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I am an independent student trying to work out a proof. I have followed steps to get to a case where I have $c \vec v \cdot \vec y > d\vec v \cdot \vec x$ -- where c and d are scalars, and v,y and x are vectors. In simple algebra I would factor out the v's -- but I am still learning linear algebra. Is there an analogous operation? It seems like vectors cannot be divided like scalars.

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    $\begingroup$ Just to be clear; $c$ and $d$ are scalars, $\bf v$, $\bf x$ and $\bf y$ are vectors and $>$ is an inequality sign? It's not true that you can "cancel out" the $v$'s, but what you can do is say $c {\bf v} \cdot {\bf y} > d {\bf v} \cdot {\bf x} \iff {\bf v} \cdot (c{\bf y} - d{\bf x}) > 0$. Try to work with the bilinearity of the dot product. $\endgroup$ Jun 17, 2015 at 14:44
  • $\begingroup$ What ultimately would you want to prove if you could make some statement about $y$ relative to $x$? $\endgroup$
    – Muphrid
    Jun 17, 2015 at 14:46

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You can't in general factor out a vector. Remember that scalar multiplication and the dot product are completely different operations and you you can't treat them as if they are the same.

But in the above you can use that $$ c(\vec{v}\cdot\vec{y}) = (c\vec{v})\cdot \vec{y} = \vec{v}\cdot(c\vec{y}). $$ That is, you can move the scalar around.

There are other rules, as $$ \vec{v}\cdot \vec{y} + \vec{v}\cdot\vec{z} = \vec{v}\cdot (\vec{y} + \vec{z}). $$ Here you are factoring out the vector $\vec{v}$.

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  • $\begingroup$ Thanks. What about if dV * (X - Z) -- can I distribute the dot product to get dV * X - dV * Z ? $\endgroup$
    – bernie2436
    Jun 17, 2015 at 15:09
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    $\begingroup$ @bernie2436: Yes, if $V$, $X$ and $Z$ are vectors and $d$ is a scalar, you can do this. $\endgroup$
    – Thomas
    Jun 17, 2015 at 15:17
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    $\begingroup$ thanks for your help. I am learning this stuff on my own so feedback like this is really really useful $\endgroup$
    – bernie2436
    Jun 17, 2015 at 16:38

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