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I've tried putting it up as:

$$ [u_1 v_1 + \ldots + u_n v_n] = [u_1 w_1 + \ldots + u_n w_n] $$

But this doesn't make it immediately clear...I can't simply divide by $u_1 + \ldots + u_n$ as these ($u$, $v$ and $w$) are vectors...

Any hints?

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    $\begingroup$ There must be more to this. Perhaps it's supposed to say that if, for every vector $u$ one has $u\cdot v = u\cdot w$, then $v=w$? As it is, the statement is simply false. $\endgroup$
    – MJD
    Commented May 14, 2014 at 15:31
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    $\begingroup$ (As an example of what MJD's talking about) $(1,0,0)\cdot(0,1,0)=0$ and $(1,0,0)\cdot(0,0,1)=0$ but but $(0,1,0)\neq(0,0,1)$. Is the question you really want to ask the on MJD mentions? $\endgroup$
    – rschwieb
    Commented May 14, 2014 at 15:31
  • $\begingroup$ What if $u=0$?! $\endgroup$ Commented May 14, 2014 at 15:36
  • $\begingroup$ It just says: Suppose we know that $u v = u w$, does it follow that $v = w$? So I guess I could say: ''No, if u=0, then this does not need to hold''? $\endgroup$ Commented May 14, 2014 at 16:04
  • $\begingroup$ . . . or you could give examples where $u\ne0$. $\endgroup$ Commented May 14, 2014 at 16:54

4 Answers 4

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If $u\cdot v=u\cdot w$ for all $u$ (equivalently $u\cdot(v-w)=0$), then with $u=v-w$, we get $\|v-w\|^2=(v-w)\cdot(v-w)=0$. Hence $v=w$.

P.S.: Of course, if $v$ are $w$ assumed to be vectors from some inner-product space $S$ with a basis $s_1,\ldots,s_k$, then "for all $u$" can be replaced by "for $u=s_i$, $i=1,\ldots,k$".

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  • $\begingroup$ Hah, that's even better :) $\endgroup$
    – rschwieb
    Commented May 14, 2014 at 15:37
  • $\begingroup$ I'd never seen this one before; gorgeous! $\endgroup$ Commented May 14, 2014 at 15:41
  • $\begingroup$ But what does $u=s_i=v-w$ mean then? $\endgroup$
    – mavavilj
    Commented Feb 26, 2018 at 7:43
  • $\begingroup$ More can be said than this answer or @rschwieb 's equivalent one: the linear functional $u\mapsto u\cdot(v-w)$ has the same norm as the vector $v-w.$ $\endgroup$ Commented Mar 31, 2023 at 10:30
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Presumably you want to add "for all $u$" to that question.

Rearranging, you get $u\cdot (v-w)=0$.

If $v-w\neq 0$, can you see how to pick a $u$ so that $u\cdot(v-w)\neq 0$? A very simple choice of $u$ would work.

By contrapositive, you will have proved that if $u\cdot (v-w)=0$ for all $u$, then $v=w$.

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$$ u\cdot v=u\cdot w $$

Others have shown how to show that $v=w$ if one assumes the above for all values of $u$.

To show that it's now true if one just assumes $u$, $v$, $w$ are some vectors, let's look at the circumstances in which it would fail. Recall that $u\cdot v = \|u\| \|v\|\cos\theta$ where $\theta$ is the angle between the vectors $u$ and $v$.

Thus one circumstance in which the conclusion does not hold is when $v$ and $w$ are of equal lengths, i.e. $\|v\|=\|w\|$, and both are at the same angle with $u$. Just draw a picture. One can rotate $v$ about an axis in which the vector $u$ lies and get many vectors $w$ having the same length as $v$ and making the same angles with $u$.

Another circumstance in which it fails is this: picture $u$ and $v$ as arrow pointing out from the origin, and draw a plane or hyperplane at right angles to $u$ passing through the endpoint of the arrowhead of $v$. Choose an arbitrary point in that hyperplane, and draw an arrow from the origin to that point. Call that vector $w$. Then show that $u\cdot v=u\cdot w$.

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Can I not choose $u=(1,0,0)$, $u=(0,1,0)$ and $u=(0,0,1)$. When I plug them into $u\cdot v=u\cdot w$, I get three equations: $v_1=w_1$, $v_2=w_2$ and $v_3=w_3$, so $v$ must be equal to $w$.

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