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?
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?
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 $u=s_i$, $i=1,\ldots,k$".
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 $v=w$.
$$ u\cdot v=u\cdot w $$
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 $u\cdot v=u\cdot w$.
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 $w$.