Proposition 2.4.1.
The dot product is commutative. That is, if \(u, v \in
\RR^n\text{,}\)
\begin{equation*}
u \cdot v = v \cdot u\text{.}
\end{equation*}
Section 2.4 Proofs - Dot and Cross Products
Subsection 2.4.1 Prooving the Dot Product Properties
In Proposition 2.1.5, I listed four properties of the dot product. In this section, I’ll demonstrate how to prove some of these properties.
Proof.
I first write the vector in coordinates. This is the full generality of \(\RR^n\text{,}\) so I write vectors with \(n\) coordinates using ellipses
\begin{align*}
\amp u = \begin{pmatrix}
u_1 \\ u_2 \\ u_3 \\ \vdots \\ u_n
\end{pmatrix}
\amp \amp v = \begin{pmatrix}
v_1 \\ v_2 \\ v_3 \\ \vdots \\ v_n
\end{pmatrix}
\end{align*}
Then I’ll start with the left side and see if I can make a chain of equality that results in the right side. First I use the definition of the dot product.
\begin{equation*}
u \cdot v = u_1v_1 + u_2v_2 + u_3v_3 + \ldots + u_nv_n
\end{equation*}
Then each of the products \(u_iv_i\) are products of ordinary real numbers. Multiplicaton of real numbers is commutative, so I can interchange the order of each of these products.
\begin{equation*}
u \cdot v = v_1u_1 + v_2u_2 + v_3u_3 + \ldots + v_nu_n
\end{equation*}
Then I use the definition of the dot product again. The right side of the previous equation is, by definition, the dot product \(v \cdot u\text{.}\)
\begin{equation*}
u \cdot v = v \cdot u
\end{equation*}
That is the desired equality, so I have finished the proof. Notice that the desired equation only shows up at the end of the proof. I never start with what I want to prove.
Proposition 2.4.2.
If \(u \in \RR^n\text{,}\) then
\begin{equation*}
u \cdot u = |u|^2
\end{equation*}
Proof.
I write \(u\) in its coordinates.
\begin{equation*}
u = \begin{pmatrix}
u_1 \\ u_2 \\ u_3 \\ \ldots \\ u_n
\end{pmatrix}
\end{equation*}
Then I’ll start with the left side and use the definition of the dot product.
\begin{equation*}
u \cdot u = u_1u_1 + u_2u_2 + u_3u_3 + \ldots u_nu_n
\end{equation*}
I can write these products on the left as squares.
\begin{equation*}
u \cdot u = u_1^2 + u_2^2 + u_3^2 + \ldots + u_n^2
\end{equation*}
Since the right side is a sum of squares, it is a positive number (or possibly zero if all the \(u_i\) are zero). As a positive number or zero, I can square root it and then square it without effect: \(\alpha = (\sqrt{\alpha})^2\) for any non-negative \(\alpha\text{.}\)
\begin{equation*}
u \cdot u = \left( \sqrt{u_1^2 + u_2^2 + u_3^2 + \ldots +
u_n^2} \right)^2
\end{equation*}
Then the expression inside the brackets is, by definition, the length of the vector \(u\text{.}\)
\begin{equation*}
u \cdot u = \left( |u| \right)^2 = |u|^2
\end{equation*}
This is the desired equation, so the proof is completed.
Proposition 2.4.3.
The dot product distributes over vector addition. That is, if \(u,v,w \in \RR^n\text{,}\)
\begin{equation*}
u \cdot (v + w) = u \cdot v + u \cdot w
\end{equation*}
Proof.
I’ll write the three vectors in components.
\begin{align*}
\amp u = \begin{pmatrix}
u_1 \\ u_2 \\ u_3 \\ \ldots \\ u_n
\end{pmatrix}
\amp \amp v = \begin{pmatrix}
v_1 \\ v_2 \\ v_3 \\ \ldots \\ v_n
\end{pmatrix}
\amp \amp w = \begin{pmatrix}
w_1 \\ w_2 \\ w_3 \\ \ldots \\ w_n
\end{pmatrix}
\end{align*}
Then I’ll start with the left side. I’ll use the definition of vector addition and the dot product.
\begin{align*}
u \cdot (v + w) \amp \begin{pmatrix}
u_1 \\ u_2 \\ u_3 \\ \ldots \\ u_n
\end{pmatrix} \cdot
\begin{pmatrix}
v_1 + w_1 \\ v_2 + w_2 \\ v_3 + w_3 \\ \ldots \\ v_n +
w_n
\end{pmatrix}\\
\amp = u_1(v_1 + w_1) + u_2(v_2 + w_2) + u_3(v_3 + w_3)
+ \ldots + u_n(v_n + w_n)
\end{align*}
Each of these products can be distributed, since this is all arithmetic of ordinary real numbers.
\begin{align*}
u \cdot (v + w) \amp = (u_1v_1 + u_1w_1) + (u_2v_2 + u_2w_2) \\
\amp + (u_3v_3 + u_3w_3) + \ldots + (u_nv_n + u_nw_n)
\end{align*}
The right side is the sum of many products. I can rearrange this sum as I please, so I’ll put all the terms with \(v_i\) first, then all those with \(w_i\) afterwards.
\begin{align*}
u \cdot (v + w) \amp = (u_1v_1 + u_2v_2 + u_3v_3 +
\ldots + u_nv_n)\\
\amp + (u_1w_1 + u_2w_2 + u_3w_3 + \ldots + u_nw_n)
\end{align*}
Then I can use the definition of the dot product again. On the right side, the term in the first bracket is, by definition, \(u \cdot v\text{.}\) The term in the second bracket is, again by definition, \(u \cdot w\text{.}\)
\begin{equation*}
u \cdot (v + w) = u \cdot v + u \cdot w
\end{equation*}
This is the desired equation, so the proof is finished.
Proposition 2.4.4.
If \(u,v \in \RR^n\) and \(a \in \RR\text{,}\) then
\begin{equation*}
u \cdot (av) = (au) \cdot v = a (u \cdot v)
\end{equation*}
Proof.
I’ll write the two vectors in components.
\begin{align*}
\amp u = \begin{pmatrix}
u_1 \\ u_2 \\ u_3 \\ \ldots \\ u_n
\end{pmatrix}
\amp \amp v = \begin{pmatrix}
v_1 \\ v_2 \\ v_3 \\ \ldots \\ v_n
\end{pmatrix}
\end{align*}
Then I want to consider this sum:
\begin{equation*}
au_1v_1 + au_2v_2 + au_3v_3 + \ldots au_nv_n\text{.}
\end{equation*}
I’m going to organize this sum in three different ways, which will correspond to the three desired versions in the equation in the proposition. Since I will only be arranging brackets for addition and multiplication of numbers in acceptable ways, all three of these organizations are just calculating the same sum, so I can conclude that all three are equal.
First, I’ll factor the \(a\) out of the entire sum.
\begin{equation*}
a(u_1v_2 + u_2v_2 + u_3v_3 + \ldots u _nv_n)
\end{equation*}
What is in brackets is the definition of the dot product, so this is \(a (u \cdot v)\text{.}\) Next I’ll bracket the \(a\) with the \(u_i\) components.
\begin{equation*}
(au_1)v_2 + (au_2)v_2 + (au_3)v_3 + \ldots (au_n)v_n
\end{equation*}
This is, by definition, also a dot product.
\begin{equation*}
\begin{pmatrix}
au_1 \\ au_2 \\ au_3 \\ \vdots \\ au_n
\end{pmatrix} \cdot
\begin{pmatrix}
v_1 \\ v_2 \\ v_3 \\ \vdots \\ v_n
\end{pmatrix}
\end{equation*}
The first vector in this dot product is, by definition, the scalar multiple of \(u\) by \(a\text{.}\) Therefore, this organization of the sum gives \((au) \cdot v\text{.}\) Finally, I’ll change the order of the multiplications and group \(a\) with the \(v_i\) components.
\begin{equation*}
u_1(av_2) + u_2(av_2) + u_3(av_3) + \ldots u_n(av_n)
\end{equation*}
This is, by definition, also a dot product.
\begin{equation*}
\begin{pmatrix}
u_1 \\ u_2 \\ u_3 \\ \vdots \\ u_n
\end{pmatrix} \cdot
\begin{pmatrix}
av_1 \\ av_2 \\ av_3 \\ \vdots \\ av_n
\end{pmatrix}
\end{equation*}
The second vector in this dot product is, by definition, the scalar multiple of \(v\) by \(a\text{.}\) Therefore, this organization of the sum gives \(u \cdot (av)\text{.}\) Since all three organizations are valid ways of calculating the sum, I conclude that
\begin{equation*}
u \cdot (av) = (au) \cdot v = a(u \cdot v)\text{,}
\end{equation*}
which concludes the proof.
