x and y are parallel iff ∥x∥²∥y∥² = |< x, y >|².

Dependencies:

  1. Inner product space
  2. Inner product is anti-linear in second argument
  3. Zero in inner product

Let $x$ and $y$ be vectors from the inner product space $V$ and $x \neq 0$. Then $y = kx$ for some scalar $k$ iff $|\langle x, y \rangle|^2 = \|x\|^2\|y\|^2$.

Proof of 'only-if' part

Let $y = kx$ for some scalar $k$.

\begin{align} & |\langle x, y \rangle|^2 \\ &= \langle x, kx \rangle \overline{\langle x, kx \rangle} \\ &= \langle x, kx \rangle \langle kx, x \rangle \tag{by conjugate symmetry} \\ &= \overline{k}\langle x, x \rangle k\langle x, x \rangle \tag{by (anti-)linearity} \\ &= \langle x, x \rangle k\overline{k}\langle x, x \rangle \tag{multiplicative commutativity in field} \\ &= \langle x, x \rangle \langle kx, kx \rangle \tag{by (anti-)linearity} \\ &= \|x\|^2 \|y\|^2 \end{align}

Proof of 'if' part

By the positive-definiteness property of inner-product spaces, $x \neq 0 \Rightarrow \|x\|^2 \neq 0$.

Let $|\langle x, y \rangle|^2 = \|x\|^2\|y\|^2$.

\[ w = y - \frac{\langle y, x \rangle}{\|x\|^2}x \]

(Intutively, $w$ is the component of $y$ perpendicular to $x$. We have to prove that this is 0.)

\begin{align} & \langle w, x \rangle \\ &= \left\langle y - \frac{\langle y, x \rangle}{\|x\|^2}x, x \right\rangle \\ &= \langle y, x \rangle - \frac{\langle y, x \rangle}{\|x\|^2} \langle x, x \rangle \tag{by linearity in first argument} \\ &= \langle y, x \rangle - \langle y, x \rangle = 0 \end{align}

\begin{align} & \langle w, y \rangle \\ &= \left\langle y - \frac{\langle y, x \rangle}{\|x\|^2}x , y \right\rangle \\ &= \langle y, y \rangle - \frac{\langle y, x \rangle}{\|x\|^2} \langle x , y \rangle \tag{by linearity in first argument} \\ &= \frac{\|x\|^2\|y\|^2 - \langle y, x \rangle\langle x , y \rangle}{\|x\|^2} \\ &= \frac{\|x\|^2\|y\|^2 - \overline{\langle x, y \rangle}\langle x , y \rangle}{\|x\|^2} \tag{conjugate symmetry} \\ &= \frac{\|x\|^2\|y\|^2 - |\langle x, y \rangle|^2}{\|x\|^2} = 0 \end{align}

\begin{align} & \|w\|^2 = \langle w, w \rangle \\ &= \left\langle w, y - \frac{\langle y, x \rangle}{\|x\|^2}x \right\rangle \\ &= \langle w, y \rangle - \frac{\overline{\langle y, x \rangle}}{\|x\|^2} \langle w, x \rangle \tag{by anti-linearity} \\ &= 0 - \frac{\overline{\langle y, x \rangle}}{\|x\|^2} 0 = 0 \end{align}

By the positive-definiteness property of inner-product spaces, $\|w\|^2 = 0 \Rightarrow w = 0$

Therefore, $y = \frac{\langle y, x \rangle}{\|x\|^2}x \Rightarrow y = kx$ for some scalar $k$.

Dependency for: None

Info:

Transitive dependencies:

  1. /complex-numbers/conjugation-is-homomorphic
  2. Group
  3. Ring
  4. Field
  5. Vector Space
  6. Inner product space
  7. Inner product is anti-linear in second argument
  8. Zero in inner product