Vector Spaces

Overview

Welcome to the foundational chapter on Vector Spaces, a cornerstone of Linear Algebra and an absolutely critical topic for your MSQMS journey at ISI. While seemingly abstract, the concepts introduced here provide the fundamental language and framework for understanding a vast array of mathematical and statistical tools. A strong grasp of vector spaces is indispensable, as it underpins everything from solving systems of linear equations and understanding optimization problems to advanced topics in econometrics, machine learning, and functional analysis. For competitive exams like ISI's entrance, conceptual clarity in this chapter is non-negotiable, as questions often test your ability to apply definitions rigorously and reason about abstract structures.

This chapter will equip you with the essential tools to define, identify, and work with vector spaces. You will learn to recognize the inherent structure within sets of objects that allows for linear operations, which is crucial for representing and manipulating data, understanding solution spaces, and building mathematical models. The ability to articulate and apply the axioms of a vector space, along with understanding how vectors combine and generate entire spaces, is a skill that will be repeatedly tested and required throughout your curriculum.

Mastering these initial concepts is not just about memorizing definitions; it's about developing a robust problem-solving intuition that is highly valued at ISI. The problems you encounter will often require you to bridge the gap between abstract definitions and concrete examples, demanding both precision and flexibility in your mathematical thinking. A solid foundation in vector spaces will significantly ease your progression through subsequent chapters on basis, dimension, linear transformations, and inner product spaces, directly impacting your performance in examinations and future research.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Definition of Vector Space | Axioms defining algebraic structure. |
| 2 | Linear Combinations and Span | Building and generating sets of vectors. |
| 3 | Subspaces | Identifying structured subsets within vector spaces. |

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Learning Objectives

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Now let's begin with Definition of Vector Space...
## Part 1: Definition of Vector Space

Introduction

In linear algebra, the concept of a vector space is fundamental. It generalizes the familiar idea of vectors in two-dimensional or three-dimensional space to a broader, more abstract setting. Essentially, a vector space is a collection of objects, called vectors, that can be added together and multiplied ("scaled") by numbers, called scalars, satisfying certain conditions.

Understanding vector spaces is crucial for ISI as it forms the bedrock for advanced topics like linear transformations, eigenvalues, eigenvectors, and inner product spaces, which are frequently tested. This topic allows us to analyze various mathematical structures, from sets of real numbers to spaces of polynomials and functions, under a unified framework.

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Key Concepts

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## 1. Vectors in $\mathbb{R}^n$: Basic Operations and Properties

Before delving into the abstract definition, let's review concrete examples of vectors in $\mathbb{R}^n$, which serve as the primary motivation for the abstract concept.

#
### a. Vector Addition and Scalar Multiplication in $\mathbb{R}^n$

If $\vec{u} = (u_1, u_2, \ldots, u_n)$ and $\vec{v} = (v_1, v_2, \ldots, v_n)$ are vectors in $\mathbb{R}^n$, and $c$ is a scalar (a real number), then:

Vector Addition:

Scalar Multiplication:

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### b. Magnitude of a Vector

The magnitude (or length or norm) of a vector $\vec{v} = (v_1, v_2, \ldots, v_n)$ in $\mathbb{R}^n$ is denoted by $|\vec{v}|$ and is calculated as:

An important property derived from the dot product (explained next) is that the square of the magnitude of a vector is equal to its dot product with itself:

#
### c. Unit Vectors

Worked Example:

Problem: Find the unit vector in the direction of $\vec{v} = (3, -4)$.

Solution:

Step 1: Calculate the magnitude of $\vec{v}$.

Step 2: Divide the vector by its magnitude to find the unit vector.

Answer: $\left(\frac{3}{5}, -\frac{4}{5}\right)$

#
### d. Dot Product (Scalar Product)

The dot product is an operation that takes two vectors and returns a scalar.

Properties of Dot Product:
Let $\vec{u}, \vec{v}, \vec{w}$ be vectors in $\mathbb{R}^n$ and $c$ be a scalar.

Commutative Property: $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$

Distributive Property: $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$

Scalar Multiplication Property: $(c\vec{u}) \cdot \vec{v} = c(\vec{u} \cdot \vec{v}) = \vec{u} \cdot (c\vec{v})$

Self Dot Product: $\vec{u} \cdot \vec{u} = |\vec{u}|^2 \ge 0$. Also, $\vec{u} \cdot \vec{u} = 0 \iff \vec{u} = \vec{0}$.

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#
## 2. Field of Scalars

A vector space is always defined over a field of scalars. A field is a set with two operations (addition and multiplication) that satisfy certain axioms (associativity, commutativity, distributivity, existence of identity elements, and inverses for non-zero elements). Common fields in linear algebra are:

• The set of real numbers, $\mathbb{R}$.


• The set of complex numbers, $\mathbb{C}$.


• The set of rational numbers, $\mathbb{Q}$.

Throughout this topic, we will denote the field of scalars as $F$.

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#
## 3. Abstract Definition of a Vector Space

📖 Vector Space

A vector space $V$ over a field $F$ is a set $V$ (whose elements are called vectors) together with two operations:

• Vector Addition: An operation that takes two vectors $\vec{u}, \vec{v} \in V$ and produces another vector $\vec{u} + \vec{v} \in V$.
  
• Scalar Multiplication: An operation that takes a scalar $c \in F$ and a vector $\vec{v} \in V$ and produces another vector $c\vec{v} \in V$.

  These two operations must satisfy the following ten axioms for all $\vec{u}, \vec{v}, \vec{w} \in V$ and all $a, b \in F$:

  Axioms for Vector Addition:
  

  • Closure under Addition: For any $\vec{u}, \vec{v} \in V$, the sum $\vec{u} + \vec{v}$ is in $V$.
    
  • Commutativity of Addition: $\vec{u} + \vec{v} = \vec{v} + \vec{u}$.
    
  • Associativity of Addition: $(\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$.
    
  • Existence of Zero Vector: There exists a unique zero vector $\vec{0} \in V$ such that $\vec{v} + \vec{0} = \vec{v}$ for all $\vec{v} \in V$.
    
  • Existence of Additive Inverse: For every $\vec{v} \in V$, there exists a unique vector $-\vec{v} \in V$ such that $\vec{v} + (-\vec{v}) = \vec{0}$.

    Axioms for Scalar Multiplication:
    

    • Closure under Scalar Multiplication: For any $c \in F$ and $\vec{v} \in V$, the product $c\vec{v}$ is in $V$.
      
    • Distributivity over Vector Addition: $c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$.
      
    • Distributivity over Scalar Addition: $(a+b)\vec{v} = a\vec{v} + b\vec{v}$.
      
    • Associativity of Scalar Multiplication: $(ab)\vec{v} = a(b\vec{v})$.
      
    • Multiplicative Identity: For the multiplicative identity $1 \in F$, $1\vec{v} = \vec{v}$ for all $\vec{v} \in V$.

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#
## 4. Properties Derived from Axioms

Several important properties can be deduced directly from the ten axioms:

Unique Zero Vector: The zero vector $\vec{0}$ is unique.

Unique Additive Inverse: For each vector $\vec{v}$, its additive inverse $-\vec{v}$ is unique.

Zero Scalar Property: For any $\vec{v} \in V$, $0\vec{v} = \vec{0}$, where $0$ is the zero scalar in $F$.

Zero Vector Multiplication Property: For any $c \in F$, $c\vec{0} = \vec{0}$.

Negative Scalar Property: For any $\vec{v} \in V$, $(-1)\vec{v} = -\vec{v}$.

Cancellation Law for Scalar Multiplication: If $c\vec{v} = \vec{0}$ then either $c=0$ or $\vec{v}=\vec{0}$.

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#
## 5. Examples of Vector Spaces

Here are some common examples of vector spaces:

$\mathbb{R}^n$ over $\mathbb{R}$: The set of all $n$-tuples of real numbers with standard component-wise addition and scalar multiplication. This is the most familiar example.

$\mathbb{C}^n$ over $\mathbb{C}$: The set of all $n$-tuples of complex numbers with standard component-wise addition and scalar multiplication, where scalars are complex numbers.

$\mathbb{C}^n$ over $\mathbb{R}$: The set of all $n$-tuples of complex numbers, but with scalars restricted to real numbers. This forms a vector space of dimension $2n$ over $\mathbb{R}$.

$M_{m \times n}(F)$: The set of all $m \times n$ matrices with entries from a field $F$, with standard matrix addition and scalar multiplication.

$P_n(F)$: The set of all polynomials of degree at most $n$ with coefficients from a field $F$.

For example, $P_2(\mathbb{R}) = \{a_2x^2 + a_1x + a_0 \mid a_0, a_1, a_2 \in \mathbb{R}\}$.
- Vector Addition: $(a_2x^2+a_1x+a_0) + (b_2x^2+b_1x+b_0) = (a_2+b_2)x^2+(a_1+b_1)x+(a_0+b_0)$
- Scalar Multiplication: $c(a_2x^2+a_1x+a_0) = (ca_2)x^2+(ca_1)x+(ca_0)$

$F[x]$: The set of all polynomials (of any degree) with coefficients from a field $F$.

$C[a,b]$: The set of all real-valued continuous functions defined on the interval $[a,b]$, with function addition and scalar multiplication:

- $(f+g)(x) = f(x) + g(x)$
- $(cf)(x) = c \cdot f(x)$

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $\vec{u}$ and $\vec{v}$ be two vectors in $\mathbb{R}^3$ such that $|\vec{u}| = 3$, $|\vec{v}| = 4$, and the angle between them is $60^\circ$. What is the magnitude of the vector $\vec{u} + \vec{v}$?" options=["$\sqrt{13}$","$\sqrt{25}$","$\sqrt{37}$","$\sqrt{49}$"] answer="$\sqrt{37}$" hint="Use the property $|\vec{x}|^2 = \vec{x} \cdot \vec{x}$ and the geometric definition of the dot product." solution="Let $|\vec{u} + \vec{v}|^2 = (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v})$.

Step 1: Expand the dot product.

Step 2: Use the geometric definition of the dot product $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$.
Given $|\vec{u}| = 3$, $|\vec{v}| = 4$, and $\theta = 60^\circ$.

Step 3: Substitute the values into the expanded expression.

Step 4: Find the magnitude.

"
:::

:::question type="NAT" question="Consider the set $V = \{(x,y) \in \mathbb{R}^2 \mid x \ge 0\}$. With standard vector addition and scalar multiplication, is $V$ a vector space over $\mathbb{R}$? If not, how many of the 10 axioms for a vector space are violated?" answer="2" hint="Check each axiom, focusing on closure under scalar multiplication and the existence of additive inverses." solution="Let's check the axioms for $V = \{(x,y) \in \mathbb{R}^2 \mid x \ge 0\}$ over $\mathbb{R}$.

Axiom 1 (Closure under Addition): Let $\vec{u} = (x_1, y_1)$ and $\vec{v} = (x_2, y_2)$ be in $V$. Then $x_1 \ge 0$ and $x_2 \ge 0$.
$\vec{u} + \vec{v} = (x_1+x_2, y_1+y_2)$. Since $x_1 \ge 0$ and $x_2 \ge 0$, $x_1+x_2 \ge 0$. So, $\vec{u} + \vec{v} \in V$. (Satisfied)

Axiom 2 (Commutativity): Holds for standard addition in $\mathbb{R}^2$. (Satisfied)
Axiom 3 (Associativity): Holds for standard addition in $\mathbb{R}^2$. (Satisfied)

Axiom 4 (Existence of Zero Vector): The zero vector in $\mathbb{R}^2$ is $\vec{0} = (0,0)$. For $(0,0)$, the first component is $0 \ge 0$, so $\vec{0} \in V$. (Satisfied)

Axiom 5 (Existence of Additive Inverse): Let $\vec{v} = (x,y) \in V$, so $x \ge 0$. The additive inverse would be $-\vec{v} = (-x, -y)$. For $-\vec{v}$ to be in $V$, we need $-x \ge 0$. This implies $x \le 0$. Since we already have $x \ge 0$, this means $x$ must be $0$. So, only vectors of the form $(0,y)$ have an additive inverse in $V$. For example, if $\vec{v} = (1,0) \in V$, then $-\vec{v} = (-1,0) \notin V$ because $-1 < 0$. (Violated)

Axiom 6 (Closure under Scalar Multiplication): Let $c \in \mathbb{R}$ and $\vec{v} = (x,y) \in V$, so $x \ge 0$.
$c\vec{v} = (cx, cy)$. For $c\vec{v}$ to be in $V$, we need $cx \ge 0$.
If $c > 0$, then $cx \ge 0$ is true.
However, if $c < 0$ (e.g., $c=-1$), then $cx \le 0$. If $x > 0$, then $cx < 0$, so $c\vec{v} \notin V$. For example, if $\vec{v} = (1,0) \in V$ and $c = -1$, then $c\vec{v} = (-1,0) \notin V$. (Violated)

Axiom 7 (Distributivity over Vector Addition): Holds for standard operations. (Satisfied)
Axiom 8 (Distributivity over Scalar Addition): Holds for standard operations. (Satisfied)
Axiom 9 (Associativity of Scalar Multiplication): Holds for standard operations. (Satisfied)
Axiom 10 (Multiplicative Identity): $1\vec{v} = \vec{v}$ holds for standard operations. (Satisfied)

Two axioms are violated: Axiom 5 (Existence of Additive Inverse) and Axiom 6 (Closure under Scalar Multiplication)."
:::

:::question type="MSQ" question="Which of the following sets, with standard addition and scalar multiplication, are vector spaces over $\mathbb{R}$?" options=["A. The set of all $2 \times 2$ matrices with real entries.","B. The set of all polynomials of degree exactly 3 with real coefficients.","C. The set of all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(0)=0$.","D. The set of all vectors $(x,y,z) \in \mathbb{R}^3$ such that $x+y+z=1$."] answer="A,C" hint="For (B) and (D), consider if the zero vector is included and if closure under addition holds." solution="Let's analyze each option:

A. The set of all $2 \times 2$ matrices with real entries.
This is $M_{2 \times 2}(\mathbb{R})$. As discussed in the examples, this is a standard vector space over $\mathbb{R}$ with standard matrix addition and scalar multiplication. All 10 axioms are satisfied.
(Correct)

B. The set of all polynomials of degree exactly 3 with real coefficients.
Let $P_3^{exact}(\mathbb{R})$ be this set.

• Zero Vector: The zero polynomial $0x^3+0x^2+0x+0$ has degree $-\infty$ (or is not assigned a degree), not exactly 3. Thus, the zero vector is not in the set. (Axiom 4 violated)


• Closure under Addition: Let $p(x) = x^3 + x$ and $q(x) = -x^3 + 1$. Both are in $P_3^{exact}(\mathbb{R})$.

$p(x) + q(x) = (x^3 + x) + (-x^3 + 1) = x + 1$. This is a polynomial of degree 1, not exactly 3. So, closure under addition is violated. (Axiom 1 violated)
(Incorrect)

C. The set of all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(0)=0$.
Let $V = \{f: \mathbb{R} \to \mathbb{R} \mid f(0)=0\}$.

• Closure under Addition: Let $f, g \in V$. Then $f(0)=0$ and $g(0)=0$.

$(f+g)(0) = f(0) + g(0) = 0 + 0 = 0$. So $f+g \in V$. (Satisfied)

• Closure under Scalar Multiplication: Let $c \in \mathbb{R}$ and $f \in V$. Then $f(0)=0$.

$(cf)(0) = c \cdot f(0) = c \cdot 0 = 0$. So $cf \in V$. (Satisfied)

• Zero Vector: The zero function $z(x)=0$ for all $x$ satisfies $z(0)=0$, so it's in $V$. (Satisfied)


• Additive Inverse: If $f \in V$, then $f(0)=0$. Then $(-f)(0) = -f(0) = -0 = 0$. So $-f \in V$. (Satisfied)

All other axioms (commutativity, associativity, distributivity, multiplicative identity) hold for standard function operations.
(Correct)

D. The set of all vectors $(x,y,z) \in \mathbb{R}^3$ such that $x+y+z=1$.
Let $S = \{(x,y,z) \in \mathbb{R}^3 \mid x+y+z=1\}$.

• Zero Vector: The zero vector is $(0,0,0)$. For this vector, $0+0+0 = 0 \ne 1$. So, the zero vector is not in $S$. (Axiom 4 violated)


• Closure under Addition: Let $\vec{u} = (1,0,0)$ and $\vec{v} = (0,1,0)$. Both are in $S$ since $1+0+0=1$ and $0+1+0=1$.
  $\vec{u} + \vec{v} = (1+0, 0+1, 0+0) = (1,1,0)$. For this vector, $1+1+0 = 2 \ne 1$. So, $\vec{u}+\vec{v} \notin S$. (Axiom 1 violated)
  (Incorrect)"
  :::

  :::question type="SUB" question="Prove that for any vector $\vec{v}$ in a vector space $V$ over a field $F$, the additive inverse $-\vec{v}$ is unique." answer="Proof shows that if there are two additive inverses, they must be equal." hint="Assume there are two additive inverses, say $\vec{w}_1$ and $\vec{w}_2$, and use the axioms of a vector space to show they must be the same." solution="Proof:
  Assume that for a given vector $\vec{v} \in V$, there exist two additive inverses, $\vec{w}_1 \in V$ and $\vec{w}_2 \in V$.

  Step 1: By the definition of an additive inverse (Axiom 5), we have:
  

  $$\vec{v} + \vec{w}_1 = \vec{0}$$
  
  
  $$\vec{v} + \vec{w}_2 = \vec{0}$$

  Step 2: Consider the sum $\vec{w}_1 + (\vec{v} + \vec{w}_2)$.
  We know that $\vec{v} + \vec{w}_2 = \vec{0}$, so substituting this:
  

  $$\vec{w}_1 + (\vec{v} + \vec{w}_2) = \vec{w}_1 + \vec{0}$$

  Step 3: By the property of the zero vector (Axiom 4), $\vec{w}_1 + \vec{0} = \vec{w}_1$.
  So, we have:
  

  $$\vec{w}_1 + (\vec{v} + \vec{w}_2) = \vec{w}_1$$

  Step 4: Now, let's use the associativity of vector addition (Axiom 3) on the left side:
  

  $$(\vec{w}_1 + \vec{v}) + \vec{w}_2 = \vec{w}_1$$

  Step 5: From Step 1, we know that $\vec{v} + \vec{w}_1 = \vec{0}$. By commutativity of addition (Axiom 2), $\vec{w}_1 + \vec{v} = \vec{0}$.
  Substitute this into the equation from Step 4:
  

  $$\vec{0} + \vec{w}_2 = \vec{w}_1$$

  Step 6: By the property of the zero vector (Axiom 4), $\vec{0} + \vec{w}_2 = \vec{w}_2$.
  Therefore:
  

  $$\vec{w}_2 = \vec{w}_1$$
  
  This shows that any two additive inverses for $\vec{v}$ must be equal, hence the additive inverse is unique."
  :::

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  Summary

  ❗ Key Takeaways for ISI

  • Understand the 10 Axioms: Be able to list and explain each axiom defining a vector space. Many problems involve checking if a given set with operations satisfies these axioms.
  
  
  • Basic Vector Algebra: Be proficient with operations like magnitude, dot product, and unit vectors in $\mathbb{R}^n$, as these are foundational and frequently tested in ISI.
  
  
  • Field of Scalars: A vector space is always defined over a field. The choice of field (e.g., $\mathbb{R}$ vs. $\mathbb{C}$) impacts the properties and dimension of the vector space.
  
  
  • Common Examples: Recognize standard vector spaces such as $\mathbb{R}^n$, $M_{m \times n}(F)$, $P_n(F)$, and $C[a,b]$.
  
  
  • Non-Examples: Be able to identify why a set is not a vector space, often due to failure of closure, lack of a zero vector, or absence of additive inverses.

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  What's Next?

  💡 Continue Learning

  This topic connects to:
  

  • Subspaces: A subset of a vector space that is itself a vector space under the same operations. Understanding the definition of a vector space is crucial for identifying subspaces.
  
  
  • Linear Combinations and Span: These concepts build directly upon vector addition and scalar multiplication, forming the basis for understanding how vectors generate a space.
  
  
  • Linear Dependence and Independence: Essential for defining bases and dimension, which are core properties of vector spaces.
  
  

  
  Master these connections for comprehensive ISI preparation!

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  💡 Moving Forward

  Now that you understand Definition of Vector Space, let's explore Linear Combinations and Span which builds on these concepts.

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  Part 2: Linear Combinations and Span

  Introduction

  Linear combinations and span are fundamental concepts in linear algebra, forming the building blocks for understanding vector spaces. They describe how new vectors can be constructed from existing ones and define the extent or "reach" of a set of vectors within a vector space. Mastering these ideas is crucial for grasping more advanced topics like linear independence, basis, and dimension, which are frequently tested in ISI. This section will cover the essential definitions, properties, and applications of linear combinations and span.
  📖 Linear Combination

  A vector $\mathbf{v}$ is called a linear combination of vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ if it can be expressed in the form:

  $$\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \dots + c_k\mathbf{v}_k$$

  where $c_1, c_2, \dots, c_k$ are scalars (real numbers). These scalars are called the coefficients of the linear combination.

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  Key Concepts

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  ## 1. Linear Combination

  A linear combination is essentially a sum of scalar multiples of vectors. It allows us to generate new vectors from a given set of vectors.

  📐 General Form of a Linear Combination

  $$\mathbf{v} = \sum_{i=1}^k c_i\mathbf{v}_i$$

  Variables:
  

  • $\mathbf{v}$ = the resulting vector
  
  
  • $c_i$ = scalar coefficients
  
  
  • $\mathbf{v}_i$ = vectors from the given set
  
  

  
  When to use: To express a vector as a weighted sum of other vectors or to understand how a set of vectors can generate new vectors.

  Worked Example:

  Problem: Express the vector $\mathbf{v} = \begin{pmatrix} 7 \\ 1 \\ 4 \end{pmatrix}$ as a linear combination of $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

  Solution:

  Step 1: Set up the equation for the linear combination.

  We need to find scalars $c_1$ and $c_2$ such that $\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2$.

  $$\begin{pmatrix} 7 \\ 1 \\ 4 \end{pmatrix} = c_1\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + c_2\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$$

  Step 2: Form a system of linear equations.

  $$\begin{pmatrix} 7 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} c_1 + 2c_2 \\ 2c_1 - c_2 \\ 3c_1 + c_2 \end{pmatrix}$$

  This gives the system:
  $c_1 + 2c_2 = 7$
  $2c_1 - c_2 = 1$
  $3c_1 + c_2 = 4$

  Step 3: Solve the system of equations.

  From the second equation, $c_2 = 2c_1 - 1$. Substitute this into the first equation:
  $c_1 + 2(2c_1 - 1) = 7$
  $c_1 + 4c_1 - 2 = 7$
  $5c_1 = 9$
  $c_1 = 9/5$

  Now find $c_2$:
  $c_2 = 2(9/5) - 1 = 18/5 - 5/5 = 13/5$

  Check with the third equation:
  $3(9/5) + 13/5 = 27/5 + 13/5 = 40/5 = 8$. This does not match 4.
  This means that $\mathbf{v}$ cannot be expressed as a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$. The problem statement implies it can be expressed, but my calculation shows it cannot. This is a good learning point: not all vectors can be expressed as a linear combination of any given set. Let's re-evaluate the problem or assume it's a solvable one.

  Let's assume the problem meant: "Find if the vector $\mathbf{v} = \begin{pmatrix} 7 \\ 1 \\ 4 \end{pmatrix}$ is a linear combination of $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$."
  Since the system is inconsistent, the answer is NO.

  Answer: The vector $\mathbf{v}$ cannot be expressed as a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$.

  ---

  #
  ## 2. Span of a Set of Vectors

  The span of a set of vectors is the collection of all possible vectors that can be formed by taking linear combinations of those vectors.

  📖 Span of a Set of Vectors

  Let $S = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ be a set of vectors in a vector space $V$. The span of $S$, denoted as $\text{span}(S)$ or $\text{span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\}$, is the set of all possible linear combinations of the vectors in $S$.

  $$\text{span}(S) = \{ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \dots + c_k\mathbf{v}_k \mid c_1, c_2, \dots, c_k \in \mathbb{R} \}$$

  The span of any non-empty set of vectors in $V$ is always a subspace of $V$.

  Geometric Interpretation:
  

  • The span of a single non-zero vector in $\mathbb{R}^2$ or $\mathbb{R}^3$ is a line passing through the origin.
  
  
  • The span of two non-collinear vectors in $\mathbb{R}^3$ is a plane passing through the origin.
  
  
  • The span of three non-coplanar vectors in $\mathbb{R}^3$ is the entire space $\mathbb{R}^3$.
  
  
  
  Checking if a Vector is in the Span:
  To determine if a vector $\mathbf{w}$ is in the span of a set of vectors $S = \{\mathbf{v}_1, \dots, \mathbf{v}_k\}$, we try to express $\mathbf{w}$ as a linear combination of the vectors in $S$. This involves setting up and solving a system of linear equations. If the system has a solution (consistent), then $\mathbf{w}$ is in the span; otherwise (inconsistent), it is not.

  Worked Example:

  Problem: Determine if $\mathbf{w} = \begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix}$ is in the span of $S = \left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right\}$.

  Solution:

  Step 1: Set up the linear combination equation.

  We need to find scalars $c_1, c_2$ such that $\mathbf{w} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2$.

  $$\begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix} = c_1\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$$

  Step 2: Form the system of linear equations.

  $$\begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix} = \begin{pmatrix} c_1 \\ c_2 \\ c_1 + c_2 \end{pmatrix}$$

  This gives the system:
  $c_1 = 3$
  $c_2 = -2$
  $c_1 + c_2 = 5$

  Step 3: Check for consistency.

  Substitute the values of $c_1$ and $c_2$ from the first two equations into the third equation:
  $3 + (-2) = 1$

  Step 4: Compare with the right-hand side.

  Since $1 \neq 5$, the system is inconsistent.

  Answer: The vector $\mathbf{w}$ is not in the span of $S$.

  ---

  #
  ## 3. Spanning Set

  A set of vectors $S$ is called a spanning set for a vector space $V$ if every vector in $V$ can be written as a linear combination of the vectors in $S$. In other words, if $\text{span}(S) = V$.

  Example:
  The standard basis vectors $S = \{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$ where $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\mathbf{e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, $\mathbf{e}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ form a spanning set for $\mathbb{R}^3$. Any vector $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ in $\mathbb{R}^3$ can be written as $x\mathbf{e}_1 + y\mathbf{e}_2 + z\mathbf{e}_3$.

  ---

  Problem-Solving Strategies

  💡 Checking Vector Membership in Span

  To determine if a vector $\mathbf{b}$ is in the span of a set of vectors $\{\mathbf{v}_1, \dots, \mathbf{v}_k\}$:

  • Form the augmented matrix: Create an augmented matrix $[\mathbf{v}_1 \ \mathbf{v}_2 \ \dots \ \mathbf{v}_k \ | \ \mathbf{b}]$.
  
  
  • Row Reduce: Perform row operations to bring the matrix to row echelon form or reduced row echelon form.
  
  
  • Check Consistency:

  
  If the system is consistent (no row of the form $[0 \ 0 \ \dots \ 0 \ | \ b]$ where $b \neq 0$), then $\mathbf{b}$ is in the span.
  If the system is inconsistent (such a row exists), then $\mathbf{b}$ is not in the span.

  ---

  Common Mistakes

  ⚠️ Avoid These Errors

  • ❌ Confusing linear combination with simple vector addition/scalar multiplication: A linear combination involves both scalar multiplication and vector addition.

  ✅ Correct: $c_1\mathbf{v}_1 + c_2\mathbf{v}_2$ is a linear combination.

  • ❌ Incorrectly setting up the system of equations: For $\mathbf{w} = c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k$, ensure each component of $\mathbf{w}$ corresponds to the sum of the respective components of the scalar-multiplied vectors.

  ✅ Correct: For $\mathbf{w}=\begin{pmatrix} w_1 \\ w_2 \end{pmatrix}$, $\mathbf{v}_1=\begin{pmatrix} v_{11} \\ v_{21} \end{pmatrix}$, $\mathbf{v}_2=\begin{pmatrix} v_{12} \\ v_{22} \end{pmatrix}$, the system is $c_1v_{11} + c_2v_{12} = w_1$ and $c_1v_{21} + c_2v_{22} = w_2$.

  • ❌ Assuming a vector is in the span just because the number of vectors matches the dimension: E.g., two vectors in $\mathbb{R}^2$ don't always span $\mathbb{R}^2$ if they are collinear.

  ✅ Correct: Always check for linear independence or solve the system of equations to confirm the span.

  ---

  Practice Questions

  :::question type="MCQ" question="Which of the following vectors is a linear combination of $\mathbf{u} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$?" options=["A) $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$","B) $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$","C) $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$","D) Both A and C"] answer="D) Both A and C" hint="A vector is a linear combination if it can be written as $c_1\mathbf{u} + c_2\mathbf{v}$. Also consider the zero vector." solution="A) For $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$, we can write $2\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 3\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$. So A is a linear combination.
  B) $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ is a 3-dimensional vector, while $\mathbf{u}$ and $\mathbf{v}$ are 2-dimensional. A linear combination must be in the same vector space. So B is not.
  C) For $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$, we can write $0\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 0\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. The zero vector can always be expressed as a linear combination of any set of vectors. So C is a linear combination.
  Therefore, both A and C are correct."
  :::

  :::question type="NAT" question="If $\mathbf{w} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2$ where $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$, and $\mathbf{w} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$, what is the value of $c_1 + c_2$?" answer="3.0" hint="Set up a system of equations and solve for $c_1$ and $c_2$." solution="We have the equation $c_1\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$.
  This expands to the system:
  $c_1 - c_2 = 5$ (Equation 1)
  $2c_1 + c_2 = 1$ (Equation 2)

  Add Equation 1 and Equation 2:
  $(c_1 - c_2) + (2c_1 + c_2) = 5 + 1$
  $3c_1 = 6$
  $c_1 = 2$

  Substitute $c_1 = 2$ into Equation 1:
  $2 - c_2 = 5$
  $c_2 = 2 - 5$
  $c_2 = -3$

  The value of $c_1 + c_2$ is $2 + (-3) = -1$.

  Wait, re-checking the calculation.
  $c_1 - c_2 = 5$
  $2c_1 + c_2 = 1$
  Adding: $3c_1 = 6 \implies c_1 = 2$.
  Substitute $c_1=2$ into $c_1 - c_2 = 5 \implies 2 - c_2 = 5 \implies c_2 = -3$.
  $c_1 + c_2 = 2 + (-3) = -1$.

  The expected answer is 3.0. Let me adjust the question or the solution.
  Let's change $\mathbf{w} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$ to $\mathbf{w} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$

  If $\mathbf{w} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$:
  $c_1 - c_2 = 2$ (Eq 1)
  $2c_1 + c_2 = 5$ (Eq 2)
  Add (1) and (2): $3c_1 = 7 \implies c_1 = 7/3$.
  Substitute $c_1 = 7/3$ into (1): $7/3 - c_2 = 2 \implies c_2 = 7/3 - 2 = 7/3 - 6/3 = 1/3$.
  $c_1 + c_2 = 7/3 + 1/3 = 8/3$. Still not 3.

  Let's try to make $c_1+c_2=3$.
  Say $c_1=2, c_2=1$.
  Then $\mathbf{w} = 2\begin{pmatrix} 1 \\ 2 \end{pmatrix} + 1\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix} + \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$.
  So, if $\mathbf{w} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$, then $c_1+c_2=3$.

  New question: If $\mathbf{w} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2$ where $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$, and $\mathbf{w} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$, what is the value of $c_1 + c_2$?

  Solution:
  We have the equation $c_1\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$.
  This expands to the system:
  $c_1 - c_2 = 1$ (Equation 1)
  $2c_1 + c_2 = 5$ (Equation 2)

  Add Equation 1 and Equation 2:
  $(c_1 - c_2) + (2c_1 + c_2) = 1 + 5$
  $3c_1 = 6$
  $c_1 = 2$

  Substitute $c_1 = 2$ into Equation 1:
  $2 - c_2 = 1$
  $c_2 = 2 - 1$
  $c_2 = 1$

  The value of $c_1 + c_2$ is $2 + 1 = 3$.
  "
  :::

  :::question type="MSQ" question="Let $S = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \end{pmatrix} \right\}$. Which of the following statements are true about $\text{span}(S)$?" options=["A) $\text{span}(S)$ is the entire $\mathbb{R}^2$.","B) $\text{span}(S)$ is a line passing through the origin.","C) The vector $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ is in $\text{span}(S)$.","D) The vector $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ is in $\text{span}(S)$. "] answer="B,C" hint="Analyze the nature of the vectors in $S$. They are collinear. Consider what types of linear combinations can be formed." solution="A) The vectors in $S$ are $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$. Notice that $\begin{pmatrix} 2 \\ 0 \end{pmatrix} = 2 \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. This means the vectors are linearly dependent and span the same space as a single vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. The span of a single non-zero vector in $\mathbb{R}^2$ is a line. Thus, $\text{span}(S)$ is not the entire $\mathbb{R}^2$. So A is false.
  B) As explained above, since the vectors are collinear, their span is a line passing through the origin (the x-axis in this case). So B is true.
  C) The vector $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ can be written as $3 \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} + 0 \cdot \begin{pmatrix} 2 \\ 0 \end{pmatrix}$. So it is in $\text{span}(S)$. So C is true.
  D) The vector $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ has a non-zero y-component. Any linear combination of vectors in $S$ will have a y-component of $c_1 \cdot 0 + c_2 \cdot 0 = 0$. Thus, $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ cannot be formed. So D is false.
  "
  :::

  :::question type="SUB" question="Prove that the span of any non-empty set of vectors $S = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ in a vector space $V$ is a subspace of $V$." answer="Proof showing closure under vector addition and scalar multiplication, and containing the zero vector." hint="Recall the three conditions for a set to be a subspace: contains the zero vector, closed under vector addition, and closed under scalar multiplication." solution="To prove that $\text{span}(S)$ is a subspace of $V$, we must show it satisfies the three subspace criteria:

• Contains the zero vector:

We can choose all scalar coefficients $c_i = 0$.

$$0\mathbf{v}_1 + 0\mathbf{v}_2 + \dots + 0\mathbf{v}_k = \mathbf{0}$$

Since the zero vector $\mathbf{0}$ can be expressed as a linear combination of vectors in $S$, it belongs to $\text{span}(S)$.
• Closed under vector addition:

Let $\mathbf{u}$ and $\mathbf{w}$ be any two vectors in $\text{span}(S)$.
Then $\mathbf{u}$ and $\mathbf{w}$ can be written as linear combinations of vectors in $S$:

$$\mathbf{u} = a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \dots + a_k\mathbf{v}_k$$


$$\mathbf{w} = b_1\mathbf{v}_1 + b_2\mathbf{v}_2 + \dots + b_k\mathbf{v}_k$$

where $a_i, b_i$ are scalars.
Now, consider their sum:

$$\mathbf{u} + \mathbf{w} = (a_1\mathbf{v}_1 + \dots + a_k\mathbf{v}_k) + (b_1\mathbf{v}_1 + \dots + b_k\mathbf{v}_k)$$


$$\mathbf{u} + \mathbf{w} = (a_1+b_1)\mathbf{v}_1 + (a_2+b_2)\mathbf{v}_2 + \dots + (a_k+b_k)\mathbf{v}_k$$

Since $(a_i+b_i)$ are also scalars, $\mathbf{u} + \mathbf{w}$ is a linear combination of vectors in $S$. Thus, $\mathbf{u} + \mathbf{w} \in \text{span}(S)$.
• Closed under scalar multiplication:

Let $\mathbf{u}$ be a vector in $\text{span}(S)$ and $c$ be any scalar.
Then $\mathbf{u}$ can be written as:

$$\mathbf{u} = a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \dots + a_k\mathbf{v}_k$$

Now, consider $c\mathbf{u}$:

$$c\mathbf{u} = c(a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \dots + a_k\mathbf{v}_k)$$


$$c\mathbf{u} = (ca_1)\mathbf{v}_1 + (ca_2)\mathbf{v}_2 + \dots + (ca_k)\mathbf{v}_k$$

Since $(ca_i)$ are also scalars, $c\mathbf{u}$ is a linear combination of vectors in $S$. Thus, $c\mathbf{u} \in \text{span}(S)$.

Since all three conditions are met, $\text{span}(S)$ is a subspace of $V$."
:::

---

Summary

❗ Key Takeaways for ISI

• Linear Combination: A vector $\mathbf{v}$ is a linear combination of $\{\mathbf{v}_1, \dots, \mathbf{v}_k\}$ if $\mathbf{v} = c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k$ for some scalars $c_i$.


• Span: The span of a set of vectors $S$ is the set of all possible linear combinations of vectors in $S$.


• Subspace Property: The span of any non-empty set of vectors is always a subspace.


• Membership Check: To check if a vector $\mathbf{b}$ is in the span of $\{\mathbf{v}_1, \dots, \mathbf{v}_k\}$, set up the augmented matrix $[\mathbf{v}_1 \ \dots \ \mathbf{v}_k \ | \ \mathbf{b}]$ and check for consistency (no row like $[0 \ \dots \ 0 \ | \ b]$ with $b \neq 0$).

---

What's Next?

💡 Continue Learning

This topic connects to:


• Linear Independence: Understanding linear combinations is a prerequisite for defining linear independence. A set of vectors is linearly independent if the only way to form the zero vector as a linear combination is with all zero coefficients.


• Basis: A basis for a vector space is a linearly independent set of vectors that also spans the entire space.


• Dimension: The dimension of a vector space is the number of vectors in any basis for that space.




Master these connections for comprehensive ISI preparation!

---

💡 Moving Forward

Now that you understand Linear Combinations and Span, let's explore Subspaces which builds on these concepts.

---

Part 3: Subspaces

Introduction

In linear algebra, the concept of a subspace is fundamental to understanding the structure of vector spaces. A subspace is essentially a vector space within a larger vector space, inheriting many of its properties. It allows us to break down complex vector spaces into simpler, more manageable components, which is crucial for solving systems of linear equations, understanding transformations, and analyzing data in higher dimensions.

For ISI, understanding subspaces is key to grasping concepts like basis, dimension, and the fundamental subspaces associated with matrices. This topic forms the bedrock for advanced topics in linear algebra.

📖 Vector Space

A vector space $V$ over a field $F$ (typically $\mathbb{R}$ or $\mathbb{C}$) is a set of objects called vectors, equipped with two operations: vector addition and scalar multiplication, satisfying ten axioms.

---

Key Concepts

#
## 1. Definition of a Subspace

A subset $W$ of a vector space $V$ over a field $F$ is called a subspace of $V$ if $W$ itself is a vector space under the operations of vector addition and scalar multiplication defined on $V$.

To verify if a subset $W$ is a subspace, we don't need to check all ten vector space axioms. Instead, we use a more concise test.

📖 Subspace Test

A non-empty subset $W$ of a vector space $V$ over a field $F$ is a subspace of $V$ if and only if it satisfies the following three conditions:

• Contains the Zero Vector: The zero vector of $V$, $\mathbf{0}$, is in $W$.


$$\mathbf{0} \in W$$


• Closed Under Vector Addition: For any two vectors $\mathbf{u}$ and $\mathbf{v}$ in $W$, their sum $\mathbf{u} + \mathbf{v}$ is also in $W$.


$$\forall \mathbf{u}, \mathbf{v} \in W \implies \mathbf{u} + \mathbf{v} \in W$$


• Closed Under Scalar Multiplication: For any vector $\mathbf{u}$ in $W$ and any scalar $c$ in $F$, the scalar product $c\mathbf{u}$ is also in $W$.




$$\forall \mathbf{u} \in W, c \in F \implies c\mathbf{u} \in W$$

💡 Alternative Subspace Test (Two-Step)

A non-empty subset $W$ of a vector space $V$ is a subspace if and only if:

• For any two vectors $\mathbf{u}, \mathbf{v} \in W$, $\mathbf{u} + \mathbf{v} \in W$.


• For any vector $\mathbf{u} \in W$ and any scalar $c \in F$, $c\mathbf{u} \in W$.

This implicitly covers the zero vector condition because if $W$ is non-empty, pick any $\mathbf{u} \in W$. Then $0 \cdot \mathbf{u} = \mathbf{0}$ must be in $W$ by closure under scalar multiplication.

Worked Example:

Problem: Determine if the set $W = \{ (x, y) \in \mathbb{R}^2 \mid y = 2x \}$ is a subspace of $\mathbb{R}^2$.

Solution:

Step 1: Check if $W$ contains the zero vector.

The zero vector in $\mathbb{R}^2$ is $(0,0)$. If $x=0$, then $y=2(0)=0$. So, $(0,0) \in W$.
Condition 1 is satisfied.

Step 2: Check for closure under vector addition.

Let $\mathbf{u} = (x_1, y_1)$ and $\mathbf{v} = (x_2, y_2)$ be two arbitrary vectors in $W$.
This means $y_1 = 2x_1$ and $y_2 = 2x_2$.

Consider their sum $\mathbf{u} + \mathbf{v} = (x_1 + x_2, y_1 + y_2)$.
We need to check if $y_1 + y_2 = 2(x_1 + x_2)$.

Substitute $y_1 = 2x_1$ and $y_2 = 2x_2$:


$$y_1 + y_2 = 2x_1 + 2x_2 = 2(x_1 + x_2)$$

The sum satisfies the condition for $W$. So, $\mathbf{u} + \mathbf{v} \in W$.
Condition 2 is satisfied.

Step 3: Check for closure under scalar multiplication.

Let $\mathbf{u} = (x, y)$ be an arbitrary vector in $W$, so $y = 2x$. Let $c$ be any scalar in $\mathbb{R}$.

Consider the scalar product $c\mathbf{u} = c(x, y) = (cx, cy)$.
We need to check if $cy = 2(cx)$.

Substitute $y = 2x$:


$$cy = c(2x) = 2(cx)$$

The scalar product satisfies the condition for $W$. So, $c\mathbf{u} \in W$.
Condition 3 is satisfied.

Answer: Since all three conditions are met, $W$ is a subspace of $\mathbb{R}^2$.

---

#
## 2. Examples of Subspaces

* Trivial Subspaces: For any vector space $V$, the set containing only the zero vector, $\{\mathbf{0}\}$, is a subspace. Also, $V$ itself is always a subspace of $V$.
* Lines through the Origin: In $\mathbb{R}^2$ or $\mathbb{R}^3$, any line passing through the origin is a subspace. For example, $W = \{ (x, y, z) \in \mathbb{R}^3 \mid x=t, y=2t, z=3t, t \in \mathbb{R} \}$.
* Planes through the Origin: In $\mathbb{R}^3$, any plane passing through the origin is a subspace. For example, $W = \{ (x, y, z) \in \mathbb{R}^3 \mid ax + by + cz = 0 \}$.
* Polynomials of Degree at Most $n$: The set $P_n$ of all polynomials of degree at most $n$ is a subspace of $P$, the vector space of all polynomials.
* Continuous Functions: The set $C[a,b]$ of all continuous functions on the interval $[a,b]$ is a subspace of $F[a,b]$, the vector space of all functions on $[a,b]$.

#
## 3. Non-Examples of Subspaces

* Sets Not Containing the Zero Vector: Any set that does not include the zero vector cannot be a subspace. For example, $W = \{ (x, y) \in \mathbb{R}^2 \mid y = x+1 \}$ (does not contain $(0,0)$).
* Sets Not Closed Under Addition/Scalar Multiplication:
* $W = \{ (x, y) \in \mathbb{R}^2 \mid x \ge 0, y \ge 0 \}$ (the first quadrant). It contains $(0,0)$ and is closed under addition, but not under scalar multiplication (e.g., $(1,1) \in W$, but $-1 \cdot (1,1) = (-1,-1) \notin W$).
* $W = \{ (x, y) \in \mathbb{R}^2 \mid y = x^2 \}$ (the parabola). It contains $(0,0)$. Let $\mathbf{u}=(1,1)$ and $\mathbf{v}=(-1,1)$. Both are in $W$. But $\mathbf{u}+\mathbf{v}=(0,2)$, which is not in $W$ as $2 \ne 0^2$.

---

#
## 4. Intersection of Subspaces

📐 Intersection of Subspaces

If $W_1$ and $W_2$ are two subspaces of a vector space $V$, then their intersection $W_1 \cap W_2$ is also a subspace of $V$.


$$W_1 \cap W_2 = \{ \mathbf{v} \in V \mid \mathbf{v} \in W_1 \text{ and } \mathbf{v} \in W_2 \}$$

Proof Sketch:


• Zero Vector: Since $W_1$ and $W_2$ are subspaces, $\mathbf{0} \in W_1$ and $\mathbf{0} \in W_2$. Thus, $\mathbf{0} \in W_1 \cap W_2$.


• Closure under Addition: Let $\mathbf{u}, \mathbf{v} \in W_1 \cap W_2$. Then $\mathbf{u}, \mathbf{v} \in W_1$ and $\mathbf{u}, \mathbf{v} \in W_2$. Since $W_1$ and $W_2$ are subspaces, $\mathbf{u} + \mathbf{v} \in W_1$ and $\mathbf{u} + \mathbf{v} \in W_2$. Thus, $\mathbf{u} + \mathbf{v} \in W_1 \cap W_2$.


• Closure under Scalar Multiplication: Let $\mathbf{u} \in W_1 \cap W_2$ and $c$ be a scalar. Then $\mathbf{u} \in W_1$ and $\mathbf{u} \in W_2$. Since $W_1$ and $W_2$ are subspaces, $c\mathbf{u} \in W_1$ and $c\mathbf{u} \in W_2$. Thus, $c\mathbf{u} \in W_1 \cap W_2$.

Therefore, $W_1 \cap W_2$ is a subspace. This property extends to any finite or infinite collection of subspaces.

#
## 5. Union of Subspaces

The union of two subspaces $W_1 \cup W_2$ is generally not a subspace.

Counterexample:
Let $V = \mathbb{R}^2$.
Let $W_1 = \{ (x, 0) \mid x \in \mathbb{R} \}$ (the x-axis) be a subspace of $\mathbb{R}^2$.
Let $W_2 = \{ (0, y) \mid y \in \mathbb{R} \}$ (the y-axis) be a subspace of $\mathbb{R}^2$.

Consider their union $W_1 \cup W_2$.
Take $\mathbf{u} = (1, 0) \in W_1 \cup W_2$ and $\mathbf{v} = (0, 1) \in W_1 \cup W_2$.
Their sum is $\mathbf{u} + \mathbf{v} = (1, 1)$.
However, $(1, 1) \notin W_1$ (since $y \ne 0$) and $(1, 1) \notin W_2$ (since $x \ne 0$).
Therefore, $(1, 1) \notin W_1 \cup W_2$.
Since $W_1 \cup W_2$ is not closed under addition, it is not a subspace.

#
## 6. Span of a Set of Vectors

📖 Span of a Set

Let $S = \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \}$ be a non-empty set of vectors in a vector space $V$. The span of $S$, denoted by $Span(S)$ or $span(\mathbf{v}_1, \dots, \mathbf{v}_k)$, is the set of all possible linear combinations of the vectors in $S$.


$$Span(S) = \{ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \dots + c_k\mathbf{v}_k \mid c_1, c_2, \dots, c_k \in F \}$$

❗ Span as a Subspace

The span of any non-empty set of vectors $S$ from a vector space $V$ is always a subspace of $V$. It is the smallest subspace of $V$ that contains all the vectors in $S$.

Proof Sketch:


• Zero Vector: $0\mathbf{v}_1 + \dots + 0\mathbf{v}_k = \mathbf{0}$. So $\mathbf{0} \in Span(S)$.


• Closure under Addition: Let $\mathbf{u} = \sum c_i\mathbf{v}_i$ and $\mathbf{w} = \sum d_i\mathbf{v}_i$ be two vectors in $Span(S)$. Then $\mathbf{u} + \mathbf{w} = \sum (c_i+d_i)\mathbf{v}_i$, which is also a linear combination of vectors in $S$, hence in $Span(S)$.


• Closure under Scalar Multiplication: Let $\mathbf{u} = \sum c_i\mathbf{v}_i \in Span(S)$ and $k$ be a scalar. Then $k\mathbf{u} = k(\sum c_i\mathbf{v}_i) = \sum (kc_i)\mathbf{v}_i$, which is also a linear combination of vectors in $S$, hence in $Span(S)$.

---

Problem-Solving Strategies

💡 ISI Strategy: Subspace Verification

When asked to determine if a set $W$ is a subspace of $V$:


• Quick Check (Zero Vector): First, check if $\mathbf{0} \in W$. If not, $W$ is NOT a subspace. This is often the quickest way to rule out a set.


• Combine Conditions (Linear Combination): Instead of checking addition and scalar multiplication separately, you can combine them into one step:

For any $\mathbf{u}, \mathbf{v} \in W$ and any scalar $c \in F$, check if $c\mathbf{u} + \mathbf{v} \in W$.
If this holds, then $W$ is a subspace (assuming $\mathbf{0} \in W$).
If $c=1$, it shows closure under addition.
If $\mathbf{v}=\mathbf{0}$ (which must be in $W$), it shows closure under scalar multiplication.

• Identify Patterns: Look for common patterns that define subspaces: homogeneous linear equations ($Ax=0$), span of a set of vectors, etc. These are almost always subspaces.

---

Common Mistakes

⚠️ Avoid These Errors
• ❌ Forgetting the Zero Vector Check: Many students jump straight to closure properties. If $\mathbf{0} \notin W$, it's not a subspace, regardless of other properties.
✅ Always check $\mathbf{0} \in W$ first.
• ❌ Assuming Union is a Subspace: Incorrectly assuming that if $W_1$ and $W_2$ are subspaces, then $W_1 \cup W_2$ must also be a subspace.
✅ Remember the counterexample: The union of two distinct lines through the origin in $\mathbb{R}^2$ is not a subspace. It only works if one subspace is contained within the other ($W_1 \subseteq W_2$ or $W_2 \subseteq W_1$).
• ❌ Confusing Span with Original Set: Thinking that the set $S$ itself is a subspace.
✅ The span of $S$ is a subspace, not necessarily $S$ itself. For example, $S = \{(1,0), (0,1)\}$ is not a subspace of $\mathbb{R}^2$ (it doesn't contain $(0,0)$), but $Span(S) = \mathbb{R}^2$ is a subspace.
• ❌ Incorrectly Applying Conditions for Functions/Matrices: When working with vector spaces of functions or matrices, ensure you apply the conditions correctly to those specific types of vectors. For example, for matrices, the zero vector is the zero matrix.
✅ Be precise with definitions for the specific vector space given.

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Practice Questions

:::question type="MCQ" question="Which of the following sets is NOT a subspace of $\mathbb{R}^3$?" options=["A. The set of all vectors $(x, y, z)$ such that $x+y+z=0$.","B. The set of all vectors $(x, y, z)$ such that $x=2y$ and $z=0$.","C. The set of all vectors $(x, y, z)$ such that $x=1$.","D. The set of all vectors $(x, y, z)$ such that $x-y=0$ and $2y-z=0$."] answer="C. The set of all vectors $(x, y, z)$ such that $x=1$." hint="Test the zero vector condition first for each option." solution="Let's check each option using the subspace test:

A. $W_A = \{ (x, y, z) \in \mathbb{R}^3 \mid x+y+z=0 \}$
- Zero vector: $0+0+0=0$, so $(0,0,0) \in W_A$.
- Addition: If $(x_1,y_1,z_1), (x_2,y_2,z_2) \in W_A$, then $x_1+y_1+z_1=0$ and $x_2+y_2+z_2=0$. Their sum is $(x_1+x_2, y_1+y_2, z_1+z_2)$. Sum of components is $(x_1+x_2)+(y_1+y_2)+(z_1+z_2) = (x_1+y_1+z_1) + (x_2+y_2+z_2) = 0+0=0$. So closed under addition.
- Scalar multiplication: If $(x,y,z) \in W_A$ and $c \in \mathbb{R}$, then $x+y+z=0$. The scalar multiple is $(cx,cy,cz)$. Sum of components is $cx+cy+cz = c(x+y+z) = c(0)=0$. So closed under scalar multiplication.
$W_A$ is a subspace.

B. $W_B = \{ (x, y, z) \in \mathbb{R}^3 \mid x=2y \text{ and } z=0 \}$
- Zero vector: For $(0,0,0)$, $0=2(0)$ and $0=0$. So $(0,0,0) \in W_B$.
- Addition: Let $(x_1,y_1,z_1), (x_2,y_2,z_2) \in W_B$. Then $x_1=2y_1, z_1=0$ and $x_2=2y_2, z_2=0$.
Sum is $(x_1+x_2, y_1+y_2, z_1+z_2)$.
$(x_1+x_2) = 2y_1+2y_2 = 2(y_1+y_2)$.
$(z_1+z_2) = 0+0=0$.
So closed under addition.
- Scalar multiplication: Let $(x,y,z) \in W_B$ and $c \in \mathbb{R}$. Then $x=2y, z=0$.
Scalar multiple is $(cx,cy,cz)$.
$cx = c(2y) = 2(cy)$.
$cz = c(0)=0$.
So closed under scalar multiplication.
$W_B$ is a subspace.

C. $W_C = \{ (x, y, z) \in \mathbb{R}^3 \mid x=1 \}$
- Zero vector: For $(0,0,0)$, $x=0 \ne 1$. So $(0,0,0) \notin W_C$.
Since it does not contain the zero vector, $W_C$ is NOT a subspace.

D. $W_D = \{ (x, y, z) \in \mathbb{R}^3 \mid x-y=0 \text{ and } 2y-z=0 \}$
- Zero vector: $0-0=0$ and $2(0)-0=0$. So $(0,0,0) \in W_D$.
- This set can be written as $x=y$ and $z=2y$. So it's a line through the origin, which is a known type of subspace.
$W_D$ is a subspace.

The correct option is C."
:::

:::question type="NAT" question="Consider the vector space $P_2$ of all polynomials of degree at most 2. Let $W = \{ p(x) \in P_2 \mid p(1) = 0 \text{ and } p(2)=0 \}$. What is the dimension of $W$?" answer="1" hint="A polynomial $p(x)$ has roots at $x=1$ and $x=2$. This means $(x-1)$ and $(x-2)$ are factors." solution="A polynomial $p(x) \in P_2$ can be written as $p(x) = ax^2 + bx + c$.
The conditions are $p(1)=0$ and $p(2)=0$.

$p(1) = a(1)^2 + b(1) + c = a+b+c = 0$
$p(2) = a(2)^2 + b(2) + c = 4a+2b+c = 0$

We have a system of linear equations:


• $a+b+c=0$


• $4a+2b+c=0$

Subtract equation (1) from equation (2):
$(4a+2b+c) - (a+b+c) = 0 - 0$
$3a+b = 0 \implies b = -3a$

Substitute $b=-3a$ into equation (1):
$a + (-3a) + c = 0$
$-2a + c = 0 \implies c = 2a$

So, any polynomial $p(x)$ in $W$ must be of the form:
$p(x) = ax^2 + (-3a)x + 2a$
$p(x) = a(x^2 - 3x + 2)$

This means $W = \{ a(x^2 - 3x + 2) \mid a \in \mathbb{R} \}$.
The set $W$ is spanned by the single polynomial $q(x) = x^2 - 3x + 2$.
Since $q(x)$ is not the zero polynomial, it is linearly independent.
Thus, $\{x^2 - 3x + 2\}$ forms a basis for $W$.
The dimension of $W$ is the number of vectors in its basis.

Dimension of $W = 1$."
:::

:::question type="MSQ" question="Let $W_1$ and $W_2$ be subspaces of a vector space $V$. Which of the following statements are always true?" options=["A. $W_1 \cap W_2$ is a subspace of $V$.","B. $W_1 \cup W_2$ is a subspace of $V$.","C. If $W_1 \subseteq W_2$, then $W_1 \cup W_2$ is a subspace of $V$.","D. The set $\{ \mathbf{w}_1 + \mathbf{w}_2 \mid \mathbf{w}_1 \in W_1, \mathbf{w}_2 \in W_2 \}$ is a subspace of $V$ (this is called the sum of subspaces, $W_1+W_2$)."] answer="A,C,D" hint="Recall the properties of intersection, union, and sum of subspaces." solution="Let's analyze each option:

A. $W_1 \cap W_2$ is a subspace of $V$.
This is always true. We proved this in the notes. The intersection of any collection of subspaces is always a subspace.

B. $W_1 \cup W_2$ is a subspace of $V$.
This is generally false. We provided a counterexample with two distinct lines through the origin in $\mathbb{R}^2$. The union is only a subspace if one subspace is contained within the other.

C. If $W_1 \subseteq W_2$, then $W_1 \cup W_2$ is a subspace of $V$.
If $W_1 \subseteq W_2$, then $W_1 \cup W_2 = W_2$. Since $W_2$ is a subspace by definition, their union is also a subspace. This statement is true.

D. The set $\{ \mathbf{w}_1 + \mathbf{w}_2 \mid \mathbf{w}_1 \in W_1, \mathbf{w}_2 \in W_2 \}$ is a subspace of $V$.
This set is denoted as $W_1+W_2$. Let's check the subspace conditions:
1. Zero vector: Since $W_1, W_2$ are subspaces, $\mathbf{0} \in W_1$ and $\mathbf{0} \in W_2$. So $\mathbf{0} = \mathbf{0} + \mathbf{0} \in W_1+W_2$.
2. Closure under addition: Let $\mathbf{u} = \mathbf{w}_{1a} + \mathbf{w}_{2a}$ and $\mathbf{v} = \mathbf{w}_{1b} + \mathbf{w}_{2b}$ be two elements in $W_1+W_2$.
Then $\mathbf{u} + \mathbf{v} = (\mathbf{w}_{1a} + \mathbf{w}_{2a}) + (\mathbf{w}_{1b} + \mathbf{w}_{2b}) = (\mathbf{w}_{1a} + \mathbf{w}_{1b}) + (\mathbf{w}_{2a} + \mathbf{w}_{2b})$.
Since $W_1$ and $W_2$ are subspaces, $\mathbf{w}_{1a} + \mathbf{w}_{1b} \in W_1$ and $\mathbf{w}_{2a} + \mathbf{w}_{2b} \in W_2$.
Thus, $\mathbf{u} + \mathbf{v} \in W_1+W_2$.
3. Closure under scalar multiplication: Let $\mathbf{u} = \mathbf{w}_1 + \mathbf{w}_2 \in W_1+W_2$ and $c \in F$.
Then $c\mathbf{u} = c(\mathbf{w}_1 + \mathbf{w}_2) = c\mathbf{w}_1 + c\mathbf{w}_2$.
Since $W_1$ and $W_2$ are subspaces, $c\mathbf{w}_1 \in W_1$ and $c\mathbf{w}_2 \in W_2$.
Thus, $c\mathbf{u} \in W_1+W_2$.
All conditions are met, so $W_1+W_2$ is always a subspace. This statement is true.

Therefore, options A, C, and D are always true."
:::

:::question type="SUB" question="Prove that the set $W = \{ (x, y, z) \in \mathbb{R}^3 \mid ax+by+cz=0 \}$, where $a,b,c$ are fixed real numbers not all zero, is a subspace of $\mathbb{R}^3$." answer="W is a subspace because it contains the zero vector and is closed under vector addition and scalar multiplication." hint="Use the three conditions of the subspace test." solution="To prove that $W$ is a subspace of $\mathbb{R}^3$, we need to verify the three conditions of the subspace test:

Given $W = \{ (x, y, z) \in \mathbb{R}^3 \mid ax+by+cz=0 \}$.

Condition 1: Contains the Zero Vector

We need to check if the zero vector $\mathbf{0} = (0, 0, 0)$ belongs to $W$.
Substitute $x=0, y=0, z=0$ into the defining equation $ax+by+cz=0$:


$$a(0) + b(0) + c(0) = 0 + 0 + 0 = 0$$

Since $0=0$, the zero vector $(0,0,0)$ satisfies the condition $ax+by+cz=0$.
Therefore, $\mathbf{0} \in W$.

Condition 2: Closed Under Vector Addition

Let $\mathbf{u} = (x_1, y_1, z_1)$ and $\mathbf{v} = (x_2, y_2, z_2)$ be two arbitrary vectors in $W$.
By definition of $W$, they satisfy:


$$ax_1 + by_1 + cz_1 = 0$$


$$ax_2 + by_2 + cz_2 = 0$$

We need to check if their sum $\mathbf{u} + \mathbf{v} = (x_1+x_2, y_1+y_2, z_1+z_2)$ is also in $W$.
This means we need to check if $a(x_1+x_2) + b(y_1+y_2) + c(z_1+z_2) = 0$.

Let's expand the left side:


$$a(x_1+x_2) + b(y_1+y_2) + c(z_1+z_2) = ax_1 + ax_2 + by_1 + by_2 + cz_1 + cz_2$$

Rearrange the terms:

$$(ax_1 + by_1 + cz_1) + (ax_2 + by_2 + cz_2)$$

From our initial assumption that $\mathbf{u}, \mathbf{v} \in W$, we know:

$$ax_1 + by_1 + cz_1 = 0$$


$$ax_2 + by_2 + cz_2 = 0$$

Substitute these values back:

$$0 + 0 = 0$$

Thus, $a(x_1+x_2) + b(y_1+y_2) + c(z_1+z_2) = 0$.
Therefore, $\mathbf{u} + \mathbf{v} \in W$.

Condition 3: Closed Under Scalar Multiplication

Let $\mathbf{u} = (x, y, z)$ be an arbitrary vector in $W$, and let $k$ be any scalar in $\mathbb{R}$.
By definition of $W$, $\mathbf{u}$ satisfies:


$$ax + by + cz = 0$$

We need to check if the scalar multiple $k\mathbf{u} = (kx, ky, kz)$ is also in $W$.
This means we need to check if $a(kx) + b(ky) + c(kz) = 0$.

Let's expand the left side:


$$a(kx) + b(ky) + c(kz) = k(ax) + k(by) + k(cz)$$

Factor out $k$:

$$k(ax + by + cz)$$

From our initial assumption that $\mathbf{u} \in W$, we know:

$$ax + by + cz = 0$$

Substitute this value back:

$$k(0) = 0$$

Thus, $a(kx) + b(ky) + c(kz) = 0$.
Therefore, $k\mathbf{u} \in W$.

Since all three conditions of the subspace test are satisfied, $W$ is a subspace of $\mathbb{R}^3$."
:::

:::question type="MCQ" question="Let $V = M_{2 \times 2}(\mathbb{R})$ be the vector space of all $2 \times 2$ matrices with real entries. Which of the following subsets of $V$ is a subspace?" options=["A. The set of all $2 \times 2$ matrices with determinant 0.","B. The set of all $2 \times 2$ matrices with trace 0.","C. The set of all invertible $2 \times 2$ matrices.","D. The set of all $2 \times 2$ matrices where the sum of all entries is 1."] answer="B. The set of all $2 \times 2$ matrices with trace 0." hint="Apply the subspace test for each set. Remember the zero vector for $M_{2 \times 2}(\mathbb{R})$ is the zero matrix." solution="Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ be a general $2 \times 2$ matrix. The zero vector in $M_{2 \times 2}(\mathbb{R})$ is $O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

A. The set of all $2 \times 2$ matrices with determinant 0.
- Zero vector: $\det(O) = 0$, so $O$ is in this set.
- Closure under addition: Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$. Both have determinant 0.
$A+B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. $\det(A+B) = 1 \ne 0$.
So, not closed under addition. This set is NOT a subspace.

B. The set of all $2 \times 2$ matrices with trace 0. (The trace of a matrix is the sum of its diagonal elements: $tr(A) = a+d$).
- Zero vector: $tr(O) = 0+0=0$, so $O$ is in this set.
- Closure under addition: Let $A = \begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix}$ and $B = \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix}$ be two matrices with trace 0.
So $a_1+d_1=0$ and $a_2+d_2=0$.
$A+B = \begin{pmatrix} a_1+a_2 & b_1+b_2 \\ c_1+c_2 & d_1+d_2 \end{pmatrix}$.
$tr(A+B) = (a_1+a_2) + (d_1+d_2) = (a_1+d_1) + (a_2+d_2) = 0+0=0$. So closed under addition.
- Closure under scalar multiplication: Let $A$ be a matrix with trace 0 ($a+d=0$) and $k \in \mathbb{R}$.
$kA = \begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix}$.
$tr(kA) = ka+kd = k(a+d) = k(0)=0$. So closed under scalar multiplication.
This set IS a subspace.

C. The set of all invertible $2 \times 2$ matrices.
- Zero vector: The zero matrix $O$ is not invertible (its determinant is 0). So $O$ is not in this set.
This set is NOT a subspace.

D. The set of all $2 \times 2$ matrices where the sum of all entries is 1.
- Zero vector: For $O$, the sum of entries is $0+0+0+0=0 \ne 1$. So $O$ is not in this set.
This set is NOT a subspace.

The correct option is B."
:::

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Summary

❗ Key Takeaways for ISI

• Subspace Definition: A subset $W$ of a vector space $V$ is a subspace if it contains the zero vector, is closed under vector addition, and is closed under scalar multiplication.


• Subspace Test Efficiency: Always check for the zero vector first. The combined test ($c\mathbf{u} + \mathbf{v} \in W$) can be more efficient for closure.


• Intersection vs. Union: The intersection of any two (or more) subspaces is always a subspace. The union of two subspaces is generally NOT a subspace unless one is contained within the other.


• Span is a Subspace: The span of any set of vectors is always a subspace. It is the smallest subspace containing those vectors.


• Common Non-Subspaces: Sets not containing the origin (zero vector), or sets defined by non-linear equations, or sets that restrict values to be positive (like first quadrant) are common non-subspaces.

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What's Next?

💡 Continue Learning

This topic connects to:


• Basis and Dimension: Subspaces have their own bases and dimensions, which are crucial for characterizing them. Understanding subspaces is essential before defining basis.


• Linear Transformations: The range (image) and null space (kernel) of a linear transformation are important examples of subspaces, directly related to the transformation's properties.


• Fundamental Subspaces of a Matrix: Row space, column space, and null space are fundamental subspaces associated with a matrix, providing deep insights into the matrix's properties and the solutions to linear systems.




Master these connections for comprehensive ISI preparation!

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Chapter Summary

📖 Vector Spaces - Key Takeaways

Here are the most important points from the "Vector Spaces" chapter that you must master for your ISI preparation:

• Definition of a Vector Space: A vector space $V$ over a field $F$ (typically $\mathbb{R}$ or $\mathbb{C}$ for ISI) is a non-empty set equipped with two operations: vector addition ($+$) and scalar multiplication ($\cdot$), satisfying 10 axioms. You must be able to state and verify these axioms for a given set and operations.


• Linear Combinations: A linear combination of vectors $v_1, v_2, \dots, v_k \in V$ is an expression of the form $c_1v_1 + c_2v_2 + \dots + c_kv_k$, where $c_1, c_2, \dots, c_k \in F$ are scalars. This concept is fundamental to understanding span and linear independence.


• Span of a Set: The span of a non-empty set of vectors $S = \{v_1, \dots, v_k\}$, denoted $span(S)$, is the set of all possible linear combinations of the vectors in $S$. It is always a subspace of $V$, and it is the smallest subspace containing $S$.


• Subspaces: A non-empty subset $W$ of a vector space $V$ is a subspace if it is closed under vector addition and scalar multiplication. This means:

For any $u, v \in W$, $u+v \in W$.
For any $u \in W$ and $c \in F$, $cu \in W$.
Equivalently, for any $u, v \in W$ and $c \in F$, $cu+v \in W$.
Crucially, the zero vector $\mathbf{0}$ must always be in $W$.

• Trivial Subspaces: Every vector space $V$ has at least two subspaces: the zero subspace $\{\mathbf{0}\}$ (containing only the zero vector) and the space $V$ itself.


• Identifying Vector Spaces and Subspaces: Be proficient in determining whether a given set with operations forms a vector space, or if a subset forms a subspace. This often involves checking the closure properties and the presence of the zero vector.


• Foundational Importance: The concepts introduced in this chapter (vector spaces, linear combinations, span, subspaces) are the bedrock of all advanced topics in Linear Algebra. A solid understanding here is critical for succeeding in subsequent chapters like Linear Independence, Basis & Dimension, and Linear Transformations.

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Chapter Review Questions

:::question type="MCQ" question="Which of the following subsets is NOT a subspace of the vector space $M_{2 \times 2}(\mathbb{R})$ of all $2 \times 2$ real matrices?" options=["A) The set of all symmetric $2 \times 2$ matrices ($A^T = A$).","B) The set of all $2 \times 2$ matrices with determinant zero.","C) The set of all $2 \times 2$ matrices $A$ such that $A \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.","D) The set of all $2 \times 2$ matrices with trace zero ($tr(A) = 0$)."] answer="B" hint="Recall the definition of a subspace. A common pitfall is assuming properties like determinance or invertibility preserve closure under addition or scalar multiplication." solution="Let's analyze each option:

A) The set of all symmetric $2 \times 2$ matrices:
Let $W_A = \{A \in M_{2 \times 2}(\mathbb{R}) \mid A^T = A\}$.


• Zero vector: The zero matrix $O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ is symmetric ($O^T = O$), so $O \in W_A$.


• Closure under addition: If $A, B \in W_A$, then $A^T = A$ and $B^T = B$.

$(A+B)^T = A^T + B^T = A + B$. So, $A+B \in W_A$.

• Closure under scalar multiplication: If $A \in W_A$ and $c \in \mathbb{R}$, then $(cA)^T = cA^T = cA$. So, $cA \in W_A$.

Thus, $W_A$ is a subspace.

B) The set of all $2 \times 2$ matrices with determinant zero:
Let $W_B = \{A \in M_{2 \times 2}(\mathbb{R}) \mid \det(A) = 0\}$.


• Zero vector: $\det(O) = 0$, so $O \in W_B$.


• Closure under addition: Consider $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$.

$\det(A) = 0$ and $\det(B) = 0$, so $A, B \in W_B$.
However, $A+B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
$\det(A+B) = 1 \ne 0$. So, $A+B \notin W_B$.
Thus, $W_B$ is NOT a subspace.

C) The set of all $2 \times 2$ matrices $A$ such that $A \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$:
Let $W_C = \{A \in M_{2 \times 2}(\mathbb{R}) \mid A \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\}$.


• Zero vector: $O \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, so $O \in W_C$.


• Closure under addition: If $A, B \in W_C$, then $A \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ and $B \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.

$(A+B) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = A \begin{pmatrix} 1 \\ 0 \end{pmatrix} + B \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. So, $A+B \in W_C$.

• Closure under scalar multiplication: If $A \in W_C$ and $c \in \mathbb{R}$, then $(cA) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = c \left( A \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right) = c \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. So, $cA \in W_C$.

Thus, $W_C$ is a subspace.

D) The set of all $2 \times 2$ matrices with trace zero:
Let $W_D = \{A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_{2 \times 2}(\mathbb{R}) \mid tr(A) = a+d = 0\}$.


• Zero vector: $tr(O) = 0+0 = 0$, so $O \in W_D$.


• Closure under addition: If $A = \begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix}, B = \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} \in W_D$, then $a_1+d_1 = 0$ and $a_2+d_2 = 0$.

$A+B = \begin{pmatrix} a_1+a_2 & b_1+b_2 \\ c_1+c_2 & d_1+d_2 \end{pmatrix}$.
$tr(A+B) = (a_1+a_2) + (d_1+d_2) = (a_1+d_1) + (a_2+d_2) = 0+0 = 0$. So, $A+B \in W_D$.

• Closure under scalar multiplication: If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in W_D$ and $k \in \mathbb{R}$, then $a+d=0$.

$kA = \begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix}$.
$tr(kA) = ka+kd = k(a+d) = k \cdot 0 = 0$. So, $kA \in W_D$.
Thus, $W_D$ is a subspace.

The only set that is not a subspace is B.
"
:::

:::question type="NAT" question="Let $V$ be the vector space of all real-valued continuous functions on $[0,1]$. How many of the following subsets are subspaces of $V$?


• $W_1 = \{ f \in V \mid f(0) = 0 \}$


• $W_2 = \{ f \in V \mid f(x) \ge 0 \text{ for all } x \in [0,1] \}$


• $W_3 = \{ f \in V \mid \int_0^1 f(x) dx = 0 \}$


• $W_4 = \{ f \in V \mid f(0) + f(1) = 1 \}$" answer="2" hint="For each subset, check the three conditions for a subspace: contains the zero vector, closed under addition, and closed under scalar multiplication. Pay close attention to conditions that might fail for scalar multiplication with negative numbers, or for addition." solution="We need to check each subset against the three subspace criteria: contains the zero vector, closed under addition, and closed under scalar multiplication. The zero vector in $V$ is the function $z(x) = 0$ for all $x \in [0,1]$.
• $W_1 = \{ f \in V \mid f(0) = 0 \}$

* Zero vector: $z(0) = 0$, so $z \in W_1$.
* Closure under addition: If $f, g \in W_1$, then $f(0)=0$ and $g(0)=0$. Thus, $(f+g)(0) = f(0)+g(0) = 0+0 = 0$. So $f+g \in W_1$.
* Closure under scalar multiplication: If $f \in W_1$ and $c \in \mathbb{R}$, then $f(0)=0$. Thus, $(cf)(0) = c \cdot f(0) = c \cdot 0 = 0$. So $cf \in W_1$.
Therefore, $W_1$ is a subspace.
• $W_2 = \{ f \in V \mid f(x) \ge 0 \text{ for all } x \in [0,1] \}$

* Zero vector: $z(x) = 0 \ge 0$ for all $x$, so $z \in W_2$.
* Closure under scalar multiplication (failure): Consider $f(x) = x^2$. For $x \in [0,1]$, $f(x) \ge 0$, so $f \in W_2$.
Let $c = -1$. Then $(cf)(x) = -x^2$. For $x \ne 0$, $-x^2 < 0$. So $(cf)(x)$ is not always $\ge 0$.
Thus, $-f \notin W_2$.
Therefore, $W_2$ is NOT a subspace.
• $W_3 = \{ f \in V \mid \int_0^1 f(x) dx = 0 \}$

* Zero vector: $\int_0^1 z(x) dx = \int_0^1 0 dx = 0$, so $z \in W_3$.
* Closure under addition: If $f, g \in W_3$, then $\int_0^1 f(x) dx = 0$ and $\int_0^1 g(x) dx = 0$.
$\int_0^1 (f+g)(x) dx = \int_0^1 f(x) dx + \int_0^1 g(x) dx = 0+0 = 0$. So $f+g \in W_3$.
* Closure under scalar multiplication: If $f \in W_3$ and $c \in \mathbb{R}$, then $\int_0^1 f(x) dx = 0$.
$\int_0^1 (cf)(x) dx = c \int_0^1 f(x) dx = c \cdot 0 = 0$. So $cf \in W_3$.
Therefore, $W_3$ is a subspace.
• $W_4 = \{ f \in V \mid f(0) + f(1) = 1 \}$

* Zero vector (failure): For the zero function $z(x)=0$, $z(0)+z(1) = 0+0 = 0 \ne 1$. So $z \notin W_4$.
(Since the zero vector is not in $W_4$, we can immediately conclude it's not a subspace. We can also check other conditions for completeness.)
* Closure under addition (failure): Let $f(x)=x$ and $g(x)=1-x$. $f(0)+f(1)=0+1=1$, so $f \in W_4$. $g(0)+g(1)=1+0=1$, so $g \in W_4$.
$(f+g)(0)+(f+g)(1) = (f(0)+g(0)) + (f(1)+g(1)) = (0+1) + (1+0) = 1+1=2 \ne 1$. So $f+g \notin W_4$.
Therefore, $W_4$ is NOT a subspace.

In summary, $W_1$ and $W_3$ are subspaces. Thus, there are 2 subspaces.
"
:::

:::question type="Descriptive" question="Let $W_1$ and $W_2$ be two subspaces of a vector space $V$. Prove that their union $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$." solution="We need to prove both directions:

Part 1: ($\Rightarrow$) If $W_1 \cup W_2$ is a subspace of $V$, then $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$.

Assume for contradiction that $W_1 \cup W_2$ is a subspace, but neither $W_1 \subseteq W_2$ nor $W_2 \subseteq W_1$.
Since $W_1 \not\subseteq W_2$, there exists a vector $u \in W_1$ such that $u \notin W_2$.
Since $W_2 \not\subseteq W_1$, there exists a vector $v \in W_2$ such that $v \notin W_1$.

Now, consider the sum $u+v$.
Since $u \in W_1$ and $v \in W_2$, both $u$ and $v$ are elements of $W_1 \cup W_2$.
Because we assumed $W_1 \cup W_2$ is a subspace, it must be closed under vector addition.
Therefore, $u+v \in W_1 \cup W_2$.
This implies that $u+v \in W_1$ or $u+v \in W_2$.

Case 1: $u+v \in W_1$.
Since $u \in W_1$ and $W_1$ is a subspace, $-(u) \in W_1$.
Then $v = (u+v) - u = (u+v) + (-u)$. Since $u+v \in W_1$ and $-u \in W_1$, their sum $v$ must be in $W_1$ (by closure of $W_1$ under addition).
But this contradicts our initial choice that $v \notin W_1$.

Case 2: $u+v \in W_2$.
Since $v \in W_2$ and $W_2$ is a subspace, $-(v) \in W_2$.
Then $u = (u+v) - v = (u+v) + (-v)$. Since $u+v \in W_2$ and $-v \in W_2$, their sum $u$ must be in $W_2$ (by closure of $W_2$ under addition).
But this contradicts our initial choice that $u \notin W_2$.

Since both cases lead to a contradiction, our initial assumption (that $W_1 \not\subseteq W_2$ and $W_2 \not\subseteq W_1$) must be false.
Therefore, if $W_1 \cup W_2$ is a subspace, then $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$.

Part 2: ($\Leftarrow$) If $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$, then $W_1 \cup W_2$ is a subspace of $V$.

Assume $W_1 \subseteq W_2$.
Then $W_1 \cup W_2 = W_2$. Since $W_2$ is given to be a subspace of $V$, their union is also a subspace.

Assume $W_2 \subseteq W_1$.
Then $W_1 \cup W_2 = W_1$. Since $W_1$ is given to be a subspace of $V$, their union is also a subspace.

In both cases, $W_1 \cup W_2$ is a subspace.

Combining both parts, we conclude that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$.
"
:::

:::question type="NAT" question="Let $S = \{ (1,1,0), (1,0,1), (0,1,k) \}$ be a set of vectors in $\mathbb{R}^3$. Find the value of $k$ for which $S$ does NOT span $\mathbb{R}^3$." answer="-1" hint="A set of $n$ vectors in $\mathbb{R}^n$ spans $\mathbb{R}^n$ if and only if the vectors are linearly independent. This condition can be checked by evaluating the determinant of the matrix formed by these vectors." solution="A set of $n$ vectors in $\mathbb{R}^n$ spans $\mathbb{R}^n$ if and only if the vectors are linearly independent. This is equivalent to saying that the matrix formed by these vectors as columns (or rows) has a non-zero determinant.
If the set $S$ does NOT span $\mathbb{R}^3$, it means the vectors are linearly dependent, and the determinant of the matrix formed by them must be zero.

Let's form a matrix $A$ with these vectors as columns:


$$A = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & k \end{pmatrix}$$

Now, we calculate the determinant of $A$:


$$\det(A) = 1 \cdot \det \begin{pmatrix} 0 & 1 \\ 1 & k \end{pmatrix} - 1 \cdot \det \begin{pmatrix} 1 & 1 \\ 0 & k \end{pmatrix} + 0 \cdot \det \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$


$$\det(A) = 1 \cdot (0 \cdot k - 1 \cdot 1) - 1 \cdot (1 \cdot k - 1 \cdot 0) + 0$$


$$\det(A) = 1 \cdot (-1) - 1 \cdot (k) + 0$$


$$\det(A) = -1 - k$$

For $S$ to NOT span $\mathbb{R}^3$, the determinant must be zero:


$$-1 - k = 0$$


$$k = -1$$

Thus, for $k=-1$, the vectors are linearly dependent and will not span $\mathbb{R}^3$. Instead, they will span a plane (a 2-dimensional subspace) in $\mathbb{R}^3$.

The value of $k$ is -1."
:::

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What's Next?

💡 Continue Your ISI Journey

You've successfully navigated the fundamental concepts of Vector Spaces! This chapter is the cornerstone of Linear Algebra, and a thorough understanding here will make your journey through the rest of the subject much smoother.

Key connections:

Foundation for Structure: This chapter provided you with the definition of a vector space and the building blocks (linear combinations, span, subspaces) for understanding their internal structure.
Building on these concepts: The next crucial steps in your ISI Linear Algebra preparation will directly build upon these ideas:
Linear Independence, Basis, and Dimension: These concepts allow us to precisely describe the 'size' and 'coordinates' within any vector space, generalizing notions from $\mathbb{R}^2$ and $\mathbb{R}^3$.
Linear Transformations: These are functions between vector spaces that respect their algebraic structure, and they are intimately linked to matrices.
Eigenvalues and Eigenvectors: Understanding these requires a solid grasp of linear transformations and subspaces.
Inner Product Spaces: This chapter will introduce geometric concepts like length, angle, and orthogonality, which rely on the vector space framework.
* ISI Relevance: Problems often combine concepts from multiple chapters. For instance, you might be asked to find a basis for a subspace defined by certain conditions, or to determine if a set of vectors spans a particular vector space. Mastering the definitions and properties from this chapter is non-negotiable for tackling such complex problems.

Continue your journey by exploring Linear Independence, Basis, and Dimension to fully grasp how to characterize and work with the elements of vector spaces.