# I Why is the dot product equivalent to matrix multiplication?

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1. May 16, 2017

### rdgn

Why is the dot product equivalent to the matrix multiplication of its components?

I've seen some proofs using Pythagorean and cosine law but they don't give you an intuitive feel as to why matrix multiplication works.

The geometric definition ($ab cosθ$) is very easy to understand. To a certain extent, I can understand why matrix multiplication works when either vector $a$ or $b$ is parallel to the x or y-axis since the product would ultimately simplify to the geometric definition, but I don't understand why it works for any arbitrary vector.

Can anyone give an intuition of why this works? (not proof/s, since I've already seen some)

Last edited: May 16, 2017
2. May 16, 2017

### Staff: Mentor

What is the definition of a dot product, that is acceptable for you? This is important since you basically ask, why two definitions are equivalent. So what's yours?

3. May 16, 2017

### rdgn

Well, based on the intuition of the geometric definition, it's simply how much of a certain vector goes into another vector. (or more formally), how much of a vector is projected into another vector multiplied by the magnitude of that vector. Also, a measure of how much they point in the same direction.

Btw, I found this video by 3b1b (youtube.com/watch?v=LyGKycYT2v0), I think it's great but it might take a while before the idea sinks into my head (and i should familiarize myself more with linear transformations).

4. May 16, 2017

### Staff: Mentor

A major point is, as soon as you write vectors $v$ as $(v_1,v_2,\ldots)$ and perform matrix multiplications you will have to chose some basis first to make sense of the components. Thus you will have to bridge the gap between a purely geometric point of view and the algebraic view in terms of numbers, components first.

An interesting read about the various products can be found on the first pages of this pdf:
https://arxiv.org/pdf/1205.5935.pdf
It stresses the geometry behind those products (dot-product, $\wedge$-product).

5. May 16, 2017

### WWGD

You can also argue that the dot product is bilinear and so it is a 2-tensor. Tensors of total (meaning sum of covariant and contravariant indices) order $\leq 2$ can be represented as a standard matrix ( order 1) or as a quadratic form (order 2 ).

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