I'm wondering where the following equality came from: $$ \langle x , y \rangle = \|x \| \| y \| \cos \theta$$
where the thing on the LHS is the inner product and $\|\cdot\|$ is the norm induced by $\langle \cdot, \cdot \rangle$. Do we need the Cauchy Schwarz inequality to prove this? I'm asking because I'm reading my notes and there is a proof of the C. S. – inequality. It's fairly short but longer than the following:
Claim: $|\langle x,y \rangle | \leq \|x \| \|y \|$
Proof: Since $\langle x , y \rangle = \|x \| \| y \| \cos \theta$ we have $-\|x \| \| y \| \leq \langle x , y \rangle \leq \|x \| \| y \|$.
Thanks.
Best Answer
My favorite interpretation of the identity $$ \langle x,y \rangle = \|x\| \|y\| \cos \theta \tag{1} $$ is that (1) defines the angle $\theta \in [-\pi/2,\pi/2]$ between the vectors $x$ and $y$. Since $$ \langle x,y \rangle =\sum_{j=1}^n x_j y_j $$ is the standard inner product in $\mathbb{R}^n$, the elementary Cauchy-Schwarz inequality implies that $$ -1 \leq \frac{\langle x,y \rangle}{\|x\| \|y\|} \leq 1, $$ and therefore there exists a unique angle $\theta \in \left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$ such that (1) holds.
However, once you have fixed $x$ and $y$, the question becomes two-dimensional in the plane spanned by $x$ and $y$. There you can apply stadard elementary geometry to show that the cosine of the angle between the two vectors is given by (1).