Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x,y$ be arbitrary vectors where $\mathbf{y} \cdot \mathbf{y} = 1$ and $c$ be a real valued scalar. If

$\mathbf{x} \cdot \mathbf{y} = c = c (\mathbf{y} \cdot \mathbf{y} ) = (c \mathbf{y} ) \cdot \mathbf{y} $, then

$\mathbf{x} = c \mathbf{y}$

Is this true? Feels like a sloppy conclusion no?

share|cite|improve this question
up vote 3 down vote accepted

Now that we have counterexamples, let's look where the fault in the reasoning is.

When you say $x \cdot y = c y \cdot y$, you can't "eliminate" the common $y$ because that would require invertibility of the dot product. By itself, the dot product is not invertible, and you cannot reconstruct $x$ without additional information: in particular, information from the cross product.

Given a vector $x$, and a unit vector $y$, you can decompose $x$ as

$$x = (x \cdot y) y + y \times (x \times y) = x_\parallel + x_\perp$$

With the information you have, you know that $x_\parallel = cy$, but you don't know $x \times y$, and because of that, you don't know $x_\perp$. If you knew that $x \times y =0$, then $x = x_\parallel$, and you'd know you're done.

share|cite|improve this answer
So in Algebra sense, the vector space with the operation of dot product does not form a group? – Hawk Feb 3 '13 at 19:49
Yes, that's true, in part because the dot product takes two elements of $\mathbb R^n$ and returns an element of $\mathbb R$, concerns of invertibilty aside. – Muphrid Feb 3 '13 at 20:01
Oh so closure failed from the start – Hawk Feb 3 '13 at 20:13

$\mathbf{x} = c \mathbf{y} + $ some vector perpendicular to $\mathbf{y}$

share|cite|improve this answer
So if $\mathbf{x}$ and $\mathbf{y}$ are not orthogonal, then it is okay? – Hawk Feb 3 '13 at 0:27
@sizz: If $\rm x=y+z$ where $\rm z,y$ are orthogonal, then $\rm x,y$ are not orthogonal. – Asaf Karagila Feb 3 '13 at 1:23
Are there circumstances in which my result could be true? Under what conditions? – Hawk Feb 3 '13 at 1:46
@sizz: Your result is true under the condition that $\mathbf x$ is a multiple of $\mathbf y$. It really doesn't get any simpler than that. – Rahul Feb 3 '13 at 2:42
Does it have to be a linear (constant) multiple? – Hawk Feb 3 '13 at 23:27

$x=(17,1)$, $y=(0,1)$. ${}{}{}{}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.