Equality in Cauchy-Schwarz I have question about equality in Cauchy-Schwarz inequality:
I have this proof with me:

In proof, they only show that equality holds if $x$ is a specific scalar multiple of $y$, namely $\frac{||x||}{||y||}$, but in the statement of theorem, they claim that equality iff one of them scalar multiple of the other.

So, is that any scalar, or a specific scalar?
 A: If $y:=cx$, then taking norms of both sides gives $|c|=\|y\|/\|x\|$. So it holds for any scalar multiple $c$. 
A: First note that, if $x = \lambda y$, then $||x||^2 \cdot ||y||^2 = \langle x, x \rangle \langle \lambda x, \lambda x \rangle =  \langle x, y \rangle^2$. 

Here is a different proof which is a bit simpler:
Let $P_y$ be the projection onto $y$, sending $x$ to $\frac{\langle x, y \rangle}{\langle y, y \rangle} y$. You can check that $\langle P_y(x), x - P_y(x) \rangle = 0$, so that this is indeed the orthogonal projection. 

$$||x||^2 \stackrel{\text{pythagorean theorem}}{=} ||P_y(x)||^2 + ||x - P_y (x)||^2 \geq ||P_y(x)||^2 = \frac{|\langle x, y \rangle|^2}{|\langle y, y \rangle|^2}||y||^2 = | \langle x , y \rangle|^2 / ||y||^2$$
So $||x||^2 \cdot ||y||^2 \geq |\langle x, y \rangle|^2$ so $||x|| \cdot ||y|| \geq |\langle x, y \rangle|$. 
Note that we get equality if and only if $||x - P_y (x) ||^2 = 0$, if and only if $x = P_y(x)$. 
Since $P_y$ is a projection, $P_y \circ P_y = P_y$. From this one can show that  $P_y (x) = x$ if and only if $x$ is in the image of $P_y$. This happens if and only if $x = \lambda y$ for some $\lambda \in \mathbb{R}$, since the image of $P_y$ is $\langle y \rangle$. 
In conclusion, we have equality if and only if $||x - P_y(x)||^2 = 0$, if and only if $P_y (x) = x$, if and only if $x$ is in the image of $P_y$, if and only if $x = \lambda y$ for some $\lambda \in \mathbb{R}$. 
