Is this proof correct? $|u \cdot v| = |u | |v |$ IFF one vector is a scalar multiple of the other. I am working my way through a linear algebra book and would appreciate some help verifying my proof.
Prove that $|u \cdot v| = |u | |v |$ if and only if one vector is a scalar multiple of the
other.
PROOF:
Let $k ∈ ℝ$ and $u ,v \in\mathbb R^n$ and $~u =k~v$
ASSUME: $|u\cdot v| = |u | |v |$
our assumption holds IFF  $|kv \cdot v| = |kv | |v |$
which again holds IFF $k|v \cdot v| = k|v | |v |$
and, by definition of the dot product, holds IFF $k|v|^2 = k|v |^2$
Q.E.D.
 A: It is not right.
I don't see any attempt to check what happens if $u$ is not a multiple of $v$.
It would be helpful if you can use $u.v = \|u\|\|v\|\cos \theta$.
A: Let's make the proof of CS explicit to show @CSquared's answer doesn't require circularity. In fact, it's simpler to run through the proof of CS rather than invoking it, as we don't need to check two directions separately.
Write $f(k):=u-kv$ so$$0\le|f(k)|^2=f(k)\cdot f(k)=|u|^2+k^2|v|^2-2ku\cdot v,$$with equality iff $f(k)=0$ i.e. $u=kv$. (You can see where this is going: it involves the convention for $k$ used in the OP, which CSquared reverses.) The special case $k:=v\cdot u/|v|^2$ gives $0\le|u|^2-|u\cdot v|^2/|v|^2$, which rearranges to $|u\cdot v|^2\ge|u|^2|v|^2$, again with equality iff $u=kv$. Now just take the square root.
(The above proof actually works even on complex spaces, due to the careful use of $v\cdot u$ at one point instead of $u\cdot v$, and of $|u\cdot v|^2$ instead of $(u\cdot v)^2$.)
A: In your proof, you make the assumption that $\vec{u}=k\vec{v}$ and $|\vec{u}\cdot\vec{v}|=||\vec{u}||\,||\vec{v}||$ and then claim that this holds when $$|\vec{u}\cdot\vec{v}|=||\vec{u}||\,||\vec{v}|| \Longleftrightarrow |k\vec{v}\cdot\vec{v}|=||k\vec{v}||\,||\vec{v}||\Longleftrightarrow k|\vec{v}\cdot\vec{v}|=k||\vec{v}||\,||\vec{v}||=k||\vec{v}||^2$$ This demonstrates the $\Leftarrow $ direction, that is, assuming the far right conditions, then we can arrive at the far left condition, but you have not shown that the far left condition implies the far right conditions. This proof uses an assumption which is supposed to be proved when you suggest that $\vec{u}=k\vec{v}$ and $|\vec{u}\cdot\vec{v}|=||\vec{u}||\,||\vec{v}||\Longrightarrow |k\vec{v}\cdot\vec{v}|=||k\vec{v}||\,||\vec{v}||$, which is the statement you are trying to prove.
Here is how I went about proving this:
$\Rightarrow$ direction: Suppose $|\vec{u}\cdot \vec{v}|=||\vec{u}||\,||\vec{v}||$. We want to show that one vector is a scalar multiple of the other. Let  $$\vec{u}=\begin{bmatrix}u_1\\u_2\\\vdots\\u_n \end{bmatrix}, \vec{v}=\begin{bmatrix}v_1\\v_2\\\vdots\\v_n \end{bmatrix}$$
Then we have that $$\begin{align}|\vec{u}\cdot\vec{v}|&=|u_1v_1+u_2v_2+...+u_nv_n|\end{align}$$
and
$$\begin{align}||\vec{u}||\,||\vec{v}||&=\sqrt{(u_1^2+u_2^2+...+u_n^2)(v_1^2+v_2^2+...+v_n^2)}\end{align} $$
This implies that $$(u_1v_1+u_2v_2+...+u_nv_n)^2=(u_1^2+u_2^2+...+u_n^2)(v_1^2+v_2^2+...+v_n^2)$$  but by the Cauchy-Schwarz inequality, $$(u_1v_1+u_2v_2+...+u_nv_n)^2\leq(u_1^2+u_2^2+...+u_n^2)(v_1^2+v_2^2+...+v_n^2)$$ with equality if and only if $\vec{u}$ and $\vec{v}$ are linearly dependent, or in other words, without loss of generality, $\vec{v}=k\vec{u}$.
$\Leftarrow$ direction: Assume vectors $\vec{u}$ and $\vec{v}$ are linearly dependent, that is, without loss of generality, $\vec{v}=k\vec{u}$ for some $k\in\mathbb{R}$. Then, $$|\vec{u}\cdot\vec{v}|=|\vec{u}\cdot k\vec{u}|=|k|\,||\vec{u}||\,||\vec{u}||=||\vec{v}||\,||\vec{u}||$$
Others have suggested using $|\vec{u}\cdot\vec{v}|=||\vec{u}||\,||\vec{v}||\,\cos\theta$ where $\theta$ represents the angle between the two vectors and demonstrating that $\cos\theta=1$ when $\theta =0+2\pi r$ for some $r\in\mathbb{Z}$, which implies that the vectors are parallel and that they are linearly dependent.
