Vectors in \( \mathbb{R}^n \)

In linear algebra, \( \mathbb{R}^n \) serves as a foundational vector space. Understanding how n-tuple vectors are defined, represented as column matrices, and manipulated via vector addition and scalar multiplication is essential for advanced linear spaces.

Table of Contents

Definition of Vectors in \( \mathbb{R}^n \)

A vector in \( \mathbb{R}^n \) is an \( n \)-tuple (an ordered list of \( n \) elements) written as \( (x_1, x_2, \dots, x_n) \), where each \( x_i \) is a real number. For any positive integer \( n \), \( \mathbb{R}^n \) forms a vector space.

Examples:

A vector \( \mathbf{u} \) in \( \mathbb{R}^n \) can also be represented as a column matrix (an \( n \times 1 \) matrix):

\[ \mathbf{u} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{bmatrix} \]

The real numbers \( x_1, x_2, \dots, x_n \) are called the components (or entries) of the vector.

Vector Operations and Axioms

Vector Operations in \( \mathbb{R}^n \):

For any vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) in \( \mathbb{R}^n \) and real numbers \( r, s \):

  1. \( \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} \) (Commutativity)
  2. \( (\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w}) \) (Associativity of addition)
  3. \( r \cdot (s \cdot \mathbf{u}) = (rs) \cdot \mathbf{u} \) (Associativity of scalars)
  4. \( \mathbf{u} + \mathbf{0} = \mathbf{u} \) (Zero vector identity)
  5. \( \mathbf{u} + (-\mathbf{u}) = \mathbf{0} \) (Negative vector inverse)
  6. \( r \cdot (\mathbf{u} + \mathbf{v}) = r\mathbf{u} + r\mathbf{v} \) (Distributivity over vector addition)
  7. \( (r + s) \cdot \mathbf{u} = r\mathbf{u} + s\mathbf{u} \) (Distributivity over scalar addition)
  8. \( 1 \cdot \mathbf{u} = \mathbf{u} \) (Scalar identity)

These eight axioms, together with the two closure properties (closure under addition and closure under scalar multiplication), confirm that \( \mathbb{R}^n \) is a vector space.

Examples with Solutions

Example 1: Evaluating Vector Expressions

Show Solution

Evaluate the following expressions and express the answers as single column vectors:

a) \( 2 \begin{bmatrix} 2 \\ -3 \\ 0 \\ 8 \\ -9 \end{bmatrix} + 3 \begin{bmatrix} -7 \\ 0 \\ -4 \\ 2 \\ 1 \end{bmatrix} \)
b) \( -2 \left( \begin{bmatrix} 0 \\ 1 \\ -1 \\ 4 \\ -5 \end{bmatrix} - 2 \begin{bmatrix} -1 \\ 2 \\ 7 \\ 0 \\ -1 \end{bmatrix} \right) + 2 \begin{bmatrix} 0 \\ -2 \\ -2 \\ 1 \\ -1 \end{bmatrix} \)

Solution

a) Apply scalar multiplication first:

\[ = \begin{bmatrix} 4 \\ -6 \\ 0 \\ 16 \\ -18 \end{bmatrix} + \begin{bmatrix} -21 \\ 0 \\ -12 \\ 6 \\ 3 \end{bmatrix} \]

Now add the corresponding components:

\[ = \begin{bmatrix} 4 - 21 \\ -6 + 0 \\ 0 - 12 \\ 16 + 6 \\ -18 + 3 \end{bmatrix} = \begin{bmatrix} -17 \\ -6 \\ -12 \\ 22 \\ -15 \end{bmatrix} \]

b) Simplify inside the parentheses first (scalar multiply and vector subtract):

\[ = -2 \left( \begin{bmatrix} 0 \\ 1 \\ -1 \\ 4 \\ -5 \end{bmatrix} + \begin{bmatrix} 2 \\ -4 \\ -14 \\ 0 \\ 2 \end{bmatrix} \right) + 2 \begin{bmatrix} 0 \\ -2 \\ -2 \\ 1 \\ -1 \end{bmatrix} \]

Add vectors inside the parentheses:

\[ = -2 \begin{bmatrix} 2 \\ -3 \\ -15 \\ 4 \\ -3 \end{bmatrix} + 2 \begin{bmatrix} 0 \\ -2 \\ -2 \\ 1 \\ -1 \end{bmatrix} \]

Apply scalar multiplication and add the resulting vectors:

\[ = \begin{bmatrix} -4 \\ 6 \\ 30 \\ -8 \\ 6 \end{bmatrix} + \begin{bmatrix} 0 \\ -4 \\ -4 \\ 2 \\ -2 \end{bmatrix} = \begin{bmatrix} -4 \\ 2 \\ 26 \\ -6 \\ 4 \end{bmatrix} \]

Example 2: Solving Vector Equations for Unknown Vectors

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Find the unknown vector \( \mathbf{u} \) in each equation:

a) \( 2\mathbf{u} = 2 \begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix} - 3 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \)
b) \( -3 \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} - \mathbf{u} = -2 \begin{bmatrix} -1 \\ 3 \\ 4 \end{bmatrix} \)

Solution

a) Simplify the right side using scalar multiplication and vector subtraction:

\[ 2\mathbf{u} = \begin{bmatrix} 8 \\ 12 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 8 \\ 9 \\ 0 \end{bmatrix} \]

Multiply both sides by \( \frac{1}{2} \):

\[ \mathbf{u} = \frac{1}{2} \begin{bmatrix} 8 \\ 9 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 9/2 \\ 0 \end{bmatrix} \]

b) Scalar multiply the vectors:

\[ \begin{bmatrix} -3 \\ -6 \\ 3 \end{bmatrix} - \mathbf{u} = \begin{bmatrix} 2 \\ -6 \\ -8 \end{bmatrix} \]

Isolate \( \mathbf{u} \) by adding the negative of the left-hand vector to both sides:

\[ -\mathbf{u} = \begin{bmatrix} 2 \\ -6 \\ -8 \end{bmatrix} - \begin{bmatrix} -3 \\ -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 - (-3) \\ -6 - (-6) \\ -8 - 3 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \\ -11 \end{bmatrix} \]

Multiply by \( -1 \) to solve for \( \mathbf{u} \):

\[ \mathbf{u} = \begin{bmatrix} -5 \\ 0 \\ 11 \end{bmatrix} \]

Verification: \( -3\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} - \begin{bmatrix} -5 \\ 0 \\ 11 \end{bmatrix} = \begin{bmatrix} -3 \\ -6 \\ 3 \end{bmatrix} + \begin{bmatrix} 5 \\ 0 \\ -11 \end{bmatrix} = \begin{bmatrix} 2 \\ -6 \\ -8 \end{bmatrix} = -2\begin{bmatrix} -1 \\ 3 \\ 4 \end{bmatrix} \) ✓

Example 3: Finding Scalar Coefficients in Linear Combinations

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Find the real numbers \( r \) and \( s \) such that:

a) \( r \begin{bmatrix} 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix} \)
b) \( r \begin{bmatrix} 1 \\ 2 \end{bmatrix} + s \begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix} \)

Solution

a) Perform scalar multiplication and vector addition on the left:

\[ \begin{bmatrix} r \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ s \end{bmatrix} = \begin{bmatrix} r \\ s \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix} \]

By vector equality, component-wise comparison yields:

\[ r = 2, \quad s = -3 \]

b) Scalar multiply and combine the left side:

\[ \begin{bmatrix} r \\ 2r \end{bmatrix} + \begin{bmatrix} 2s \\ 2s \end{bmatrix} = \begin{bmatrix} r + 2s \\ 2r + 2s \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix} \]

Set up the system of linear equations from corresponding components:

\[ \begin{cases} r + 2s = -1 \\ 2r + 2s = 3 \end{cases} \]

Subtract the first equation from the second equation to eliminate \( s \):

\[ (2r + 2s) - (r + 2s) = 3 - (-1) \implies r = 4 \]

Substitute \( r = 4 \) into the first equation:

\[ 4 + 2s = -1 \implies 2s = -5 \implies s = -\frac{5}{2} \]

The constants are \( r = 4 \) and \( s = -\frac{5}{2} \).

Verification: \( 4\begin{bmatrix} 1 \\ 2 \end{bmatrix} - \frac{5}{2}\begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 8 \end{bmatrix} - \begin{bmatrix} 5 \\ 5 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix} \) ✓

Frequently Asked Questions

What is a vector in \( \mathbb{R}^n \)?

A vector in \( \mathbb{R}^n \) is an ordered n-tuple \( (x_1, x_2, \dots, x_n) \) where each \( x_i \) is a real number. It can also be represented as a column matrix with \( n \) rows.

How do you add two vectors in \( \mathbb{R}^n \)?

To add two vectors in \( \mathbb{R}^n \), add their corresponding components: \( (x_1, x_2, \dots, x_n) + (y_1, y_2, \dots, y_n) = (x_1+y_1, x_2+y_2, \dots, x_n+y_n) \).

How do you multiply a vector in \( \mathbb{R}^n \) by a scalar?

To multiply a vector by a scalar \( k \), multiply each component by \( k \): \( k(x_1, x_2, \dots, x_n) = (kx_1, kx_2, \dots, kx_n) \).

What is the zero vector in \( \mathbb{R}^n \)?

The zero vector in \( \mathbb{R}^n \) is the vector with all components equal to zero: \( \mathbf{0} = (0, 0, \dots, 0) \). It acts as the additive identity, so \( \mathbf{u} + \mathbf{0} = \mathbf{u} \) for any vector \( \mathbf{u} \in \mathbb{R}^n \).

Is \( \mathbb{R}^n \) a vector space?

Yes. For any positive integer \( n \), \( \mathbb{R}^n \) forms a vector space under standard vector addition and scalar multiplication, satisfying all ten vector space axioms.

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