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Question taken from the book Analytical Geometry, Paulo Boulos. No vector space was specified, but $\overrightarrow{u}$ is a vector itself, the question statement is written exactly like this. In that book it is said that the geometric segments that ensure the algebraic proofs of analytic geometry. So I always wonder if this kind of question is proved by geometric or algebraic arguments.

Prove that $\overrightarrow{u}=-\overrightarrow{u} \iff \overrightarrow{u}=\overrightarrow{0}$

user113581321
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  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Shaun Apr 08 '22 at 14:56
  • Well, you ought to be able to prove one of the two implications, at least. – lulu Apr 08 '22 at 14:56
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    Note that the question (or at least one of the two implications) can't be answered as stated. More information is needed. What is $\vec u$? Presumably it is a vector in some vector space. But over what field? – lulu Apr 08 '22 at 14:57
  • This question was taken from the book Analytical Geometry, Paulo Boulos. The question is written exactly like this, he didn't specify any vector space, but rather $\overrightarrow{u}$ is a vector. – user113581321 Apr 08 '22 at 14:59
  • Then use the different axioms of a vector space – Lelouch Apr 08 '22 at 15:03
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    @lulu you mean like in $\mathbb{Z}/2\mathbb{Z}$ is a field, and in it $1+1 = 0$, but $1\neq 0\ ?$ – Adam Rubinson Apr 08 '22 at 15:16
  • See What's so special about characteristic 2? – JMoravitz Apr 08 '22 at 15:23
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    While I have not read the specific book and don't have access to it, I wouldn't be surprised if it specifies somewhere near the start or where it begins introducing vector spaces that it may have specified that it will only ever consider vector spaces over the scalar field $\Bbb R$ or similar, thus making any time that "vector space" is mentioned it really means "real vector space." – JMoravitz Apr 08 '22 at 15:26
  • @AdamRubinson Exactly. In characteristic $2$, it is always true that $\vec u=-\vec u$. – lulu Apr 08 '22 at 16:57

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Given the subject of Analytic Geometry and the discussion of vectors as geometric objects, I doubt the books starts off with introducing the concept of vector spaces in abstract. It most likely is concerned only with a very specific $3$-dimensional vector space over the real numbers.

Unfortunately, those who have lived for years in the canopy of the tree of mathematics exploring its myriad twisted pathways and the bright green foliage all around tend to forget that from the bottom, its a big brown trunk with a few large branches barely within reach.

The result can be shown in two ways at this level.

Algebraically, $-\vec 0 = \vec 0$ by definition, and adding $\vec u$ to both sides of $\vec u = -\vec u$ gives $$\begin{align}\vec u + \vec u&= -\vec u + \vec u\\2\vec u &= \vec 0\\\vec u &= \frac 12 \vec 0\\\vec u &= \vec 0\end{align}$$

Geometrically, if $\vec u \ne \vec 0$, then $\vec u$ is a line segment with $\vec 0$ as one endpoint, and by definition $-\vec u$ is the line segment extending from $\vec 0$ in the other direction on the same line and having the same length. Because these two line segments extend in opposite directions, they are not the same. So if $\vec u \ne 0$, then $\vec u \ne -\vec u$. But also by definition $-\vec 0 = \vec 0$.

At least that is one way of seeing it geometrically. Not everybody defines the geometric vector space in exactly the same way, so your definitions may be a bit different, in which case, my description is not strictly by your definition, but instead is a fairly easy conclusion from your definition. But without knowing your definition, I cannot say more.

Paul Sinclair
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