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4.2 The Vector Space \(\mathbb{R}^n\) and Subspaces

\(\mathbb{R}^n\): vector space containing vectors with \(n\) components

\(\mathbb{R}^1\): real numbers

A horizontal line representing the 1D real number line with an arrow pointing right labeled x.

\(\mathbb{R}^2\): vectors w/ 2 components

A 2D Cartesian coordinate system with x and y axes intersecting at an origin.

\(\mathbb{R}^3\): \(x, y, z\) coordinate system

\(\vdots\)

\(\mathbb{R}^n\): same idea

The reason we can operate with them (e.g. adding them) the same way we operate with numbers is because they are all in vector spaces.

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Vector space is a space that has the following 8 fundamental properties:

  1. \(\vec{u} + \vec{v} = \vec{v} + \vec{u}\) (e.g. in \(\mathbb{R}^1\), \(1 + 2 = 2 + 1\))
  2. \(\vec{u} + (\vec{v} + \vec{w}) = (\vec{u} + \vec{v}) + \vec{w}\) (e.g. in \(\mathbb{R}^1\), \(1 + (2 + 3) = (1 + 2) + 3\))
  3. There is a zero vector \(\vec{0}\) such that \(\vec{u} + \vec{0} = \vec{u}\)
  4. \(\vec{u} + (-\vec{u}) = (-\vec{u}) + \vec{u} = \vec{0}\) vector has its own additive inverse
  5. \(a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v}\)
  6. \((a + b)\vec{u} = a\vec{u} + b\vec{u}\)
  7. \(a(b\vec{u}) = (ab)\vec{u}\)
  8. There is a 1 (one) such that \(1 \cdot \vec{u} = \vec{u}\)

Other things that are not vectors in traditional sense can also be in vector spaces.

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Examples of Vector Spaces

For example, M2: vector space containing all 2×2 matrices. "I" is the 2×2 identity.

P2: all second degree polynomials

\[ ax^2 + bx + c \]

and lots of others

A lot of times we work with subspaces of known vector spaces

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What is a subspace?

It is a part of a vector space (has all 8 properties) such that it is closed under addition and scalar multiplication.

Closed under addition:

Sum of "vectors" in subspace remains in the same subspace.

Closed under scalar multiplication:

Scalar multiple remains in the same subspace.

Example: \( \mathbb{R}^2 \) is contained in \( \mathbb{R}^3 \)

Two arbitrary \( \mathbb{R}^2 \) vectors in \( \mathbb{R}^2 \) remains in \( \mathbb{R}^2 \)

\[ \vec{u} = \begin{bmatrix} a \\ b \end{bmatrix} \]\[ \vec{v} = \begin{bmatrix} c \\ d \end{bmatrix} \]\[ \vec{u} + \vec{v} = \begin{bmatrix} a+c \\ b+d \end{bmatrix} \]Still some \( \mathbb{R}^2 \) vector

a, b, c, d are numbers

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Vector Spaces and Subspaces

Scalar Multiplication

\[ \vec{u} = \begin{bmatrix} a \\ b \end{bmatrix} \quad k\vec{u} = \begin{bmatrix} ka \\ kb \end{bmatrix} \]

Where \( ka \) and \( kb \) are numbers. The result remains an \( \mathbb{R}^2 \) vector.

Linear Combinations

Equivalently: linear combinations remain in the same space.

→ This implies the zero vector MUST be in a subspace.

ANY subspace MUST have ALL of these:

  1. Closed under addition
  2. Closed under scalar multiplication
  3. Has zero vector
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Example: Subspaces of \( \mathbb{R}^3 \)

Another example: is part of \( \mathbb{R}^3 \) such that \( y = 1 \) a subspace?

3D coordinate axes with a blue plane drawn parallel to the x-z plane, passing through y=1.

Closed under addition?

Let \( \vec{u} = \begin{bmatrix} a \\ 1 \\ b \end{bmatrix} \) and \( \vec{v} = \begin{bmatrix} c \\ 1 \\ d \end{bmatrix} \), where \( a, b, c, d \) are numbers.

\[ \vec{u} + \vec{v} = \begin{bmatrix} a+c \\ 2 \\ b+d \end{bmatrix} \]

This is NOT a vector from that plane!

  • It left the plane!
  • It is NOT closed under addition.

So that plane is NOT a subspace of \( \mathbb{R}^3 \).

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Null Space of a Matrix

An important subspace we often work with is the null space of a matrix.

  • All possible solutions to \( A\vec{x} = \vec{0} \)
  • (All possible \( \vec{x} \) that get sent to \( 0 \))

Example

\[ \begin{aligned} x_1 - 4x_2 - 3x_3 - 7x_4 &= 0 \\ 2x_1 - x_2 + x_3 + 7x_4 &= 0 \\ x_1 + 2x_2 + 3x_3 + 11x_4 &= 0 \end{aligned} \]

This system can be written as \( A\vec{x} = \vec{0} \):

\[ \underbrace{\begin{bmatrix} 1 & -4 & -3 & -7 \\ 2 & -1 & 1 & 7 \\ 1 & 2 & 3 & 11 \end{bmatrix}}_{A} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

(Null space of \( A \) contains ALL vectors the matrix transforms into \( \vec{0} \))

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Solving the System

To find the null space, we solve the homogeneous system using an augmented matrix:

\[ \begin{bmatrix} 1 & -4 & -3 & -7 & 0 \\ 2 & -1 & 1 & 7 & 0 \\ 1 & 2 & 3 & 11 & 0 \end{bmatrix} \]

Performing row reduction (RREF):

\[ \dots \to \begin{bmatrix} \boxed{1} & 0 & 1 & 5 & 0 \\ 0 & \boxed{1} & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \begin{matrix} \leftarrow \text{zero row} \\ \text{no pivots} \end{matrix} \]

Analysis:

  • 4 variables, 2 pivots \( \to \) 2 free variables
  • \( x_3, x_4 \) have no pivots, so they are free

Parametric Solution:

\( x_3 = r \)

\( x_4 = t \)

Row 2: \( \dots \to x_2 = -r - 3t \)

Row 1: \( \dots \to x_1 = -r - 5t \)

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Null Space and Linear Combinations

The vectors the matrix sent to \( \vec{0} \) look like:

\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} -r - 5t \\ -r - 3t \\ r \\ t \end{bmatrix} \]
\[ = r \begin{bmatrix} -1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -5 \\ -3 \\ 0 \\ 1 \end{bmatrix} \]

The matrix \( A \) sends all linear combos of \( \begin{bmatrix} -1 \\ -1 \\ 1 \\ 0 \end{bmatrix} \) and \( \begin{bmatrix} -5 \\ -3 \\ 0 \\ 1 \end{bmatrix} \) to \( \vec{0} \).

Null space holds all these.