How to expand $\langle \textbf{x}-\textbf{y},\textbf{x}-\textbf{y}\rangle$ using the Cauchy–Schwarz inequality

Question

The Question

\begin{align}||\textbf{x}+\textbf{y}||^2+||\textbf{x}-\textbf{y}||^2&=\langle\textbf{x}+\textbf{y},\textbf{x}+\textbf{y} \rangle+\langle\textbf{x}-\textbf{y},\textbf{x}-\textbf{y} \rangle\\&=\langle\textbf{x},\textbf{x}+\textbf{y}\rangle+\langle\textbf{y},\textbf{x}+\textbf{y}\rangle+\langle\textbf{x},\textbf{x}-\textbf{y}\rangle-\langle\textbf{y},\textbf{x}-\textbf{y}\rangle\\&=\langle\textbf{x},\textbf{x}\rangle+\langle\textbf{x},\textbf{y}\rangle+\langle\textbf{y},\textbf{x}\rangle+\langle\textbf{y},\textbf{y}\rangle+\langle\textbf{x},\textbf{x}\rangle-\langle\textbf{x},\textbf{y}\rangle-\langle\textbf{y},\textbf{x}\rangle+\langle\textbf{y},\textbf{y}\rangle\\&=2\langle\textbf{x},\textbf{x}\rangle+2\langle\textbf{y},\textbf{y}\rangle\\&=2||\textbf{x}||^2+2||\textbf{y}||^2\end{align}

My Understanding

I understood how $\langle \textbf{x}+\textbf{y},\textbf{x}+\textbf{y}\rangle$ was expanded. I believe you take the $\textbf{x}$'s from each side and make them into a vector. Then the $\textbf{y}$'s. And then the $\textbf{x}$ from the first vector and $\textbf{y}$ from the second. And lastly the $\textbf{y}$ from the first vector and $x$ from the second. I repeated this same process for $\langle \textbf{x}-\textbf{y},\textbf{x}-\textbf{y}\rangle$ and got: $\langle \textbf{x},\textbf{x}\rangle+\langle -\textbf{y},-\textbf{y}\rangle+\langle \textbf{x},-\textbf{y}\rangle+\langle -\textbf{y},\textbf{x}\rangle$. How did they get a negative in front of the $\textbf{x}$? Also, why does $\langle -\textbf{y},-\textbf{y}\rangle$ become $\langle \textbf{y},\textbf{y}\rangle$?

Answer 1

This is due to two properties of the dot product:

1. linearity in the first argument $\langle\alpha{x} + \beta{y},z\rangle=\alpha\langle{x}, z\rangle+\beta\langle{y}, z\rangle$, where $\alpha$ and $\beta$ are complex numbers, $x$, $y$ and $z$ are vectors;
2. $\langle x, y\rangle=\langle y, x\rangle^{*}$, where the symbol $*$ is a complex conjugation.

So, $\langle x, \alpha y + \beta z\rangle=\langle \alpha y + \beta z, x\rangle^{*} = \alpha^{*}\langle y, x\rangle^{*} + \beta^{*}\langle z, x\rangle^{*} = \alpha^{*}\langle x, y\rangle + \beta^{*}\langle x, z\rangle$.

Answer 2

Hint

The Cauchy–Schwarz inequality is used to prove that the inner product is a continuous function with respect to the topology induced by the inner product itself. Therefore, using the dot product, the following hold for any vectors $\textbf{x}, \textbf{y}$, and $\textbf{z}$ in $\mathbb{R}^n$, and scalars $c \in \mathbb{R}$

1. $\textbf{x}\cdot\textbf{y}=\textbf{y}\cdot\textbf{x}.$
2. $\textbf{x}\cdot\textbf{x}=||\textbf{x}||^2.$
3. $\textbf{x}\cdot(\textbf{y}+\textbf{z})=\textbf{x}\cdot\textbf{y}+\textbf{x}\cdot\textbf{z}$ and $(\textbf{x}+\textbf{y})\cdot\textbf{z}=\textbf{x}\cdot\textbf{z}+\textbf{y}\cdot\textbf{z}.$
4. $(c\textbf{x})\cdot\textbf{y}=\textbf{x}\cdot c\textbf{y}=c(\textbf{x}\cdot\textbf{y}).$