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I'm always having difficulties with what actually suffices as a proof, and what is obvious enough to not have to prove it. Here some I have those problems with.

Let $S=\{u_1,u_2,...,u_n\}$ be a finite set of vectors. Prove that $S$ is linearly dependent if and only if $u_1=0$ or $u_{k+1} \in <\{u_1,u_2,...,u_k\}>$ for some $k (1 \leq k \lt n)$.

I think it's obvious since 1. ${0}$ is linearly dependent and 2. if $u_{k+1}$ is in the mentioned span, than it is a linear combination of other vectors of $S$ and thus $S$ is linearly dependent. So much to my thinking, but how do I appropriately express something like this?

Let $M$ be a square upper triangular Matrix with nonzero diagonal entries. Prove that the columns of $M$ are linearly independent.

I think: If you regard every column as a vector, each has a different direction. But again, I guess this doesn't really count as a mathematical proof. How to express it then?

Let $V$ be a vector space over a field of characteristic not equal to two. a) Let $u$ and $v$ be distinct vectors in V. Prove that $\{u,v\}$ is linearly independent if and only if $\{u+v,u-v\}$ is linearly independent. ( b) - the same with $\{u,v,w\}$ and $\{u+v,u+w,v+w\}$.

My proof for a):

$a_1u + a_2v$ implies $a_1=a_2=0$

If $$b_1(u+v) + b_2(u-v) = (b_1+b_2)u+(b_1-b_2)v = 0$$ then $$b_1+b_2 = b_1-b_2 = 0$$ thus $b_1 = b_2=0$

Now do I need to write additional stuff to prove the other direction (something like:

Since $$b_1(u+v) + b_2(u-v) = (b_1+b_2)u+(b_1-b_2)v = 0$$ only if $b_1 = b_2 = 0$ I can choose no $a_1,a_2 \neq 0$ for I could always split it into $b_1$ and $b_2$ which were not zero, there fore the equation couldn't be zero either.),

or would the first part already be enough?

Thanks a lot ! I have a lot to get used to :P

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up vote 1 down vote accepted

In the future, please limit yourself to one problem in your post. You're more likely to get helpful responses than if you list several problems.

For #1, you have given an argument for the "if" direction, but you haven't addressed the "only if" direction. You must show that if $S$ is linearly dependent, then one of the two cases stated holds.

For #2, you should take a linear combination of the columns and argue that each of the coefficients is zero. Your argument about directions is not valid. You can select three vectors in $\mathbb{R}^2$ that all have different directions. Those three vectors will be linearly dependent (why?).

For #3, your definitely need to say something for the other direction. Otherwise, you will have only proven the "only if" direction. For the "if" direction, you assume that $\{u+v, u-v\}$ are linearly independent (i.e. $b_1 (u+v) + b_2(u-v) = 0 \Rightarrow b_1 = b_2 = 0$. Then, to prove that $\{u, v\}$ is linearly independent, you assume that $a_1 u + a_2 v = 0$ and must show that $a_1 = a_2 = 0$. Your proof will be similar to the other direction, using an algebraic calculation to get equations on $a_1$ and $a_2$ whose only solution is $a_1 = a_2 = 0$.

Hope that helps. Keep practicing and asking questions - that's the best way to learn how to do proofs.

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Hey Michael. Thank you for your response. I thought the problem is the kinda the same for all - the proofwriting. That's why I wrote one question. For #2: I still think the argument would be valid, since I'm taking n vectors out of an n-dimensional space (all upper triangular matrices and nonzero diagonal entries $\in M_{n\times n}(F)$) with different directions, but your example where three vectors out a 2-dimensional space ($\mathbb{R}^2$). Thanks, and I'll definately keep asking :P – foaly Sep 29 '12 at 18:45
@foaly: It's better to pick one specific problem about proofwriting and you can always ask the others in other threads. For #2, those three vectors could have all been in a plane in $\mathbb{R}^3$, so you can't conclude that a set of $n$ vectors with different directions in an $n$-dimensional vector space are linearly independent. – Michael Joyce Sep 29 '12 at 19:01
For the if-direction of #3, would this be legit?: $(b_1+b_2)u+(b_1-b_2)v=0$ for $b_1=b_2=0$, if $a_1u+a_2v=0$ $a_1=c_1+c_2$, $a_2=c_1-c_2$ and thus $(c_1+c_2)u+(c_1-c_2)v=0$ hence $c_1=c_2=0$ hence $a_1=a_2=0$ – foaly Sep 29 '12 at 19:04
@foaly: I think the clearest approach is to show that you can write $a_1 u + a_2 v = c_1(u + v) + c_2(u - v)$ (solve for $c_1$ and $c_2$), then use the linear independence of $u+v, u-v$ to conclude $c_1 = c_2 = 0$ and then argue that $a_1 = a_2 = 0$ (based on your calculation relating $a_1, a_2$ to $c_1, c_2$). – Michael Joyce Sep 29 '12 at 20:11

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