Counterexample? Equality holds in the Cauchy-Schwarz Inequality iff $\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$

Question

I found this way of expressing that equality holds in Cauchy Schwarz inequality iff $\textbf{x},\textbf{y}$ and are linearly dependent in a book (as an exercise), but I think it is wrong, so there's a counterexample for it. I'm not sure if it is a valid counterexample.

Statement:

$| \langle\textbf{x} ,\textbf{y}\rangle| = \| \textbf{x}\|\| \textbf{y}\|$ iff $\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$

Counterexample:

Let $\textbf{y} = (0,0) \in \mathbb{R^2}$, $\textbf{x} = (1,1) \in \mathbb{R^2}$. Then we have:

$$|\langle \textbf{x},\textbf{y}\rangle| = |\langle \textbf{x},\textbf{0}\rangle| = 0 = 1\cdot0 = \| \textbf{x}\|\| \textbf{y}\|$$ But $$\textbf{x} = (1,1) \neq (\alpha\cdot0,\alpha\cdot0) = \alpha(0,0) = \alpha\textbf{y}, \forall \alpha \in \mathbb{R}$$

I know that if we let $\textbf{y} = (1,1)$ and $\textbf{x} = (0,0)$ the statement is true, but if the equality must work for all values of $\textbf{y}$ and $\textbf{x}$, then the instance gives is a counterexample, right?

Answer 1

The statement

$\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$

does not cover all possibilities for the two vectors to be linearly dependent. It fails precisely when $\mathbf y$ is the zero vector but $\mathbf x$ isn't. So for such a case, the Caychy-Schwarz equality will hold, but the above property doesn't.

Instead, the true characterisation of linear dependence is usually phrased as

$\alpha\textbf{x} + \beta\textbf{y} = \mathbf 0$ for some $\alpha, \beta \in \mathbb{R}$, not both $0$

This version allows for either of the vectors to be the zero without the other one. And the Cauchy-Schwarz equality will hold iff this holds.

Answer 2

You are correct, that's a counterexample. I don't know the exact statement of the book, but the statement can be fixed replacing it by "One of them is a scalar multiple of the other" (may be that's what the book says?). That way, if $\textbf{y}=\textbf{0}$, then $\textbf{y} = 0 \textbf{x}$.