Question about inner product and Cauchy Schwarz inequality

Question

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.

Answer 1

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).

Answer 2

I have seen that proof given before, so it's not out of the ordinary.

If you want to learn a proof that is independent of the cosine, there is another nice one that even for complex inner products.

It begins this way: $0\leq \langle x-\lambda y,x-\lambda y\rangle$. After expanding this into $0\leq \|x\|^2+|\lambda|^2\|y\|^2-\lambda\langle y,x\rangle-\overline{\lambda}\langle x,y\rangle$ you can use a judicious choice of $\lambda$ to get the result. (If you are satisfied with learning this for real symmetric forms then you can start with $0\leq \|x\|^2+|\lambda|^2\|y\|^2-2\lambda\langle x,y\rangle$ instead.)

If you need some help finding the judicious choice of $\lambda$, see this.

Answer 3

This is generalization of $.$(dot product) in 2-D or 3-D vector space. Given two vectors $x,y$, their inner product is the product of first vector with the projection of the second vector along the direction of first vector which is equal to $||x||.||y||.\cos\theta$ where $\theta $ represents the angle between vectors in n-D vector space.