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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.

Thanks in advance.

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  • $\begingroup$ $$A \cdot B = (a_1u_1+\dots+a_nu_n)\cdot (b_1u_1+\dots+b_nu_n) = \sum_{i,j} a_ib_j u_i\cdot u_j$$, 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. $\endgroup$
    – Andrew
    Nov 13, 2023 at 19:49
  • $\begingroup$ 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. $\endgroup$ Nov 13, 2023 at 20:41

2 Answers 2

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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.
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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 $v,w$.

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

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