# Homework Help: Linear Dependence and Subsets

1. Jan 14, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

Suppose that E,F are sets of vectors in V with $E \subseteq F$. Prove that if E is linearly dependent, then so is F.

3. The attempt at a solutionRead post #2. This proof, I think, was incorrect.

If we suppose that E is linearly dependent, then we know that there exists $E_1,...,E_n$ distinct vectors such that,
$e_1 E_1 + ... + e_n E = \vec0$, where e1,...,en are numbers. Thus, we know $\vec0 \in E$. Since $E \subseteq F$, $\vec0 \in F$. Any set containing the zero-vector must certainly be linearly dependent since $n \vec0 = \vec0$, where n is any number.

Thus, if E is linearly dependent, then F is linearly dependent.

Last edited: Jan 14, 2012
2. Jan 14, 2012

### TranscendArcu

Actually, I have to revise my proof. I don't think I can write that $\vec0 \in E$. In any case, we know that $e_1 E_1 + ...+ e_n E_n = \vec0$. We can therefore say (I think) that $\vec0 \in span{E}$. $E \subseteq F$ implies $spanE \subseteq spanF$. Thus, $\vec0 \in spanF$. Since E has a nontrivial solution to $e_1 E_1 + ...+ e_n E_n = \vec0$ and since $\vec0 \in spanF$, F must also have a nontrivial solution to get the zero vector. In one such case, F's nontrivial solution is identical to E's.

Is that better?

3. Jan 14, 2012

### Dick

Nope. The zero vector is in ANY span. F is a set of vectors. E is a subset of those vectors. Let's list the elements of F as $F=\{F_1,F_2,...,F_n\}$ and let's say that the first m of those vectors are the ones that also belong to E, so $E=\{F_1,F_2,...,F_m\}$ where m<=n. Try expressing a proof using that notation.

4. Jan 14, 2012

### TranscendArcu

So if I write $F=\{F_1,F_2,...,F_n\}$ and $E=\{F_1,F_2,...,F_m\}$ then in E there exists $F_1,...,F_m$ distinct vectors such that $f_b F_b + ... + f_m F_m = \vec0$, where $f_1,...,f_m$ are not all zero vectors. Because $E \subseteq F$, F also contains these distinct vectors. Thus, in F we can say that there exist $F_1,...,F_m$ distinct vectors such that $f_b F_b + ... + f_m F_m = \vec0$. Hence, we see that F is also linearly dependent.

5. Jan 14, 2012

### Dick

The phrasing is still clumsy. Why are you starting with $f_b$? Why not start with $f_1$? And the f's aren't vectors, they are scalars. You've got the right idea though.

6. Jan 14, 2012

### TranscendArcu

I guess it doesn't matter to start with F1 or b. I thought Fb gave the impression that E didn't necessarily have to start with the first element of F. But I seem to recall that ordering doesn't matter in sets anyway (isn't that right?)

Besides these problems (which I think will be pretty trivial to fix) is there anything else that is making the proof clumsy?

7. Jan 14, 2012

### Dick

In setting it up we designated that the first m elements of F were to be the elements of F n E. Sure, sets are unordered, but our labeling of the elements of the sets has some useful order. You kept the last element of E as F_m after all. Don't ponder this deeply, just change the 'b' to '1' and see if it reads more simply.

8. Jan 15, 2012

### TranscendArcu

Okay. Thanks!

I won't rewrite the work here since there isn't much need if the changes you recommend are so easy to make.

On the other hand, I also want to ask another question regarding sets. If I have a statement like $spanE + spanF$, is this just the same as $spanE \cap spanF$? I ask because I was doing some practice problems and one of them asked to show that $L(E \cap F) \subseteq L(E) \cap L(F)$. The answer to this problem showed that $L(E \cap F) \subseteq spanE + spanF$, but stopped there. It isn't immediately obvious to me how the stuff on the right hand side of the subset sign relates to what I was asked to show.

9. Jan 15, 2012

### Dick

No. $spanE + spanF$, is defined as the set of all vectors e+f where e is in span(E) and f is in span(F). Pick a basis for $spanE \cap spanF$ and extend it to a basis for span(E) and span(F) to see the relation.