Prove the following three conditions are equivalent. 1 $u=cv,c>0$. 2 $(u|v)=||u||||v||$. 3 $||u+v||=||u||+||v||$.

I am reading "Introduction to Linear Algebra" (in Japanese) by Kazuo Matsuzaka.

The following exercise (Exercise 3 on p.342) is in this book.

Let $V$ be an inner product space over $K$.
Let $u,v\in V$ such that $u\ne 0, v\ne 0$.
Prove the following three conditions are equivalent.

1. $u=cv,c>0$.
2. $(u|v)=||u||||v||$.
3. $||u+v||=||u||+||v||$.

My attempt is here:
We prove that $||u||||v||=|(u|v)|$ holds if and only if $u$ and $v$ are linerly dependent.
Assume that $||u||||v||=|(u|v)|$.
If $v=0$, then $u$ and $v$ are linearly dependent.
We consider the case $v\ne 0$.
Let $a:=(v|v)$.
Let $b:=-\overline{(u|v)}$.
$$\begin{eqnarray}(au+bv|au+bv)&=&\bar{a}a(u|u)+\bar{a}b(u|v)+a\bar{b}\overline{(u|v)}+\bar{b}b(v|v)\\&=&(v|v)^2(u|u)-(v|v)|(u|v)|^2-(v|v)|(u|v)|^2+|(u|v)|^2(v|v)\\&=&(v|v)^2(u|u)-(v|v)|(u|v)|^2\\&=&(v|v)((v|v)(u|u)-|(u|v)|^2)\\&=&0\\ \end{eqnarray}.$$
So, $au+bv=0$ and $a\ne 0$.
So, $u$ and $v$ are linearly dependent.
Conversely, assume that $u$ and $v$ are linearly dependent.
Then, we can write $u=cv$ or $v=cu$.
When $u=cv$, $||u||||v||=|c|||v||^2=|c|(v|v)=|c||(v|v)|=|\bar{c}||(v|v)|=|\bar{c}(v|v)|=|(cv|v)|=|(u|v)|.$
When $v=cu$, $||u||||v||=||v||||u||=|(v|u)|=|\overline{(u|v)}|=|(u|v)|.$

$1. \implies 2.$
Assume that $u=cv, c>0$.
$(u|v)=\bar{c}(v|v)=c(v|v)=c||v||||v||=||cv||||v||=||u||||v||$.

$2. \implies 3.$
Assume that $(u|v)=||u||||v||$.
$||u+v||^2=(u+v|u+v)=(u|u)+(u|v)+(v|u)+(v|v)=(u|u)+(u|v)+\overline{(u|v)}+(v|v)=(u|u)+(u|v)+(u|v)+(v|v)=(u|u)+2(u|v)+(v|v)=||u||^2+2||u||||v||+||v||^2=(||u||+||v||)^2$.
So, $||u+v||=||u||+||v||$.

$2. \implies 1.$
Assume that $(u|v)=||u||||v||$.
$(u|v)=||u||||v||\geq 0$.
So, $|(u|v)|=||u||||v||$.
So, $u$ and $v$ are linearly dependent.
So, we can write $u=cv$ or $v=cu$.
When $u=cv$, $(u|v)=(cv|v)=\bar{c}(v|v)=\bar{c}||v||^2$ and $||u||||v||=||cv||||v||=|c|||v||^2$.
So, $\bar{c}=|c|$.
So, $c\in\mathbb{R}$ and $c>0$.
So, $u=cv,c>0$.
When $v=cu$, similarly $c\in\mathbb{R}$ and $c>0$.
So, $u=(\frac{1}{c})v, \frac{1}{c}>0$.

I cannot show $3. \implies 1.$ or $3. \implies 2.$.

Please show that $3. \implies 1.$ or $3. \implies 2.$.

Solution 1:

$\lVert u+v\rVert^2=\lVert u\rVert^2+\lVert v\rVert^2+2\lVert u\rVert\lVert u\rVert$

$\Rightarrow (u|v)+(v|u)=2\sqrt{(u|u)(v|v)}$

$\Rightarrow |(u|v)|\geq \text{Re}(u|v)= \frac{(u|v)+\overline{(u|v)}}{2}=\sqrt{(u|u)(v|v)}$.

From the Cauchy-Schwarz inequality, this implies that $|(u|v)|=\sqrt{(u|u)(v|v)}$

This implies that $u=cv$ for some $c\in\mathbb{C}$. Now, observe that

$\lVert u+v \rVert=\lVert cv+v \rVert=|c+1|\lVert v \rVert$

And, $\lVert u\rVert+\lVert v\rVert=(1+|c|)\lVert v\rVert$

Since $v\neq 0$, $\lVert v\rVert>0$ and $|c+1|=|c|+1$. This implies that $c\in\mathbb{R}_{>0}$. Thus, $u=cv$ for some $c>0$.

This shows (3)$\Rightarrow$(1)

Solution 2:

$3\Rightarrow 2$: Square both sides to get $$ ||u+v||^2=(u+v|u+v)=||u||^2+||v||^2+2(u|v)=||u||^2+||v||^2+2||u||||v|| $$ from which $3$ follows. You already know that $2\Rightarrow 1$ so you are done.