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Suppose that $V$ is a vector space over $\mathbb R$ (for simplicity) with addition denoted by $\oplus$ and scalar multiplication denoted by $\otimes$. Let $\mathbf u, \mathbf w \in V$ and let $\lambda \in \mathbb R$ and suppose we are asked to compute $$ \lambda \otimes \mathbf u \oplus \mathbf v $$ I was wondering how to do this. I would tend to say it is $$ (\lambda \otimes \mathbf u) \oplus \mathbf w $$ but it might as well be $$ \lambda \otimes (\mathbf u \oplus \mathbf w) $$ since, as far as I understand, none of the axioms for a vector space settles the precedence of scalar multiplication over vector addition. Is there a general rule for this ?

EDIT: I "know" that the cultural rule is precedence of scalar multiplication over vector addition. This is hinted at from shortcuts like $$ \lambda(\mathbf u + \mathbf v) = \lambda \mathbf u + \lambda \mathbf v $$ which (I think) should be written as $$ \lambda(\mathbf u+ \mathbf v) = (\lambda \mathbf u) + (\lambda \mathbf v) \,. $$ unless everyone agrees beforehand that scalar multiplication takes precedence. The thing I am wondering about is whether a computer (say) would be able to deduce this precedence from the axioms alone.

Thanks a lot for your help!

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    $\begingroup$ $\lambda \otimes u\oplus v$ is not well-defined. In order to avoid such dilemmas, you either write $\lambda\otimes (u\oplus w)$ or $(\lambda\otimes u)\oplus w$. $\endgroup$ – Levent Jun 18 '16 at 21:05
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    $\begingroup$ @Levent That has to be wrong. Otherwise why would the distributive axiom be written the way it always is. We always write $\lambda(u+v)=\lambda u+\lambda v$. Are you claiming the RHS is not well-defined? $\endgroup$ – almagest Jun 18 '16 at 21:10
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    $\begingroup$ No, what I mean is $\lambda u+v$ is not well-defined, not $\lambda(u+v)=\lambda u+\lambda v$. But commonly we just use $\lambda u+v$ to denote $(\lambda u)+v$. $\endgroup$ – Levent Jun 18 '16 at 21:11
  • $\begingroup$ @Levent. You miss the point. If $\lambda u+v$ was not well-defined, then $\lambda u+\lambda v$ would not be well-defined, because it might mean $(\lambda u)+(\lambda^2v)$. But $\lambda u+\lambda v$ obviously is well-defined because it appears in the axioms. So $\lambda u+v$ is also well-defined - it means $(\lambda u)+v$. $\endgroup$ – almagest Jun 18 '16 at 21:19
  • $\begingroup$ @almagest your argument is invalid in $F_2$. $\endgroup$ – Daniel Valenzuela Jun 18 '16 at 21:20
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This is just a convention. And the usual convention is to continue the shortcuts used for operations among numbers (and be it in order not to confuse the reader when we already reuse the notation from numeric operations for vectors). In other words, the convention is that $\lambda\otimes u\oplus v$ is a shorthand writing for $(\lambda\otimes u)\oplus v$. Note that it is in fact common to (re-)use the multiplication and addition symbols from numeric operations and in fact write $\lambda\cdot u+v$ or just $\lambda u+v$.

In case of a vector space with a scalar product, you may even see $\lambda \cdot u \cdot v$ (or $\lambda u v$) with the dot standing for two completely different operations. Fortunately, it is true that $(\lambda\cdot u)\cdot v=\lambda\cdot(u\cdot v)$ (an equation that even has three distinct operations denoted by the same dot symbol!).

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