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Prove an identity for the angle between vectors in Cn

by RoRi

The angle \theta between two vectors non-zero A, B \in \mathbb{C}^n is defined by the equation

    \[ \theta = \arccos \frac{\frac{1}{2}(A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert}. \]

The inequality

    \[ -2 \leq \frac{A \cdot B + \overline{A \cdot B}}{\lVert A \rVert \lVert B \rVert} \leq 2 \]

we established in the previous exercise (Section 12.17, Exercise #6) show that there is a unique \theta \in [0, \pi] satisfying this equation. Prove that we have

    \[ \lVert A - B \rVert^2 = \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \lVert A \rVert \lVert B \rVert \cos \theta. \]

Proof. From the definition of \theta we have

    \begin{align*} && \theta &= \arccos \frac{\frac{1}{2} (A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert} \\[9pt] \implies && \cos \theta &= \frac{A \cdot B + \overline{A \cdot B}}{2 \lVert A \rVert \lVert B \rVert} \\[9pt] \implies && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= A \cdot B + \overline{A \cdot B}. \end{align*}

But then we know from this exercise (Section 12.17, Exercise #3) that

    \[ A \cdot B + \overline{A \cdot B} = \lVert A+B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2. \]

And, we know from this exercise (Section 12.17, Exercise #5) that

    \[ \lVert A+ B\rVert^2 = 2\lVert A \rVert^2 + 2 \lVert B \rVert^2 - \lVert A - B \rVert^2. \]

Therefore,

    \begin{align*} && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= \lVert A + B\rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \\ \implies && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= 2 \lVert A \rVert^2 + 2 \lVert V \rVert^2 - \lVert A - B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \\[9pt] \implies && \lVert A - B \rVert^2 &= \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \lVert A \rVert \lVert B \rVert \cos \theta. \qquad \blacksquare \end{align*}

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