# A question about the dot product (beginner mathematics)

In my physics textbook (but I consider this to be a matter of linear algebra) there is a chapter about vectors.

In this chapter I've found the following sentence: "If we know the dot product between the basis vectors, then we know the dot product between two arbitrary vectors". I don't understand the proof. The proof is given by writing out algebraically what the dot product between two vectors is (a sum with 6 components, where 3 components dissapear because the basis vectors of those components are orthogonal). Can somebody explain this sentence to me? I know this formula of the dot product between two vectors $$\vec{A}, \vec{B}$$: $$\vec{A}\cdot\vec{B}$$ $$=\mid \vec{A}\mid\mid\vec{B}\mid \cos \theta$$. To be clear: $$\mid \vec{R}\mid$$ denotes the length of a vector $$\vec{R}$$ and $$\theta$$ is the angle between $$\vec{A}$$ and $$\vec{B}$$.

Extra_1: this is a first year's course in mechanics and I haven't seen the dot product (yet) in my linear algebra course.

• $$A \cdot B = (a_1u_1+\dots+a_nu_n)\cdot (b_1u_1+\dots+b_nu_n) = \sum_{i,j} a_ib_j u_i\cdot u_j$$, using the bilinearity of the dot product. Thus to define an inner product, you only need specify its action on basis vectors. In a similar vein, a linear transformation is entirely specified by its action on basis vectors. Nov 13, 2023 at 19:49
• It is nontrivial to prove the distributive property $(\vec A+\vec B)\cdot\vec C = \vec A\cdot\vec C+\vec B\cdot\vec C$ from the geometric definition. This is why math texts typically start with the algebraic definition and then deduce the geometry from it. Nov 13, 2023 at 20:41

If you have the standard basis vectors $$e_1,\cdots,e_n$$ for the vector space $$V$$, then given any vector $$x,y \in V$$, you can find a unique set of $$\alpha_i, \beta_i$$ such that $$x = \sum_{i=1}^n \alpha_i e_i \qquad y = \sum_{i=1}^n \beta_i e_i$$ Then from the definition of the inner product, assuming we are working over the complex numbers, we can write $$\langle x,y \rangle = \sum_{i,j=1}^n \alpha_i \overline{\beta_i} \langle e_i , e_j\rangle$$ The dot product $$\langle x,y \rangle := x \cdot y$$ is an example of an inner product, and it satisfies (by definition) the identities \begin{align*} \langle \alpha x, y \rangle &= \alpha \langle x,y \rangle \\ \langle x, \alpha y \rangle &= \overline{\alpha} \langle x,y \rangle \\ \langle x+y,z \rangle &= \langle x,z \rangle + \langle y,z\rangle \\ \langle x, y \rangle &= \overline{ \langle y,x \rangle } \end{align*}

But in short:

• If you know two vectors, you know their decomposition in terms of the standard basis.
• If you know their decomposition, you know their inner product, provided you know the inner products of the basis vectors.

Or in reverse:

• If you know the inner products of the basis vectors, you know the inner product of any linear combination of them.
• If you know the inner products of linear combinations of the basis vectors, you ... well, you know the inner product of any two vectors in your space.

Given two arbitrary vectors, you can write them (uniquely) as a linear combination of the basis of the vector space and then just use bilinearity of the dot product.

Suppose $$V$$ is a vector space over $$\mathbb{R}$$ (if $$V$$ is a vector space over $$\mathbb{C}$$ then a similar argument works, see @PrincessEev answer for details). If $$\{e_1,...,e_n\}$$ is a basis for your vector space, given two arbitrary vectors $$v$$ and $$w$$ you can write $$v=\sum\limits_{i=1}^{n} v_{i} e_{i}$$ and $$w=\sum\limits_{i=1}^{n} w_{i} e_{i}$$.

Then

$$v\cdot w= (\sum\limits_{i=1}^{n} v_{i} e_{i})\cdot (\sum\limits_{j=1}^{n} w_{j} e_{j})=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} v_{i}w_j (e_i\cdot e_j)$$.

So if you know the values of $$e_i\cdot e_j$$ for $$i,j=1,...,n$$, you know the value of $$v\cdot w$$ for any pair of vectors $$v,w$$.

• Perhaps worth noting that this answer assumes you're working over $\mathbb{R}$ as opposed to $\mathbb{C}$. Nov 13, 2023 at 19:58
• yes sorry, I will add this in the answer Nov 13, 2023 at 20:01