Math 136 - Week 1

Chapter 1 - Vectors in Euclidean Space

1.1 Vector Addition and Scalar Multiplication

Instead of points, we work with vectors.

Definition 1.1.1 -

R^{n}

The set $R^{n}$ is defined by $R^{n} = {\vec{x} = [\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix}] | x_{1}, \dots, x_{n} \in R} .$ An element $\vec{v}$ of $R^{n}$ is called a vector.
Two vectors $\vec{x} = [\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix}] \in R^{n}$ and $\vec{y} = [\begin{matrix} y_{1} \\ ⋮ \\ y_{m} \end{matrix}] \in R^{n}$ are said to be equal if $n = m$ and $x_{i} = y_{i}$ for $1 \leq i \leq n .$

In other words, they are said to be equal if the vectors are both in the same space ( $R^{2}$ , $R^{3}$ , etc.) AND if each dimension of each vector is the same.

For example,

[\begin{matrix} 1 \\ 2 \end{matrix}] \neq [\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}]

And,

[\begin{matrix} 1 \\ 2 \end{matrix}] \neq [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]

But,

[\begin{matrix} 1 \\ 2 \end{matrix}] = [\begin{matrix} 1 \\ 2 \end{matrix}]

Remark!

Just because you are working in $R^{6}$ doesn't mean you are working in the $6^{t h}$ dimension. For example, an engineer analyzing the motion of a particle would require $3$ variables for a particle's position and $3$ variables for a particle's velocity. In Economics, one could work with $1000$ different variables and hence working in $R^{1000}$ .

Geometrically, a vector in $R^{n}$ always has its tail at the origin and its head at the corresponding point.

Definition 1.1.2 - Addition and Scalar Multiplication

Let $\vec{x} = [\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix}], \vec{y} = [\begin{matrix} y_{1} \\ ⋮ \\ y_{m} \end{matrix}] \in R^{n}$ and let $c \in R$ (called a scalar).
Addition, $\vec{x}$ , and scalar multiplication, $c \vec{x}$ , are defined by

\vec{x} + \vec{y} = [\begin{matrix} x_{1} + y_{1} \\ ⋮ \\ x_{n} + y_{n} \end{matrix}] and c \vec{x} = [\begin{matrix} c x_{1} \\ ⋮ \\ c x_{n} \end{matrix}]

In linear algebra, a scalar is just a number in $R$ . In the above definition, each $x$ is being scaled by $c$ times.

Remark!

Subtraction is just defined as the addition of a negative. When we write $\vec{x} - \vec{y}$ , we mean

\vec{x} - \vec{y} = \vec{x} + (- 1) \vec{y}

Example 1.1.3

For $\vec{x} = [\begin{matrix} 1 \\ 2 \\ - 3 \end{matrix}]$ and $\vec{y} = [\begin{matrix} - 1 \\ 3 \\ 0 \end{matrix}]$ in $R^{3}$ , we have

\vec{x} + \vec{y} = [\begin{matrix} 1 + (- 1) \\ 2 + 3 \\ - 3 + 0 \end{matrix}] = [\begin{matrix} 0 \\ 5 \\ - 3 \end{matrix}]

\sqrt{2} \vec{y} = \sqrt{2} [\begin{matrix} (- 1) \\ 3 \\ 0 \end{matrix}] = [\begin{matrix} - \sqrt{2} \\ 3 \sqrt{2} \\ 0 \end{matrix}]

2 \vec{x} - 3 \vec{y} = 2 [\begin{matrix} 1 \\ 2 \\ - 3 \end{matrix}] + (- 3) [\begin{matrix} - 1 \\ 3 \\ 0 \end{matrix}] = [\begin{matrix} 2 \\ 4 \\ - 6 \end{matrix}] + [\begin{matrix} 3 \\ - 9 \\ 0 \end{matrix}] = [\begin{matrix} 5 \\ - 5 \\ - 6 \end{matrix}]

Exercise 1.1.5

Add $\vec{x} = [\begin{matrix} 3 \\ 2 \end{matrix}]$ and $\vec{y} = [\begin{matrix} - 4 \\ - 1 \end{matrix}]$ in $R^{2}$ . Then, express the results geometrically.

\vec{x} + \vec{y} = [\begin{matrix} 3 \\ 2 \end{matrix}] + [\begin{matrix} - 4 \\ - 1 \end{matrix}] = [\begin{matrix} - 1 \\ 1 \end{matrix}]

Geometrically, We can express it as:
{8EBDB53A-B178-43E5-BBA2-EEDF58DE8EF8}.png

Definition 1.1.6 - Linear Combination

For ${\vec{v}}_{1}, \dots, {\vec{v}}_{k} \in R^{n}$ and $c_{1}, \dots, c_{k} \in R,$ we call the expression

c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} + \dots + c_{k} {\vec{v}}_{k}

a linear combination of ${\vec{v}}_{1}, \dots, {\vec{v}}_{k}$ .

This definition IS EXTREMELY IMPORTANT!!!!

Theorem 1.1.7

If $\vec{x}, \vec{y}, \vec{w} \in R^{n}$ and $c, d \in R$ , then:

\begin{aligned} V 1 & \vec{x} + \vec{y} \in R^{n} \\ V 2 & (\vec{x} + \vec{y}) + \vec{w} = \vec{x} + (\vec{y} + \vec{w}) \\ V 3 & \vec{x} + \vec{y} = \vec{y} + \vec{x} \\ V 4 & There exists \vec{0} \in R^{n} s.t. \vec{x} + \vec{0} = \vec{x} for all \vec{x} \in R^{n} \\ V 5 & For every \vec{x} \in R^{n} there exists (- \vec{x}) \in R^{n} s.t. \vec{x} + (- \vec{x}) = \vec{0} \\ V 6 & c \vec{x} \in R^{n} \\ V 7 & c (d \vec{x}) = (c d) \vec{x} \\ V 8 & (c + d) \vec{x} = c \vec{x} + d \vec{x} \\ V 9 & c (\vec{x} + \vec{y}) = c \vec{x} + c \vec{y} \\ V 10 & 1 \vec{x} = \vec{x} \end{aligned}

Proofs:
Let $\vec{x} = [\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix}], \vec{y} = [\begin{matrix} y_{1} \\ ⋮ \\ y_{n} \end{matrix}], \vec{w} = [\begin{matrix} w_{1} \\ ⋮ \\ w_{n} \end{matrix}] \in R^{n}$ and $c, d, \in R$ .

$V1:$

\begin{aligned} \vec{x} + \vec{y} & = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + [\begin{array}{c} y_{1} \\ ⋮ \\ y_{n} \end{array}] \\ = [\begin{array}{c} x_{1} + y_{1} \\ ⋮ \\ x_{n} + y_{n} \end{array}] \end{aligned}

Since $x_{1} + y_{1}, \dots, x_{n} + y_{n} \in R$ , by definition, $\vec{x} + \vec{y} \in R^{n}$ , as desired.

$V2:$
$L H S :$

\begin{aligned} (\vec{x} + \vec{y}) + \vec{w} & = ([\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + [\begin{array}{c} y_{1} \\ ⋮ \\ y_{n} \end{array}]) + [\begin{array}{c} w_{1} \\ ⋮ \\ w_{n} \end{array}] \\ = ([\begin{array}{c} x_{1} + y_{1} \\ ⋮ \\ x_{n} + y_{n} \end{array}]) + [\begin{array}{c} w_{1} \\ ⋮ \\ w_{n} \end{array}] \\ = [\begin{array}{c} (x_{1} + y_{1}) + w_{1} \\ ⋮ \\ (x_{n} + y_{n}) + w_{n} \end{array}] \\ = [\begin{array}{c} x_{1} + y_{1} + w_{1} \\ ⋮ \\ x_{n} + y_{n} + w_{n} \end{array}] \end{aligned}

$R H S :$

\begin{aligned} \vec{x} + (\vec{y} + \vec{w}) & = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + ([\begin{array}{c} y_{1} \\ ⋮ \\ y_{n} \end{array}] + [\begin{array}{c} w_{1} \\ ⋮ \\ w_{n} \end{array}]) \\ = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + ([\begin{array}{c} y_{1} + w_{1} \\ ⋮ \\ y_{n} + w_{n} \end{array}]) \\ = [\begin{array}{c} x_{1} + (y_{1} + w_{1}) \\ ⋮ \\ x_{n} + (y_{n} + w_{n}) \end{array}] \\ = [\begin{array}{c} x_{1} + y_{1} + w_{1} \\ ⋮ \\ x_{n} + y_{n} + w_{n} \end{array}] \end{aligned}

$L H S = R H S$ as desired.

$V3:$

\begin{aligned} \vec{x} + \vec{y} & = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + [\begin{array}{c} y_{1} \\ ⋮ \\ y_{n} \end{array}] \\ = [\begin{array}{c} x_{1} + y_{1} \\ ⋮ \\ x_{n} + y_{n} \end{array}] \\ = [\begin{array}{c} y_{1} + x_{1} \\ ⋮ \\ y_{n} + x_{n} \end{array}] since addition of real numbers is commutative \\ = \vec{y} + \vec{x} \end{aligned}

$V4:$
Let $\vec{0} \in R^{n}$ . Then,

\begin{aligned} \vec{x} + \vec{0} & = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + [\begin{array}{c} 0 \\ ⋮ \\ 0 \end{array}] \\ = [\begin{array}{c} x_{1} + 0 \\ ⋮ \\ x_{n} + 0 \end{array}] \\ = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] \\ = \vec{x} \end{aligned}

As desired.

$V5:$
Let $(- \vec{x} \in R^{n})$ . Then,

\begin{aligned} \vec{x} + (- \vec{x}) & = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + (- 1) [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] \\ = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + [\begin{array}{c} - x_{1} \\ ⋮ \\ - x_{n} \end{array}] \\ = [\begin{array}{c} x_{1} - x_{1} \\ ⋮ \\ x_{n} - x_{n} \end{array}] \\ = [\begin{array}{c} 0 \\ ⋮ \\ 0 \end{array}] \\ = \vec{0} \end{aligned}

As desired.

$V6:$

\begin{aligned} c \vec{x} & = c [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] \\ = [\begin{array}{c} c x_{1} \\ ⋮ \\ c x_{n} \end{array}] \\ \in R^{n} since c x_{i} \in R for all 1 \leq i \leq n . \end{aligned}

$V7:$

\begin{aligned} c (d \vec{x}) & = c (d [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}]) \\ = c ([\begin{array}{c} d x_{1} \\ ⋮ \\ d x_{n} \end{array}]) \\ = c [\begin{array}{c} d x_{1} \\ ⋮ \\ d x_{n} \end{array}] \\ = [\begin{array}{c} c (d x_{1}) \\ ⋮ \\ c (d x_{n}) \end{array}] \\ = [\begin{array}{c} d (c x_{1}) \\ ⋮ \\ d (c x_{n}) \end{array}] \\ = d [\begin{array}{c} c x_{1} \\ ⋮ \\ c x_{n} \end{array}] \\ = d (c [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}]) \\ = d (c \vec{x}) \end{aligned}

As desired.

$V8:$

\begin{aligned} (c + d) \vec{x} & = (c + d) [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] \\ = [\begin{array}{c} (c + d) x_{1} \\ ⋮ \\ (c + d) x_{n} \end{array}] \\ = [\begin{array}{c} c x_{1} + d x_{1} \\ ⋮ \\ c x_{n} + d x_{n} \end{array}] \\ = [\begin{array}{c} c x_{1} \\ ⋮ \\ c x_{n} \end{array}] + [\begin{array}{c} d x_{1} \\ ⋮ \\ d x_{n} \end{array}] \\ = c [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + d [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] \\ = c \vec{x} + d \vec{x} \end{aligned}

As desired.

$V9:$

\begin{aligned} c (\vec{x} + \vec{y}) & = c [\begin{array}{c} x_{1} + y_{1} \\ ⋮ \\ x_{n} + y_{n} \end{array}] \\ = [\begin{array}{c} c (x_{1} + y_{1}) \\ ⋮ \\ c (x_{n} + y_{n}) \end{array}] \\ = [\begin{array}{c} c x_{1} + c y_{1} \\ ⋮ \\ c x_{n} + c y_{n} \end{array}] \\ = [\begin{array}{c} c x_{1} \\ ⋮ \\ c x_{n} \end{array}] + [\begin{array}{c} c y_{1} \\ ⋮ \\ c y_{n} \end{array}] \\ = c [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] + c [\begin{array}{c} y_{1} \\ ⋮ \\ y_{n} \end{array}] \\ = c \vec{x} + c \vec{y} \end{aligned}

As desired.

$V10:$

\begin{aligned} 1 \vec{x} & = 1 [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] \\ = [\begin{array}{c} (1) x_{1} \\ ⋮ \\ (1) x_{n} \end{array}] & by addition \\ = [\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}] & rules of multiplication on R \\ = x_{n} \end{aligned}

As desired.

1.1 - Problems

Compute the following linear combinations.
1. $2 [\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix}] + 3 [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}]$ .
  Solution:
$\begin{aligned} 2 [\begin{array}{c} 2 \\ 1 \\ - 1 \end{array}] + 3 [\begin{array}{c} - 1 \\ 0 \\ 1 \end{array}] & = [\begin{array}{c} 4 \\ 2 \\ - 2 \end{array}] + [\begin{array}{c} - 3 \\ 0 \\ 3 \end{array}] \\ = [\begin{array}{c} 1 \\ 2 \\ 1 \end{array}] \end{aligned}$
1. $\frac{1}{3} [\begin{matrix} 3 \\ 1 \\ - 2 \end{matrix}] - [\begin{matrix} 4 \\ \frac{1}{2} \\ - \frac{2}{3} \end{matrix}]$ .
  Solution:
\displaystyle\frac{1}{3}\begin{bmatrix}3 \ 1 \ -2\end{bmatrix} - \begin{bmatrix}4 \ \frac{1}{2} \-\frac{2}{3}\end{bmatrix} &= \begin{bmatrix}1 \ \frac{1}{3} \ -\frac{2}{3}\end{bmatrix} + \begin{bmatrix}-4 \ -\frac{1}{2} \\frac{2}{3}\end{bmatrix} \
&= \begin{bmatrix} -3 \ -\frac{1}{6} \ 0
\end{bmatrix}
\end{aligned} \Solution:\sqrt{ 2 }\begin{bmatrix}\sqrt{ 2 } \ 0 \ \sqrt{ 2 }\end{bmatrix} + \displaystyle\frac{2}{3}\begin{bmatrix}6 \ 0 \ 0 \end{bmatrix} &= \begin{bmatrix} 2 \ 0 \ 2\end{bmatrix} + \begin{bmatrix}4 \ 0 \ 0 \end{bmatrix} \
&= \begin{bmatrix}
6 \ 0 \ 2
\end{bmatrix}
\endSolution:(a+b) \begin{bmatrix} 1 \ 1 \ -1\end{bmatrix} + (-a+b) \begin{bmatrix} 1 \ 2 \ -3\end{bmatrix} + (-a+2b) \begin{bmatrix} -1 \ -1 \ 2\end{bmatrix} &= \begin{bmatrix} (a+b) \ (a+b) \ -(a+b)\end{bmatrix} + \begin{bmatrix} (-a+b) \ 2(-a+b) \ -3(-a+b)\end{bmatrix} + \begin{bmatrix} -(-a+2b) \ -(-a+2b) \ 2(-a+2b)\end{bmatrix} \
&= \begin{bmatrix} a-a+a+b+b-2b \ a -2a+a+b+2b-2b \ -a+3a-2a-b-3b+4b\end{bmatrix} \
&= \begin{bmatrix}
a \ b \ 0
\end{bmatrix}
\end
1. $2 \vec{x} + 2 \vec{v} = 4 \vec{x}$ .
  Solution:
2\vec{x} + 2\vec{v} &= 4\vec{x} \
2\vec{x} + 2(\vec{-x}) + 2\vec{v} &= 4\vec{x} + 2(-\vec{x}) && \text{by V5 and addition} \
2(\vec{x} + (-\vec{x})) + 2\vec{v} &= 2(2\vec{x} + (-\vec{x})) && \text{by V9} \
2(\vec{0}) + 2\vec{v} &= 2(\vec{x}) && \text{by V5 and addition of vectors}\
\vec{0} + 2\vec{v} &= 2\vec{x} && \text{by scalar multiplication} \
2\vec{v}\left( \frac{1}{2} \right)&=2\vec{x}\left( \frac{1}{2} \right) && \text{by V4} \
1\vec{v} &=1\vec{x} && \text{by V7} \
\vec{x} &= \vec{v} && \text{by V10}
\endSolution:\vec{x} + 2\vec{v} &= \vec{v} + (-\vec{x}) \
\vec{x} + \vec{x} + 2\vec{v} + 2(-\vec{v}) &= \vec{v} + 2(-\vec{v}) + -\vec{x} + \vec{x} \
2\vec{x} + 2(\vec{v} + (-\vec{v})) &= \vec{v} + (-2\vec{v}) + \vec{0} && \text{by vector addition, V9, V5} \
2\vec{x} + 2(\vec{0}) &= (-\vec{v}) && \text{by V5, vector addition, V4} \
2\vec{x} + \vec{0} &= (-\vec{v}) && \text{by scalar multiplication} \
2\vec{x} \left( \frac{1}{2} \right) &= (-\vec{v})\left( \frac{1}{2} \right) && \text{by V4} \
1\vec{x} &= \left( -\frac{1}{2} \right) (\vec{v}) && \text{by V7} \
\vec{x} &= \left( -\frac{1}{2} \right)(\vec{v}) && \text{by V10}
\end
Looking at the area of the parallelogram formed by the parallelogram rule for addition on $\vec{x}, \vec{y} \in R^{2}$ .
1. Calculate each of the following and show the result geometrically. Find the area of each parallelogram formed by the parallelogram rule for addition.
  1. $[\begin{matrix} - 1 \\ 1 \end{matrix}] + [\begin{matrix} 0 \\ 2 \end{matrix}] .$
    Solution:
  $b a s e \times h = 1 \times 2 = 2$
  
  2. $[\begin{matrix} 2 \\ - 1 \end{matrix}] + [\begin{matrix} - 2 \\ 1 \end{matrix}]$ .
  Solution: $0$
2. What condition must be placed on $\vec{x}$ and $\vec{y}$ to ensure that the parallelogram has non-zero area?
  Solution: $\vec{x}, \vec{y} \neq 0$ and geometrically, $\vec{x} and \vec{y}$ can't be on the same line together.
3. Create two more pairs $\vec{x}, \vec{y} \in R^{2}$ that form a parallelogram with non-zero area. Find the area of each parallelogram.
  Solution:
  1. $[\begin{matrix} 2 \\ 0 \end{matrix}], [\begin{matrix} 0 \\ 1 \end{matrix}] .$
    Area: $2 \times 1 = 2$
  2. $[\begin{matrix} - 0.5 \\ 0 \end{matrix}], [\begin{matrix} 1 \\ 1 \end{matrix}] .$
    Area:
  $2 [\frac{(- 0.5 + 1) \times 1}{2}] = 0.5$
4. Find a general formula for the area of the parallelogram determined by $\vec{x}, \vec{y} \in R^{2}$ . Prove that your formula is correct.
  Solution:
  Let $\vec{x} = [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] and \vec{y} = [\begin{matrix} y_{1} \\ y_{2} \end{matrix}]$ .
  Area = $| x_{1} y_{1} - x_{2} y_{2} |$