# Proof-Check: Linear Algebra and Linear Independence

My proof is a solution to part a.) of a question from Friedberg, previously answered here on this site.

The question is:
"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."

My solution:
"First, suppose $\{u, v\}$ is linearly independent. It follows that $u$ and $v$ are not multiples of one another and that $0\notin \{u, v\}$. In other words, $u\neq 0$ and $v\neq 0$. Given these properties, and noting $u\neq v$, we see that $$u+v\neq 0 \text{ and } u-v\neq 0$$Hence,$$a(u+v)+b(u-v)=0 \text{,   s.t. a,b\in \mathbb F}$$ only when $a=b=0$. Thus the linearly independent set $\{u, v\}$ implies $\{u+v, u-v\}$ is linearly independent. Next, suppose $\{u+v, u-v\}$ is linearly independent. It follows that $$u+v\neq 0 \text{ and } u-v\neq 0$$ $$u\neq -v \text{              } u\neq v$$ Hence $au+bv=0$ only when $a=b=0$. Thus, the linearly independent set $\{u+v, u-v\}$ implies $\{u, v\}$ is linearly independent. Therefore $\{u, v\}$ is linearly indepdendent if and only if $\{u+v, u-v\}$ is linearly independent.

Unlike the other solutions, I feel I've answered mine correctly using little more than the properties of linearly dependent/ independent sets and vectors, as opposed to the tricky algebraic manipulation the others used. Aside from its correctness, I'm mainly wondering if the properties I used were enough to write a correct solution (although even if correct, I'm sure my proof could be polished). If there's anybody familiar with this type of stuff that would be willing to check it at their leisure, I'd be very grateful.

## 1 Answer

There are two jumps in your proof that require an explanation. That's when you go from “$u+v\neq0$ and $u-v\neq0$” to the non-existence of scalars $a$ and $b$ such that $a(u+v)+b(u-v)=0$ (unless $a=b=0$, of course). And also (but similar to the previous one) when you jump from $u\neq v$ and $u\neq-v$ to the non-existence of scalars $a$ and $b$ such that $au+bv=0$ (again, unless $a=b=0$).

• Thank you for taking your time and for the helpful feedback, José. In these instances, do you feel it might have been enough to say something like "the linear combination a(u+v)+b(u-v=0 only when a=b=0"? Again, any help and time spent is greatly appreciated – greycatbird May 30 '18 at 16:54
• @greycatbird No. You have to justify that assertion. – José Carlos Santos May 30 '18 at 16:55