2

I am just really curious, how the equivalence between definitions of inner product can be established?

$$\mathbf{x \cdot y} = \sum_{k=1}^n x_k y_k = \mathbf{|x| |y|} \cos \theta $$

1LiterTears
  • 4,803
  • One way to do this is to draw a triangle with sides given by x, y, and x-y, and then apply the law of cosines to this triangle (where $\theta$ is the angle between x and y). – user84413 Aug 21 '13 at 17:58
  • Apply properties of inner product in the space $\mathbb{R}^n$ over $\mathbb{R}$ – Supriyo Aug 21 '13 at 18:00
  • In a more general context, this is the definition of an angle between two vectors in a pre-Hilbert space is defined by $$\cos \theta = \frac{\langle x,y \rangle}{\Vert x\Vert \cdot \Vert y \Vert}$$ – AlexR Aug 21 '13 at 18:03

1 Answers1

1

The easiest way to do this for me is as follows: Firstly notice that the $\sum x_k y_k$ is invariant under rotations. Why? Because they are orthogonal matrices $O_{ij}$ so $$(O\mathbf x)^T O\mathbf y= \mathbf x^T O^T O\mathbf y= \mathbf x^T \mathbf y$$

So rotate the vectors to make $\mathbf x= (x,0,0,\ldots,0)$ and $\mathbf y= (y_1,y_2,0,\ldots,0)$. Then drawing a very simple picture should convince you of the result! What matters is that $y_1= |\mathbf y| \cos\theta$.

(This relies on you understanding why rotations are orthogonal, though. I'd explain this by just pointing out that it is true for rotations in the simplest coordinate planes just by checking, and therefore holds for other rotations.)

The above are the sort of arguments you use when you know the answers, so they might not be great for you now, but hopefully they show you you one useful picture you might come to appreciate!

not all wrong
  • 16,568
  • 2
  • 37
  • 58