Context & Problem:

I am working through proving some of the basic properties of vector spaces, and came across a question that asks us to prove that $$(-k)\vec{v} = -(k\vec{v}) = k(-\vec{v}) \color{gray}{ \; = (-k\vec{v})},$$

where the last equality, $\color{gray}{\textrm{in grey}}$, is one that I added, as it seemed important / interesting.

Here $k \in \mathbb{R}$ and $\vec{v} \; \in \; \mathbb{S}$, where $\mathbb{S}$ is a vector space.

The Issue:

My issue comes down to brackets.

Am I correct in assuming that the expression $-(k\vec{v})$ should be interpreted as a scalar multiple of the element $(k\vec{v}$), $\; i.e.$

$$-(k\vec{v}) = -1\odot(k\vec{v}),$$

where $\odot$ represents scalar-multiplication in the vector space, and $(k\vec{v}) \, \in \, \mathbb{S}$?

Or would this be understood as the additive inverse of the element $(k\vec{v}) \in \mathbb{S}$? $\; i.e.$:

$$-(k\vec{v}) + (k\vec{v}) := \vec{0}.$$

Personally, I would expect the additive inverse of $(k\vec{v})$ to be denoted $(-k\vec{v})$. I'm not sure if it's a question of convention. It seems to me there is clearly a correct interpretation (scalar multiplication); however, the author has been a bit fast-and-loose with brackets elsewhere (it is an informal resource I'm working from), which has made me doubt the interpretation.

In particular, the omission of the last equality, in grey, above, which seems the most important link in the chain, has made me wonder if the author intended the "additive inverse" interpretation.

My Question:

Does the use of brackets when discussing a vector space follow a (mostly) universal convention? If so, does my interpretation of $-(k\vec{v})$ as a scalar multiple of an element align with this convention?

I appreciate that much of notation is convention, and that one can, in theory, establish any convention, so long as it is clearly defined; I am concerned here with what would be a "normal" interpretation.

  • 1
    $\begingroup$ Personally, I would take $-(k\vec{v})$ to be the additive inverse of $k\vec{v}$, and I would write the scalar multiple as $(-1)(k\vec{v})$. Of course, these are equal, so in reality, I never write either of these expressions, but simply write $-k\vec{v}$, and don't worry about which interpretation it has. $\endgroup$
    – saulspatz
    Mar 14 '20 at 17:39
  • $\begingroup$ Basically you have $\vec{v}$ and its inverse $-\vec{v}$. Than you have $k \vec{v}$ and you prove that $(-k) \vec{v}= - (k \vec{v})$. $\endgroup$ Mar 14 '20 at 18:01
  • $\begingroup$ @saulspatz -- thanks for weighing in. Given that the additive inverse was defined $(-\vec{v}) \, : \, \vec{v} + (-\vec{v}) = \vec{0}$, I assumed, that the negative should also be inside the brackets. But I too initially read it in this way. Having reflected on the above definition, it now seems to me it should be written $\big(-(kv)\big)$. Lastly, I realize that they are all equivalent, so it eventually doesn't matter, but within the context of proving the results, formally, this distinction seems of fundamental importance (& this is the context I'm working in). Thanks v. much for weighing in! $\endgroup$
    – Rax Adaam
    Mar 14 '20 at 18:21
  • $\begingroup$ @MauroALLEGRANZA In order to prove the equivalence, one needs to know what the meaning of the statement is. If $-(k\vec{v})$ is the scalar multiple, the proof is simply $(-k)\vec{v}=((-1)k)\vec{v}=(-1)(k\vec{v}) := -(k\vec{v})$. By contrast, if $-(k\vec{v})$ is the additive inverse of $(k\vec{v})$, one must show that $(k\vec{v})+(-(k\vec{v}))=\vec{0}$. Although, we can prove that these two are equivalent, until we have shown that (which is the given exercise) the distinction is important. Do you read $-(k\vec{v})$ as the additive inverse or as a scalar multiple, from this perspective? $\endgroup$
    – Rax Adaam
    Mar 14 '20 at 18:27
  • 1
    $\begingroup$ No, thanks. It's fine the way it is. $\endgroup$
    – saulspatz
    Mar 14 '20 at 19:45

A summary of the conclusion drawn from the comments (thanks to @saulspatz)

Given that we write $k \odot \vec{v} = k\vec{v}$, in the case of the scalar $-1$, this would be either $$-1 \odot \vec{v} = -1\vec{v},$$ or, for additional precision / clarity: $$(-1)\odot\vec{v} = (-1)\vec{v}.$$

Although one can prove that $\color{blue}{-1\vec{v}} := \color{green}{(-1)\vec{v}} = -\vec{v}$, using only the properties of a vector space, if scalar multiplication is meant, then either the blue or green expressions would be appropriate.

In summary, "$- \,\square$" would conventionally be understood as the additive inverse of $\square$, whereas "$-1\square$" or "$(-1)\square$" would be used to represent scalar multiplication.


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