the quotient).

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@Mark van Leeuwen: Thanks for your answer.What do you mean by $\mathbb{F}_2$. Field of characteristic 2? –  user54755 Feb 14 '13 at 5:23
    
Yes. ${}{}{}{}$ –  Gerry Myerson Feb 14 '13 at 5:33
    
@user54755: In fact, it denotes the unique field with $2$ elements, which has characterisitic $2$, but is not the only field of characteristic $2$. For other fields of characteristic $2$, even finite ones, the result should be OK. –  Marc van Leeuwen Feb 14 '13 at 5:49
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