If $x,y \in \mathbb{C}^m$ prove $x^*y = \|x\|_2\|y\|_2 \cos \alpha$

by Zduff   Last Updated June 19, 2018 22:20 PM

If $x,y \in \mathbb{C}^m$ prove

$$x^*y = \|x\|_2\|y\|_2 \cos \alpha $$


Attempt

\begin{align} LHS = \left(x^*y\right)^2 &= \left(\bar x_1y_1 + \bar x_2y_2 + \cdots + \bar x_my_m \right)^2 \\ &= (\bar x_1y_1)^2 + \cdots + (\bar x_my_m)^2+ 2\sum_{j = 1}^m\sum_{i \neq j}^m \bar x_iy_i\bar x_j y_j \dots \end{align}


Don't know where I'm going with that



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