I'm attempting to write a rather trivial (for most math people) proof for practice in my linear algebra class and I would greatly appreciate feedback on the accuracy of my proof as well as suggestions to improve it's overall argument.

My proof writing skills are fairly rough but I'm doing my best to improve:

Prove that $\overrightarrow u$ $\cdot$ c$\overrightarrow v$ = c($\overrightarrow u$ $\cdot$ $\overrightarrow v$) for all vectors $\overrightarrow u$ and $\overrightarrow v$ in $\mathbb R^n$ multiplied by any scalars c.


Let vectors $\overrightarrow v$ and $\overrightarrow u$ be any arbitrary vectors in $\mathbb R^n$ and c be any scalar where c $\in$ $\mathbb R$.

Want to show that $\overrightarrow u$ $\cdot$ c$\overrightarrow v$ = c($\overrightarrow u \cdot \overrightarrow v$) for all vectors $\in$ $\mathbb R^n$ multiplied by any scalar c.

By the definition of the dot product and scalar multiplication, the left hand side yields:

$\overrightarrow u \cdot c\overrightarrow v = \overrightarrow u_1 c \overrightarrow v_1 + \overrightarrow u_2 c \overrightarrow v_2 + ... \overrightarrow u_n c \overrightarrow v_n$

$= c(\overrightarrow u_1 \overrightarrow v_1 + \overrightarrow u_2 \overrightarrow v_2 +... \overrightarrow u_n \overrightarrow v_n)$

$= c \overrightarrow u_1 \overrightarrow v_1 + c \overrightarrow u_2 \overrightarrow v_2 + ... c \overrightarrow u_n \overrightarrow v_n$

It follows, using the same properties, that the right hand side yields:

$c(\overrightarrow u \cdot \overrightarrow v) = c(\overrightarrow u_1 \overrightarrow v_1 + \overrightarrow u_2 \overrightarrow v_2 + ... \overrightarrow u_n \overrightarrow v_n)$

$= c \overrightarrow u_1 \overrightarrow v_1 + c \overrightarrow u_2 \overrightarrow v_2 + ... c \overrightarrow u_n \overrightarrow v_n$

It is enough to show that the left hand side equals the right hand side

$\therefore \overrightarrow u \cdot c \overrightarrow v = c(\overrightarrow u \cdot \overrightarrow v)$ for all vectors in $\mathbb R^n$ and any scalar c. $\tag*{$\blacksquare$}$

I feel like this proof is incomplete, so if someone could review it and give me some advice I would sincerely appreciate it! Thanks.

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    $\begingroup$ A (somewhat) minor comment on notation: It's pretty common to write $\overrightarrow u = (u_1, u_2, \ldots, u_n)$, with the components $u_i$ lacking the $\overrightarrow{}$ hat. The purpose of the hat (I've always assumed) is to distinguish between vectors and scalars, which is lost writing $\overrightarrow u = (\overrightarrow{u_1}, \ldots, \overrightarrow{u_n})$. $\endgroup$ – pjs36 Jun 13 '17 at 4:21
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    $\begingroup$ I was actually contemplating that as I typed it out lol. Would have saved me a lot of typing! thanks for the advice for next time :) $\endgroup$ – FuegoJohnson Jun 13 '17 at 4:23
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    $\begingroup$ You shouldn't have arrows on $u_1, v_1, \dots$. Those are scalars, not vectors. Also, you should say what you mean by these numbers through a definition like "Let $\overrightarrow{u} = (u_1, \dots, u_n)$." $\endgroup$ – user49640 Jun 13 '17 at 4:23

It's fine - just remember to remove the arrows on top once your start using subscripts, since the components of a vector aren't vectors (they're simple scalars). Here's an approach that directly goes from the LHS to the RHS:

Let $\vec u = (u_1, \ldots, u_n) \in \mathbb R^n$ and $\vec v = (v_1, \ldots, v_n) \in \mathbb R^n$ and $c \in \mathbb R$ be arbitrary. Then observe that:

\begin{align*} \vec u \cdot c \vec v &= (u_1, \ldots, u_n) \cdot (cv_1, \ldots, cv_n) \\ &= u_1cv_1 + \cdots + u_ncv_n \\ &= c(u_1v_1 + \cdots + u_nv_n) \\ &= c(\vec u \cdot \vec v) \end{align*}

as desired. $~~\blacksquare$

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  • $\begingroup$ Much simpler, and you are absolutely correct about the vector components; I suppose I was so caught up in the semantics of the proof I failed to recognize that. Thanks for verifying that my base logic was correct. Sure will save me a lot of typing in the future! $\endgroup$ – FuegoJohnson Jun 13 '17 at 4:27

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