**Read Section 1.2** for next class.
Work through homework problems.

**Lecture notes** (this page) available from course web page.
Also look for **announcements** there.

**No tutorials this week.** There is a quiz in tutorials next week.

Please **read over syllabus**, especially
before e-mailing me with questions, as it covers all of the main points.

Let me know if the bookstore runs out of **texts or combo packs**.

We write $\R^n$ for the set of all vectors with $n$ real components, e.g. $[1, 2, 3, 4, 5, 6, 7]$ is in $\R^7$.

We also often write vectors as column vectors, e.g. $\coll 1 2$.

**Vector addition:** $[u_1, \ldots, u_n] + [v_1, \ldots, v_n] :=
[u_1 + v_1, \ldots, u_n + v_n]$.

E.g. $[3, 2, 1] + [1, 0, -1] = [4, 2, 0]$.

**Scalar multiplication:** $c [u_1, \ldots, u_n] := [c u_1, \ldots, c u_n]$.

E.g. $2 [ 1 , 2, 3, 4, 5] = [2, 4, 6, 8, 10]$.

**Zero vector:** $\vec{0} := [0, 0, ..., 0]$.

The picture to the right shows geometrically that vector addition is *commutative*:
$\vec{u} + \vec{v} = \vec{v} + \vec{u}$.

In this true in $\R^n$? Let's check: $$ \begin{aligned} \vec{u} + \vec{v}\ \ &= [u_1 + v_1,\, \ldots,\, u_n + v_n] \\ &= [v_1 + u_1,\, \ldots,\, v_n + u_n] \\ &= \vec{v} + \vec{u}. \end{aligned} $$

Many other properties that hold for real numbers also hold for vectors: Theorem 1.1. But we'll see differences later.

**Example:** Simplification of an expression:
$$
\begin{aligned}
&3 \vec{b} + 2 (\vec{a} - 4 \vec{b})\\
=\ &3 \vec{b} + 2 \vec{a} - 8 \vec{b}\\
=\ &2 \vec{a} - 5 \vec{b}
\end{aligned}
$$

**True/false:**
For every vector $\vu$, we have $2 \vu = \vu + \vu$.

True, since both sides have components $[2 u_1, \ldots, 2 u_n]$.
Or, from the distributive law (Theorem 1.1(f)) and Theorem 1.1(h), we have
\[
2 \vu = (1 + 1) \vu = 1 \vu + 1 \vu = \vu + \vu .
\]

**True/false:**
For every vector $\vu$, we have $2 \vu \neq 3 \vu$.

False. If $\vu = \vec 0$, then $2 \vu = \vec 0 = 3 \vu$.

**An important real-world application:**

**Definition:** A vector $\vv$ is a **linear combination** of
vectors $\vv_1, \vv_2, \ldots, \vv_k$ if there are scalars
$c_1, c_2, \ldots, c_k$ so that
\[
\vv = c_1 \vv_1 + \cdots + c_k \vv_k .
\]
The numbers $c_1, \ldots, c_k$ are called the coefficients.
They are not necessarily unique.

**Example:** Is $\coll 1 {-1}$ a linear combination of
$\coll 1 1$, $\coll 2 {-1}$ and $\coll 0 1$?

Yes, since \[ \coll 1 {-1} = 1 \coll 1 1 + 0 \coll 2 {-1} - 2 \coll 0 1 \qqtext{(Check!)} \]

**Note:** We also have
\[
\coll 1 {-1} = -\frac{1}{3} \coll 1 1 + \frac{2}{3} \coll 2 {-1} + 0 \coll 0 1 \qqtext{(Check!)}
\]
and many more possibilities.

We will learn later how to find all solutions.

**Example:** Is $\coll 1 {-1}$ a linear combination of
$\coll 1 0$ and $\coll 2 0$?

No, since any linear combination of $\coll 1 0$ and $\coll 2 0$ has a
zero as the second component.

**Example:** Is $\coll 0 0$ a linear combination of
$\coll 1 0$ and $\coll 2 0$?

Yes. The zero vector is a linear combination of *any* set of
vectors, since you can just take $c_1 = c_2 = \cdots = c_k = 0$.

**Example:** Express $\vw_1 = \coll 3 3$ as a linear combination
of $\vu = \coll 2 1$ and $\vv = \coll {-1} 1$.

We can solve this by using $\vu$ and $\vv$ to make a new coordinate system in the plane. Use the board to show that $\vw_1 = 2 \vu + \vv$.

Similarly, show that $\vw_2 = \coll 4 {-1}$ can be expressed as $\vw_2 = \vu - 2 \vv$.

Note that in this case the coefficients are unique.
In this situation, the coefficients are called the **coordinates**
with respect to $\vu$ and $\vv$.
So the coordinates of $\vw_1$ with respect to $\vu$ and $\vv$ are $2$ and $1$,
and the coordinates of $\vw_2$ with respect to $\vu$ and $\vv$ are $1$ and $-2$.

Working in a different coordinate system is a powerful tool.

Multiplication is as usual.

Addition: $0 + 0 = 0$, $\ 0 + 1 = 1$, $\ 1 + 0 = 1$, $\ \red{1 + 1 = 0}$.

$\Z_2^n := $ vectors with $n$ components in $\Z_2$.

E.g. $[0, 1, 1, 0, 1] \in \Z_2^5$.

$[0,1,1] + [1,1,0] = \query{[1,0,1]}$ in $\Z_2^3$.

There are $\query{2^n}$ vectors in $\Z_2^n$.