Let $u$ and $v$ be vectors where $u\neq v$ in the equation: $$2u\cdot(u-v)=u\cdot(u-v)$$

I am trying to find a reason why you cant just cancel out the $(u-v)$ from both sides. My reasoning is that dividing either side by $(u-v)$ would be dividing a scalar by a vector. I'm not sure if this is the reason why or if it has to do with another rule of the dot product.

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    $\begingroup$ For example, $(1,2)\cdot (1,1) = (2,1)\cdot (1,1)$ does not imply $(1,2) = (2,1)$. $\endgroup$
    – AlvinL
    Oct 9, 2023 at 6:55
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    $\begingroup$ You can conclude $u\cdot (u-v) = 0$, which implies $|u|^2 = u\cdot v$ but that's about it. $\endgroup$
    – AlvinL
    Oct 9, 2023 at 6:57
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    $\begingroup$ Notice I edited your question to improve the formatting. It is strongly advised that you use Mathjax to format your questions on this site - it's like LaTeX for the web. I edited your question this time since you are new, but in future, please format the question yourself. See here for a quick guide: MathJax tutorial $\endgroup$
    – 5xum
    Oct 9, 2023 at 7:41
  • $\begingroup$ I don't get the question: lets call $z=u⋅(u−v)$ , then your equation says $2z=z$ which implies $2=1$ after simplification which is false, the only valid answer is $z=0$ , then $u⋅(u−v)=0$. Also, about dividing a scalar by a vector, you don't divide by vectors in matrix algebra, but instead multiply from the required one-side inverse (a left/right matrix). $\endgroup$
    – Joako
    Oct 9, 2023 at 8:22
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    $\begingroup$ Frame challenge: in math, it's not the case that we need to find reasons why we can't do operations. We need to find reasons why we are allowed to do operations, or else we can't do them. (Also, being able to articulate why we are allowed to perform operations, even familiar and uncontroversial ones, is an important part of math mastery.) $\endgroup$ Oct 9, 2023 at 17:33

1 Answer 1


You can't just cancel out $(u-v)$ for the simple reason that

1. There is no rule which would say that if $a\cdot c= b\cdot c$, then $a=b$.

This rule exists in other cases, for example for the real numbers, where it is indeed true that, if $c\neq 0$ and $ac=bc$, then $a=b$. However, there is no direct connection between the rule in $\mathbb R$ and the one in vector spaces - only the coincidental fact that they look somewhat similar. Other than that, the two "rules" are quite different.

  1. One rule applies to a field, the other "applies" to a vector space.
  2. In one rule, both $a,b,c$ and their products are elements of the same set. In the other $a,b,c$ are vectors, but $a\cdot c$ and $b\cdot c$ are scalars.
  3. One can be proven by multiplying both sides of the equation by $c^{-1}$. We know that $c^{-1}$ exists, because we are in a field and $c\neq 0$. In the other rule, the very expression $c^{-1}$ is undefined, so the same proof cannot be applied.

2. There cannot be such a rule, because we have counterexamples

We can take the vector space $V=\mathbb R^2$ and the vectors $a=(1,-1), b=(1,1), c=(1,0)$ and see that $a\cdot c = 1=b\cdot c$, however $a\neq b$, which proves that the rule does not apply in this case.

  • $\begingroup$ Note that you can cancel $ac=bc$ if $a$ and $b$ are vectors and $c$ is a non-zero scalar (real or maybe complex). So this is not really about real numbers versus vector spaces but rather about multiplication (where you can cancel) and dot products (where you cannot). It is somewhat unfortunate that the notation between the operations is very similar. $\endgroup$
    – quarague
    Oct 10, 2023 at 11:01
  • $\begingroup$ @quarague My focus was not meant to be on the reals, but on the rule itself. Real numbers were just an example of where the rule does apply, as opposed to vectors where it does not. $\endgroup$
    – 5xum
    Oct 10, 2023 at 11:26
  • $\begingroup$ My point was that the rule is not about whether you apply it to reals or vectors. The key issue is about the operation, there is a cancellation rule for the operation of multiplication and there isn't a cancellation rule for the dot product. $\endgroup$
    – quarague
    Oct 10, 2023 at 11:33
  • $\begingroup$ @quarague And my point is that the reals are just an example of where the rule applies. Not as the only example, but as one example. I added a sentence to make that more clear. $\endgroup$
    – 5xum
    Oct 10, 2023 at 11:40

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