# Is there a counterexample to the claim that if $\mathbf y\cdot\mathbf y=1$ and $\mathbf x\cdot\mathbf y=c$ then $\mathbf x=c\mathbf y$?

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?

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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.

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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 $\mathbf{y}$

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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

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

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