3: Vectors in Euclidian Space

Basic Definitions

(Euclidean Space)

The Euclidean space $\mathbb{R}^n$ is the set of all ordered $n$-tuples of real numbers:

Vectors in $\mathbb{R}^n$ can be written as column vectors or row vectors. We typically use column notation:

Vector Operations

(Vector Addition)

Given vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$, their sum is:

Vector addition is commutative and associative:

• $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
• $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$

(Scalar Multiplication)

Given a scalar $c \in \mathbb{R}$ and vector $\mathbf{v} \in \mathbb{R}^n$:

Scalar multiplication stretches or shrinks vectors (and reverses direction if $c < 0$).

Linear Combinations

(Linear Combination)

A linear combination of vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ in $\mathbb{R}^n$ is any vector of the form:

where $c_1, c_2, \ldots, c_k \in \mathbb{R}$ are scalars (called coefficients).

Example: In $\mathbb{R}^2$, let $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

Then $\begin{bmatrix} 3 \\ -2 \end{bmatrix} = 3\mathbf{v}_1 - 2\mathbf{v}_2$ is a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$.

(Expressing as Linear Combinations)

Given vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ and a target vector $\mathbf{b}$, determining if $\mathbf{b}$ can be written as a linear combination means solving:

This is equivalent to solving the linear system $A\mathbf{c} = \mathbf{b}$, where:

• $A = [\mathbf{v}_1 \ \mathbf{v}_2 \ \cdots \ \mathbf{v}_k]$ (matrix with vectors as columns)
• $\mathbf{c} = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{bmatrix}$ (coefficients)

The system has a solution ⇔ $\mathbf{b}$ is a linear combination of the vectors.

Span

(Span)

The span of vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ in $\mathbb{R}^n$ is the set of all linear combinations of these vectors:

Geometric Interpretation:

• $\text{span}\{\mathbf{v}\}$ in $\mathbb{R}^2$ or $\mathbb{R}^3$ is a line through the origin
• $\text{span}\{\mathbf{v}_1, \mathbf{v}_2\}$ (if not parallel) is a plane through the origin
• $\text{span}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ (if linearly independent) fills all of $\mathbb{R}^3$

(Properties of Span)

1. Closed under addition and scalar multiplication:

   • If $\mathbf{u}, \mathbf{v} \in \text{span}\{S\}$, then $\mathbf{u} + \mathbf{v} \in \text{span}\{S\}$
   • If $\mathbf{v} \in \text{span}\{S\}$ and $c \in \mathbb{R}$, then $c\mathbf{v} \in \text{span}\{S\}$

2. Contains the zero vector:

   • $\mathbf{0} \in \text{span}\{S\}$ for any set $S$ (set all coefficients to 0)

3. Span is a subspace:

   • $\text{span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ is always a subspace of $\mathbb{R}^n$

(Spanning Sets)

A set of vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ spans $\mathbb{R}^n$ if:

This means every vector in $\mathbb{R}^n$ can be written as a linear combination of $\mathbf{v}_1, \ldots, \mathbf{v}_k$.

Equivalently: The matrix $A = [\mathbf{v}_1 \ \cdots \ \mathbf{v}_k]$ has a solution to $A\mathbf{x} = \mathbf{b}$ for every $\mathbf{b} \in \mathbb{R}^n$.

Test: $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ spans $\mathbb{R}^n$ ⇔ $\text{rref}(A)$ has a pivot in every row.

Examples

Example 1: Span in $\mathbb{R}^2$

Let $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$.

Does $\text{span}\{\mathbf{v}_1, \mathbf{v}_2\} = \mathbb{R}^2$?

Solution: Note that $\mathbf{v}_2 = 2\mathbf{v}_1$, so they’re parallel. The span is just a line, not all of $\mathbb{R}^2$.

Example 2: Linear Combination Check

Can $\mathbf{b} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$ be written as a linear combination of $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$?

Solution: Solve $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{b}$:

This gives the system:

Subtracting: $c_2 = -4$, so $c_1 = 13$.

Yes, $\mathbf{b} = 13\mathbf{v}_1 - 4\mathbf{v}_2$.