Matrix/vector proof

Question

Let $C$ be an $m\times n$ matrix with real entries. For vectors $u,v\in\mathbb R^m$, we define $u\sim v$ if there exists a vector $x\in\mathbb R^n$ such that $Cx = u−v$. Prove that for all $u, v,w \in\mathbb R^m$:

i) $u \sim u$;

ii) if $u \sim v$ then $v \sim u$;

iii) if $u \sim v$ and $v \sim w$ then $u \sim w$.

Question from homework, but we haven't seen this in class, so I have no clue how to go about it and start.

Do I make up a matrix and prove it that way? Is that proof of this?

Answer 1

Let's just focus on (i) for now. You have to prove the claim for every matrix $C$, so making up one particular one is not the way to go. Let's first write down the definition we need. By definition, we have

$$ u\sim v\iff \exists x\in\mathbb R^n: Cx=u-v$$

write down the claim you need to prove:

For all $u\in\mathbb R^m$, the claim $u\sim u$ is true.

Clearly, as all "for all" claim proofs start, you must begin by saying:

Let $u\in\mathbb R^m$.

Now, you must prove that $u\sim u$. To do that:

1. write down what it means, by definition, that $u\sim u$.
2. using that definition, first think about what you need to do to prove the definition
3. only once you completed steps one and two: prove the definition.

In a comment (or as an edit to your quesiton), tell us: which of the steps above is giving you problems? We can help you further once we know where the problem is.

Answer 2

If it helps, you can start with a particular matrix, however in order to prove the statement you need to make it in general finding excplicty the vector $x$.

For the first point if you pick $x=0$ then you obtain $Cx = 0 = u-u$ which implies that condition i) is true.

For the second point you start from the condition $u\sim v$ which means that there exists $x$ such that $Cx=u-v$, if you consdier $-x$ you get what you need.

For the third point still your hypothesy gives you that there exist $x$ and $y$ such that $Cx=u-v$ and $Cy=v-w$, then $C(x+y) = u-v+v-w = u-w$.