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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 $-\|x \| \| y \| \leq \langle x , y \rangle \leq \|x \| \| y \|$.


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This seems a little backwards to me. Typically the Cauchy-Schwarz inequality is used to show that the angle formula is well-defined, i.e. $\frac{\langle x,y\rangle}{\|x\|\|y\|} \le 1$ and then the angle $\theta$ is defined using the given formula. – EuYu Jul 2 '12 at 7:39
@EuYu So it's like I suspected: We need the C.S.-inequality to prove $\|x \| \| y \| \cos \theta$. – Rudy the Reindeer Jul 2 '12 at 7:44
@MattN. Not necessarily. You can also define the inner product by means of the given formula. If you are working in standard euclidean spaces, the concept of angle spanned by two vectors is elementary. In a general inner product space, the angle is defined by your formula. – Siminore Jul 2 '12 at 8:13
up vote 4 down vote accepted

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).

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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.

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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.

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