Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say $V$ is a vector space with some basis $\mathcal{B}=\{b_i\}_{i\in I}$, and let $\{B_1,\dots,B_n\}$ be a partition of $\mathcal{B}$. In general, for $S$ a subspace it is not true that $$ S=\bigoplus_{i=1}^n (S\cap\langle B_i\rangle). $$ For example, take $V=\mathbb{R}^2$ as a real vector space, and $S=\langle(1,1)\rangle$, with $B_1=\{e_1\}$ and $B_2=\{e_2\}$. Then $S\cap\langle B_i\rangle=\{0\}$ for $i=1,2$, so no equality can hold.

If we add the condition that $S\cap\langle B_i\rangle\neq\{0\}$ for all $i$, can the above equality be shown to hold? Or is it still impossible? It is clear that the $\supseteq$ containment holds in all cases.

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

Take $V=K^4$ with basis $\mathcal{B}=\{e_1,e_2,e_3,e_4\}$. Consider the partition $\{\{e_1,e_2\},\{e_3,e_4\}\}$ of $\cal B$. Then we set $S=\langle e_1+e_3,e_2,e_4\rangle$. It is clear that $S\cap\langle e_1,e_2\rangle=\langle e_2\rangle$ and $S\cap\langle e_3,e_4\rangle=\langle e_4\rangle$ but $S$ is a three-dimensional subspace of $V$ whereas $\langle e_2\rangle\oplus\langle e_4\rangle$ is only a two-dimensional subspace.

share|improve this answer
add comment

The poblem with your purported equation is that the direct sum on the right hand side is in general smaller than $S$. Requiring that $S\cap\langle B_i\rangle\neq\{0\}$ is not going to help, since you can just add some external new vectors, and allot each of them to $S$ and to one of the parts $B_i$ that previously had a zero intersection with $S$; this will ensure $S\cap\langle B_i\rangle\neq\{0\}$ without doing anything to solve your problem. In your example both $S\cap B_1$ and $S\cap B_2$ are zero, so you can add two new basis vectors $e_3,e_4$, and in $W=\Bbb R^4$ consider $S=\langle (1,1,0,0),e_3,e_4\rangle$ with $B_1=\{e_1,e_3\}$ and $B_2=\{e_2,e_4\}$; now $S\cap B_1$ and $S\cap B_2$ are nonzero, but both being singletons they span a subspace $\langle e_3,e_4\rangle$ of dimension $2$, strictly contained in $S$. You see that the extension did nothing to remedy the situation restricted to (or if you want, projected onto) the subspace $\langle e_1,e_2\rangle$.

In general if you have a false statement with a counterexample, and you think of correcting it with an extra condition that is not satisfied in the example, then you should think first whether the condition has sufficient "muscle" to do this, in other words whether you can imagine this condition to be a crucial ingredient of a proof of the statement. One way to show that it does not have enough muscle is to show that you can adopt the counterexample easily to satisfy your condition without really changing the essence. And you could also guess that conditions $S\cap\langle B_i\rangle\neq\{0\}$ are not very strong for a proof; they will allow you to choose one nonzero vector in the intersection, but that's about it, and this does not seem the kind of thing that will lead to a general statement like the one you want to prove. In fact your statement is rather hopeless; the equality holds (if and) only if $S$ it actually spanned by a subset of $\mathcal B$ (just compare dimensions), which is a very exceptional situation.

share|improve this answer
    
Thanks for the more detailed criticism. –  Noomi Holloway Nov 26 '12 at 7:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.