Proposition 1.6.1.
Vector addition is commutative. That is, for any two vectors \(u, v \in \RR^n\text{,}\)
\begin{equation*}
u + v = v + u \text{.}
\end{equation*}
Section 1.6 Proofs - Vector Arithmetic
Subsection 1.6.1 Properties of Vector Arithmetic
In Section 1.3, I introduced three structures on \(\RR^n\text{:}\) vector addition, scalar multiplication, and vector length. The abstract algebraic question that mathematicians ask at this point is: what are the properties of these structures? How does vector addition resemble normal addition (which is it named after), and how does it differ? How does scalar multiplication resemble normal multiplication (which is it named after), and how does it differ? How does the length differ from the absolute value (which shares the same notation)?
These are abstract questions, but they are also practically useful. If I am in a situation where I need to use linear algebra (of which there are many), I need to know how to manipulate the objects. I have a whole collection of a habits with addition and multiplication, but I can’t assume that all those habits immediately transfer to these new structures. If the new structures resemble the old, I have to prove it to be sure. If they don’t, I have to adjust my habits. Both situation shappen in this course. In this section, I’ll mostly be proving that the properties of ordinary addition and multiplication carry over to these new structures. Later in the course, I’ll have new structures which have properties that are different from what I expect for conventional arithmetic.
In this section, I’m going to address three properties of scalar multiplication and vector addition. Some properties of the vector length operation are left for the activities.
Subsection 1.6.2 Commutativity
Addition and multiplication of numbers are commutative binary operations. This means that I can interchange the order: \(4 +
5\) is the same as \(5 + 4\text{,}\) and \(7 \times 2\) is the same as \(2 \times 7\text{.}\) I’ve defined a new addition and a new multiplication, but I can’t just assume that these new structures are commutative. I have to prove it.
Proof.
When approaching a proof, it is often very useful to return to the definition. Let me first work in \(\RR^2\) to illustrate; I’ll generalize to \(\RR^n\) afterwards. I have two vectors \(u,v \in \RR^n\text{.}\) I write them as components.
\begin{align*}
\amp u = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}
\amp \amp v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}
\end{align*}
Then I use the definition of vector addition. I’ll start with the left side of the equation and try to work towards the right.
\begin{align*}
u + v \amp = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}
+ \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} \\
\amp = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2
\end{pmatrix}
\end{align*}
That’s the definition. Now each addition in the components (\(u_1 + v_1\) and \(u_2 + v_2\)) is the addition of ordinary numbers, so I can reverse the order of this addition.
\begin{align*}
\amp = \begin{pmatrix} v_1 + u_1 \\ v_2 + u_2
\end{pmatrix}
\end{align*}
The result is precisely the definition of the right side.
\begin{align*}
\amp = v + u
\end{align*}
I started on the left side of the desired equation and ended up with the right side, thus proving the identity. This was for \(\RR^2\text{,}\) so I still need to do the general case for \(\RR^n\text{.}\) Luckily for me, the steps are exactly the same, just for a vector with \(n\) components. Here are the steps in the general case: first, I use the definition of \(u + v\text{,}\) then I switch the order of the ordinary addition of numbers in each component, then I use the definition of \(v + u\text{.}\)
\begin{align*}
u + v \amp = \begin{pmatrix} u_1 \\ u_2 \\ \vdots \\ u_n
\end{pmatrix} + \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\
v_n \end{pmatrix} \\
\amp = \begin{pmatrix} u_1 + v_2 \\ u_2 + v_2 \\
\vdots \\ u_n + v_n \end{pmatrix} \\
\amp = \begin{pmatrix} v_1 + u_2 \\ v_2 + u_2 \\
\vdots \\ v_n + u_n \end{pmatrix} \\
\amp = v + u
\end{align*}
That is the proof for the general case, so I am done.
I want to point out a couple of things about this proof. First, it is general. I didn’t try to prove by choosing specific vectors and checking. I can’t prove by example. That said, often working through some examples is extremely valuable. When starting a proof, I’ll often try some examples to get a sense of what is going on. But, in the end, the argument needs to be general — it must cover all possible cases in the statement of the proposition.
Second, I never started with \(u + v = v + u\text{.}\) I started with the left side and worked my way to the right (though I could have just as easily started from the right and worked my way to the left). I should never assume what I want to prove, so I should never write \(u + v = v + u\) until the end of the proof when it has been established. If I’m trying to figure out how to do a proof, I’ll often work backwards: I’ll start with the statement I want to get to, manipulate it and see how the piece works. But when I write up the proof, I’m never starting with my goal, only finishing with it.
This proof establishes the commutativity of vector addition. What about scalar multiplication? For real numbers, I know \(ab =
ba\text{.}\) However, scalar multiplication is something quite different. If \(a\) is a scalar and \(u\) is a vector, then \(au\) produces another vector. In defining scalar multiplication, I wrote the scalar on the left and the vector on the right. Right now, the expression \(ua\) doesn’t even mean anything. I could define it to mean the same thing so that I could do scalar multiplication on either side, but the mathematical notation convention is to always write the scalar on the left and the vector on the right. In this way, scalar multiplication isn’t commutative because of the choice of notation.
Subsection 1.6.3 Associativity
Another property of ordinary addition and multiplication is associativity. This property states that I can bracket a three-fold addition or multiplication either way, and the result is the same. For addition, this is
\begin{equation*}
a + (b + c) = (a + b) + c \text{.}
\end{equation*}
For multiplication, this is similarly
\begin{equation*}
a(bc) = (ab)c\text{.}
\end{equation*}
To be more explicit, consider the sum \(6 + 4 + 9\text{.}\) I can add this in two ways. If I start with the left sum, I get \(10 + 9 = 19\text{.}\) If I start with the right sum, I get \(6
+ 13 = 19\text{.}\) Both ways produce \(19\text{;}\) associativity says that this works for all sums of numbers. Associativity is a property that I very rarely think about. It is naturally built into my number sense, and when doing arithmetic, I use it implicitly and subconsciously all the time. However, it is a property I have to worry about for new operations. There are (strangely enough!) non-associative operations in mathematics.
Happily, both vector addition and scalar multiplication are associative.
Proposition 1.6.2.
If \(u,v,w \in \RR^n\) are vectors, then
\begin{equation*}
(u + v) + w = u + (v + w)
\end{equation*}
The proof is nearly identical to the proof of commutativity. I write the vectors with components and then use the associativity of ordinary addition in each component.
For scalar multiplication, since there are vectors and scalars involved, I have to be careful how I even state associativity.
Proposition 1.6.3.
If \(u\in \RR^n\) and \(a,b, \in \RR\) then
\begin{equation*}
a(bu) = (ab)u
\end{equation*}
This is the only statement of associativity that makes sense. I can’t multiply vectors, but I can do successive multiplication by different scalars. The property here says that I can do two scalar multiplications with the vector \(u\text{,}\) or I can first multiply the scalars normally then do one scalar multiplication. The proof is also like the proof for commutativity: I write everything in components and use the associativity of ordinary multiplication.
Subsection 1.6.4 Distribution
For real numbers, multiplication distributes over addition.
\begin{equation*}
a(b + c) = ab + ac
\end{equation*}
For vector addition and scalar multiplication, distribution still works. There are two versions of the distributive law here.
Proposition 1.6.4.
If \(u,\in \RR^n\) and \(a,b \in \RR\text{,}\) then
\begin{equation*}
(a + b)u = au + bu \text{.}
\end{equation*}
Proposition 1.6.5.
If \(u,v \in \RR^n\) and \(a \in \RR\text{,}\) then
\begin{equation*}
a(u + v) = au + av
\end{equation*}
In the first version, the addition is happening with the scalars. In the second, the addition is happening with the vectors. Since these are different kinds of addition, two different statements are required. The proofs are again like the commutativity proofs: write everything in components and use the distributive rules for ordinary numbers in each component.
