From Linear Algebra

3: Vectors in Euclidian Space

Basic Definitions

(Euclidean Space)

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

Rn={(x1,x2,…,xn)∣xi∈R}\mathbb{R}^n = \{(x_1, x_2, \ldots, x_n) \mid x_i \in \mathbb{R}\}

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

v=[v1v2⋮vn]\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}

Vector Operations

(Vector Addition)

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

u+v=[u1+v1u2+v2⋮un+vn]\mathbf{u} + \mathbf{v} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \\ \vdots \\ u_n + v_n \end{bmatrix}

Vector addition is commutative and associative:

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

(Scalar Multiplication)

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

cv=[cv1cv2⋮cvn]c\mathbf{v} = \begin{bmatrix} cv_1 \\ cv_2 \\ \vdots \\ cv_n \end{bmatrix}

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

Linear Combinations

(Linear Combination)

A linear combination of vectors v1,v2,…,vk\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k in Rn\mathbb{R}^n is any vector of the form:

c1v1+c2v2+⋯+ckvkc_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k

where c1,c2,…,ck∈Rc_1, c_2, \ldots, c_k \in \mathbb{R} are scalars (called coefficients).

Example: In R2\mathbb{R}^2, let v1=[10]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} and v2=[01]\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.

Then [3−2]=3v1−2v2\begin{bmatrix} 3 \\ -2 \end{bmatrix} = 3\mathbf{v}_1 - 2\mathbf{v}_2 is a linear combination of v1\mathbf{v}_1 and v2\mathbf{v}_2.

(Expressing as Linear Combinations)

Given vectors v1,…,vk\mathbf{v}_1, \ldots, \mathbf{v}_k and a target vector b\mathbf{b}, determining if b\mathbf{b} can be written as a linear combination means solving:

c1v1+c2v2+⋯+ckvk=bc_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{b}

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

  • A=[v1 v2 ⋯ vk]A = [\mathbf{v}_1 \ \mathbf{v}_2 \ \cdots \ \mathbf{v}_k] (matrix with vectors as columns)
  • c=[c1c2⋮ck]\mathbf{c} = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{bmatrix} (coefficients)

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

Span

(Span)

The span of vectors v1,v2,…,vk\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k in Rn\mathbb{R}^n is the set of all linear combinations of these vectors:

span{v1,v2,…,vk}={c1v1+c2v2+⋯+ckvk∣ci∈R}\text{span}\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\} = \{c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k \mid c_i \in \mathbb{R}\}

Geometric Interpretation:

  • span{v}\text{span}\{\mathbf{v}\} in R2\mathbb{R}^2 or R3\mathbb{R}^3 is a line through the origin
  • span{v1,v2}\text{span}\{\mathbf{v}_1, \mathbf{v}_2\} (if not parallel) is a plane through the origin
  • span{v1,v2,v3}\text{span}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\} (if linearly independent) fills all of R3\mathbb{R}^3

(Properties of Span)

  1. Closed under addition and scalar multiplication:

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

  2. Contains the zero vector:

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

  3. Span is a subspace:

    • span{v1,…,vk}\text{span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} is always a subspace of Rn\mathbb{R}^n

(Spanning Sets)

A set of vectors {v1,…,vk}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} spans Rn\mathbb{R}^n if:

span{v1,…,vk}=Rn\text{span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} = \mathbb{R}^n

This means every vector in Rn\mathbb{R}^n can be written as a linear combination of v1,…,vk\mathbf{v}_1, \ldots, \mathbf{v}_k.

Equivalently: The matrix A=[v1 ⋯ vk]A = [\mathbf{v}_1 \ \cdots \ \mathbf{v}_k] has a solution to Ax=bA\mathbf{x} = \mathbf{b} for every b∈Rn\mathbf{b} \in \mathbb{R}^n.

Test: {v1,…,vk}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} spans Rn\mathbb{R}^n ⇔ rref(A)\text{rref}(A) has a pivot in every row.

Examples

Example 1: Span in R2\mathbb{R}^2

Let v1=[12]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} and v2=[24]\mathbf{v}_2 = \begin{bmatrix} 2 \\ 4 \end{bmatrix}.

Does span{v1,v2}=R2\text{span}\{\mathbf{v}_1, \mathbf{v}_2\} = \mathbb{R}^2?

Solution: Note that v2=2v1\mathbf{v}_2 = 2\mathbf{v}_1, so they’re parallel. The span is just a line, not all of R2\mathbb{R}^2.

span{v1,v2}={t[12]∣t∈R}\text{span}\{\mathbf{v}_1, \mathbf{v}_2\} = \left\{t\begin{bmatrix} 1 \\ 2 \end{bmatrix} \mid t \in \mathbb{R}\right\}

Example 2: Linear Combination Check

Can b=[51]\mathbf{b} = \begin{bmatrix} 5 \\ 1 \end{bmatrix} be written as a linear combination of v1=[11]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} and v2=[23]\mathbf{v}_2 = \begin{bmatrix} 2 \\ 3 \end{bmatrix}?

Solution: Solve c1v1+c2v2=bc_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{b}:

c1[11]+c2[23]=[51]c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}

This gives the system:

c1+2c2=5c1+3c2=1\begin{align*} c_1 + 2c_2 &= 5 \\ c_1 + 3c_2 &= 1 \end{align*}

Subtracting: c2=−4c_2 = -4, so c1=13c_1 = 13.

Yes, b=13v1−4v2\mathbf{b} = 13\mathbf{v}_1 - 4\mathbf{v}_2.