Vector Spaces

This chapter introduces the fundamental concepts of vector spaces, subspaces, linear dependence, independence, basis, and dimension. These core principles form the bedrock of linear algebra, providing essential tools for abstract mathematical reasoning. Mastery of these topics is critical for the CUET PG MA examination, where they are frequently assessed and serve as prerequisites for advanced subjects.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Vector Spaces and Subspaces |
| 2 | Linear Dependence and Independence |
| 3 | Basis and Dimension |

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We begin with Vector Spaces and Subspaces.

Part 1: Vector Spaces and Subspaces

Vector spaces form the fundamental algebraic structure in linear algebra, providing a framework for operations on objects such as vectors, polynomials, and functions. We define a vector space as a set equipped with two operations—vector addition and scalar multiplication—that satisfy a specific set of axioms. Understanding vector spaces and their subspaces is crucial for solving problems involving linear transformations, eigenvalues, and basis constructions, which are frequently tested in competitive examinations.

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Core Concepts

1. Definition of a Vector Space

A real vector space $V$ is a set of objects, called vectors, together with two operations: vector addition and scalar multiplication. These operations must satisfy the following ten axioms for all vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ in $V$ and all scalars $c, d$ in $\mathbb{R}$:

Closure under Addition: $\mathbf{u} + \mathbf{v}$ is in $V$.

Commutativity of Addition: $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$.

Associativity of Addition: $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$.

Existence of Zero Vector: There exists a zero vector $\mathbf{0}$ in $V$ such that $\mathbf{u} + \mathbf{0} = \mathbf{u}$.

Existence of Negative Vector: For each $\mathbf{u}$ in $V$, there exists a vector $-\mathbf{u}$ in $V$ such that $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$.

Closure under Scalar Multiplication: $c\mathbf{u}$ is in $V$.

Distributivity (Scalar over Vector Addition): $c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}$.

Distributivity (Vector over Scalar Addition): $(c + d)\mathbf{u} = c\mathbf{u} + d\mathbf{u}$.

Associativity of Scalar Multiplication: $c(d\mathbf{u}) = (cd)\mathbf{u}$.

Identity for Scalar Multiplication: $1\mathbf{u} = \mathbf{u}$.

Quick Example:
Consider the set $V = \mathbb{R}^2$ with standard vector addition and scalar multiplication. We demonstrate closure under addition.

Step 1: Let $\mathbf{u} = (u_1, u_2)$ and $\mathbf{v} = (v_1, v_2)$ be arbitrary vectors in $\mathbb{R}^2$.

Step 2: Perform vector addition.

Answer: Since $u_1+v_1$ and $u_2+v_2$ are real numbers, $(u_1+v_1, u_2+v_2)$ is an element of $\mathbb{R}^2$. Thus, $\mathbb{R}^2$ is closed under vector addition. The other nine axioms can be verified similarly.

:::question type="MCQ" question="Which of the following sets, with standard addition and scalar multiplication, forms a vector space over $\mathbb{R}$?" options=["The set of all $2 \times 2$ matrices with integer entries.","The set of all polynomials of degree exactly 3.","The set of all continuous functions $f: [0,1] \to \mathbb{R}$.","The set of all $2 \times 2$ invertible matrices."] answer="The set of all continuous functions $f: [0,1] \to \mathbb{R}$." hint="Check the closure axioms and existence of the zero vector for each option." solution="Step 1: Analyze 'The set of all $2 \times 2$ matrices with integer entries'.
This set is not closed under scalar multiplication by all real numbers. For example, if $A$ is the matrix

which has integer entries, then the scalar multiple $\frac{1}{2}A$ is

which does not have integer entries. Thus, it is not a vector space over $\mathbb{R}$.

Step 2: Analyze 'The set of all polynomials of degree exactly 3'.
This set does not contain the zero vector (the zero polynomial has undefined degree, or degree $-\infty$). Also, it is not closed under addition. For example, $(x^3+x) + (-x^3+1) = x+1$, which has degree 1, not 3. Thus, it is not a vector space.

Step 3: Analyze 'The set of all continuous functions $f: [0,1] \to \mathbb{R}$'.
The sum of two continuous functions is continuous. A scalar multiple of a continuous function is continuous. The zero function $f(x)=0$ is continuous. The negative of a continuous function is continuous. All other vector space axioms hold for functions. Thus, this set forms a vector space.

Step 4: Analyze 'The set of all $2 \times 2$ invertible matrices'.
This set does not contain the zero matrix, which is required for a vector space. Also, it is not closed under addition. For example, $I$ is the identity matrix

which is invertible, and $-I$ is

which is also invertible. However, their sum $I + (-I)$ is

which is the zero matrix and is not invertible. Thus, it is not a vector space.
Answer: \boxed{The set of all continuous functions $f: [0,1] \to \mathbb{R}$}
"
:::

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2. Definition of a Subspace

A subspace $W$ of a vector space $V$ is a subset of $V$ that is itself a vector space under the same operations of vector addition and scalar multiplication defined on $V$. To verify if a non-empty subset $W$ is a subspace, we only need to check three conditions, often condensed into two.

Quick Example:
Determine if $W = \{(x, y) \in \mathbb{R}^2 : x - 2y = 0\}$ is a subspace of $\mathbb{R}^2$.

Step 1: Check if the zero vector is in $W$.

Since $0=0$, the zero vector is in $W$. $W$ is non-empty.

Step 2: Check closure under addition. Let $\mathbf{u} = (x_1, y_1)$ and $\mathbf{v} = (x_2, y_2)$ be in $W$.
This implies $x_1 - 2y_1 = 0$ and $x_2 - 2y_2 = 0$.

We check if $(x_1 + x_2) - 2(y_1 + y_2) = 0$.

Thus, $\mathbf{u} + \mathbf{v}$ is in $W$.

Step 3: Check closure under scalar multiplication. Let $\mathbf{u} = (x, y)$ be in $W$ and $c \in \mathbb{R}$.
This implies $x - 2y = 0$.

We check if $cx - 2(cy) = 0$.

Thus, $c\mathbf{u}$ is in $W$.

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

:::question type="MCQ" question="Which of the following is a subspace of $\mathbb{R}^3$?" options=["$W = \\{(x, y, z) \in \mathbb{R}^3 : x + 4y - 10z = -2\\}$","$W = \\{(x, y, z) \in \mathbb{R}^3 : xy = 0\\}$","$W = \\{(x, y, z) \in \mathbb{R}^3 : 2x + 3y - 4z = 0\\}$","$W = \\{(x, y, z) \in \mathbb{R}^3 : x \in \mathbb{Q}\\}$"] answer="$W = \\{(x, y, z) \in \mathbb{R}^3 : 2x + 3y - 4z = 0\\}$" hint="Apply the 3-condition subspace test. Pay attention to the zero vector and closure properties." solution="Step 1: Analyze $W = \\{(x, y, z) \in \mathbb{R}^3 : x + 4y - 10z = -2\\}$.
The zero vector $(0,0,0)$ does not satisfy the condition $0 + 4(0) - 10(0) = -2$, as $0 \neq -2$. Therefore, $W$ is not a subspace.

Step 2: Analyze $W = \\{(x, y, z) \in \mathbb{R}^3 : xy = 0\\}$.
The zero vector $(0,0,0)$ satisfies $0 \cdot 0 = 0$.
Consider closure under addition. Let $\mathbf{u} = (1, 0, 0)$ and $\mathbf{v} = (0, 1, 0)$. Both are in $W$ since $1 \cdot 0 = 0$ and $0 \cdot 1 = 0$.
However, $\mathbf{u} + \mathbf{v} = (1, 1, 0)$. For this vector, $x \cdot y = 1 \cdot 1 = 1 \neq 0$. So $\mathbf{u} + \mathbf{v}$ is not in $W$. Thus, $W$ is not closed under addition and is not a subspace.

Step 3: Analyze $W = \\{(x, y, z) \in \mathbb{R}^3 : 2x + 3y - 4z = 0\\}$.

Zero vector: $2(0) + 3(0) - 4(0) = 0$. The zero vector is in $W$.

Closure under addition: Let $\mathbf{u} = (x_1, y_1, z_1)$ and $\mathbf{v} = (x_2, y_2, z_2)$ be in $W$.

Then $2x_1 + 3y_1 - 4z_1 = 0$ and $2x_2 + 3y_2 - 4z_2 = 0$.
Their sum is $\mathbf{u} + \mathbf{v} = (x_1+x_2, y_1+y_2, z_1+z_2)$.
We check: $2(x_1+x_2) + 3(y_1+y_2) - 4(z_1+z_2) = (2x_1+3y_1-4z_1) + (2x_2+3y_2-4z_2) = 0 + 0 = 0$.
So $\mathbf{u} + \mathbf{v}$ is in $W$.

Closure under scalar multiplication: Let $\mathbf{u} = (x, y, z)$ be in $W$ and $c \in \mathbb{R}$.

Then $2x + 3y - 4z = 0$.
The scalar multiple is $c\mathbf{u} = (cx, cy, cz)$.
We check: $2(cx) + 3(cy) - 4(cz) = c(2x+3y-4z) = c(0) = 0$.
So $c\mathbf{u}$ is in $W$.
Since all conditions are met, $W$ is a subspace.

Step 4: Analyze $W = \\{(x, y, z) \in \mathbb{R}^3 : x \in \mathbb{Q}\\}$.
The zero vector $(0,0,0)$ is in $W$ since $0 \in \mathbb{Q}$.
Consider closure under scalar multiplication. Let $\mathbf{u} = (1, 0, 0)$. This is in $W$ since $1 \in \mathbb{Q}$.
Let $c = \sqrt{2}$. Then $c\mathbf{u} = (\sqrt{2}, 0, 0)$. Here, $\sqrt{2} \notin \mathbb{Q}$. So $c\mathbf{u}$ is not in $W$. Thus, $W$ is not closed under scalar multiplication and is not a subspace."
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3. Intersection and Union of Subspaces

We examine how set operations interact with the subspace property.

3.1. Intersection of Subspaces

The intersection of any two subspaces of a vector space is always a subspace itself. This is a fundamental property.

Quick Example:
Let $W_1 = \{(x, y, z) \in \mathbb{R}^3 : x=0\}$ (the $yz$-plane) and $W_2 = \{(x, y, z) \in \mathbb{R}^3 : y=0\}$ (the $xz$-plane). Both are subspaces of $\mathbb{R}^3$.
We find their intersection $W_1 \cap W_2$.

Step 1: Define the intersection.
A vector $(x,y,z)$ is in $W_1 \cap W_2$ if it satisfies both conditions: $x=0$ and $y=0$.

Step 2: Check the subspace conditions for $W_1 \cap W_2$.

Zero vector: For $(0,0,0)$, $x=0$ and $y=0$ are satisfied. So $\mathbf{0} \in W_1 \cap W_2$.

Closure under addition: Let $\mathbf{u} = (0, 0, z_1)$ and $\mathbf{v} = (0, 0, z_2)$ be in $W_1 \cap W_2$.

This vector satisfies $x=0$ and $y=0$, so it is in $W_1 \cap W_2$.

Closure under scalar multiplication: Let $\mathbf{u} = (0, 0, z)$ be in $W_1 \cap W_2$ and $c \in \mathbb{R}$.

This vector satisfies $x=0$ and $y=0$, so it is in $W_1 \cap W_2$.

Answer: $W_1 \cap W_2$ is the $z$-axis, which is a subspace of $\mathbb{R}^3$.

:::question type="MCQ" question="Let $W_1$ and $W_2$ be any two subspaces of a vector space $V$. Which of the following statements is true regarding their intersection and union?" options=["$W_1 \cup W_2$ is always a subspace, but $W_1 \cap W_2$ is not always a subspace.","$W_1 \cap W_2$ is always a subspace, and $W_1 \cup W_2$ is always a subspace.","$W_1 \cap W_2$ is always a subspace, but $W_1 \cup W_2$ is not always a subspace.","$Neither W_1 \cap W_2$ nor $W_1 \cup W_2$ are always subspaces."] answer="$W_1 \cap W_2$ is always a subspace, but $W_1 \cup W_2$ is not always a subspace." hint="Recall the properties of subspace intersection and union. For the union, consider a counterexample." solution="Step 1: Analyze the intersection $W_1 \cap W_2$.
We know that if $W_1$ and $W_2$ are subspaces of $V$, then $W_1 \cap W_2$ is always a subspace of $V$. This is a standard theorem in linear algebra.

Step 2: Analyze the union $W_1 \cup W_2$.
The union of two subspaces is not generally a subspace.
Consider a counterexample: Let $V = \mathbb{R}^2$.
Let $W_1 = \{(x, 0) : x \in \mathbb{R}\}$ (the x-axis) and $W_2 = \{(0, y) : y \in \mathbb{R}\}$ (the y-axis).
Both $W_1$ and $W_2$ are subspaces of $\mathbb{R}^2$.
Their union is $W_1 \cup W_2 = \{(x, y) : x=0 \text{ or } y=0\}$.
Consider $\mathbf{u} = (1, 0) \in W_1 \cup W_2$ and $\mathbf{v} = (0, 1) \in W_1 \cup W_2$.
Their sum is $\mathbf{u} + \mathbf{v} = (1, 1)$.
For $(1,1)$, neither $x=0$ nor $y=0$ is true. Thus, $(1,1) \notin W_1 \cup W_2$.
Therefore, $W_1 \cup W_2$ is not closed under addition, and hence is not a subspace.

Step 3: Conclude.
$W_1 \cap W_2$ is always a subspace, but $W_1 \cup W_2$ is not always a subspace. This matches the third option."
:::

3.2. Union of Subspaces

The union of two subspaces $W_1$ and $W_2$ is generally not a subspace. It is a subspace if and only if one subspace is contained within the other (i.e., $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$).

4. Linear Combinations and Span

A vector $\mathbf{v}$ is a linear combination of vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ if it can be expressed as a sum of scalar multiples of these vectors. The set of all possible linear combinations of a set of vectors is called their span.

Quick Example:
Determine if $\mathbf{v} = (7, 4, -3)$ is in the span of $S = \{\mathbf{v}_1, \mathbf{v}_2\}$, where $\mathbf{v}_1 = (1, 2, -1)$ and $\mathbf{v}_2 = (2, -1, 3)$.

Step 1: Set up the linear combination equation.
We need to find scalars $c_1, c_2$ such that $\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2$.

This leads to a system of linear equations:

Step 2: Solve the system of equations.
From the first equation, $c_1 = 7 - 2c_2$. Substitute into the second equation:

Substitute $c_2 = 2$ back into $c_1 = 7 - 2c_2$:

Check with the third equation:

The third equation requires $-c_1 + 3c_2 = -3$, but we found it equals $3$. Since $3 \neq -3$, the system is inconsistent.

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

:::question type="NAT" question="Let $\mathbf{v}_1 = (1, 0, 1)$, $\mathbf{v}_2 = (0, 1, 1)$, and $\mathbf{v}_3 = (1, 1, 0)$. If $\mathbf{w} = (2, 3, 5)$ is written as a linear combination $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3$, what is the value of $c_1 + c_2 + c_3$?" answer="5" hint="Set up a system of linear equations and solve for $c_1, c_2, c_3$. Then sum them." solution="Step 1: Set up the system of linear equations from $\mathbf{w} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3$.

This yields:

Step 2: Solve the system.
From the third equation, $c_1 = 5 - c_2$. Substitute into the first equation:

Now we have a system of two equations with $c_2$ and $c_3$:

Add these two equations:

Substitute $c_3 = 0$ into $c_2 + c_3 = 3$:

Substitute $c_2 = 3$ into $c_1 + c_2 = 5$:

So, $c_1=2, c_2=3, c_3=0$.

Step 3: Calculate $c_1 + c_2 + c_3$.

"
:::

5. Linear Independence and Dependence

The concept of linear independence is central to defining a basis for a vector space.

Quick Example:
Determine if the vectors $\mathbf{v}_1 = (1, 2, 3)$, $\mathbf{v}_2 = (0, 1, 2)$, $\mathbf{v}_3 = (-1, 0, 1)$ are linearly independent in $\mathbb{R}^3$.

Step 1: Set up the equation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}$.

This forms a homogeneous system of linear equations:

Step 2: Form the augmented matrix and reduce it to row echelon form.

Perform row operations: $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1$.

Perform row operation: $R_3 \to R_3 - 2R_2$.

Step 3: Interpret the result.
The system has infinitely many solutions because there is a free variable ($c_3$).
For example, if we let $c_3 = t$, then $c_2 = -2t$ and $c_1 = t$.
A non-trivial solution exists (e.g., $t=1 \implies c_1=1, c_2=-2, c_3=1$).

Answer: The vectors are linearly dependent.

:::question type="MCQ" question="Consider the vectors $\mathbf{u} = (1, 2, 3)$, $\mathbf{v} = (0, 1, 1)$, and $\mathbf{w} = (2, 5, 7)$ in $\mathbb{R}^3$. Which of the following statements is true?" options=["The vectors are linearly independent.","The vectors are linearly dependent, and $\mathbf{w}$ is a linear combination of $\mathbf{u}$ and $\mathbf{v}$.","The vectors are linearly dependent, but $\mathbf{w}$ is not a linear combination of $\mathbf{u}$ and $\mathbf{v}$.","The vectors span $\mathbb{R}^3$ but are not linearly independent."] answer="The vectors are linearly dependent, and $\mathbf{w}$ is a linear combination of $\mathbf{u}$ and $\mathbf{v}$." hint="Form a matrix with the vectors as columns and compute its determinant or perform row reduction. If dependent, find the linear combination." solution="Step 1: Check for linear independence using the determinant.
Form a matrix $A$ with the vectors as columns:

Calculate the determinant of $A$:

Since $\det(A) = 0$, the vectors are linearly dependent. This eliminates the first option.

Step 2: Check if $\mathbf{w}$ is a linear combination of $\mathbf{u}$ and $\mathbf{v}$.
We need to find if there exist scalars $c_1, c_2$ such that $\mathbf{w} = c_1\mathbf{u} + c_2\mathbf{v}$.

This gives the system:

From the first equation, $c_1 = 2$.
Substitute into the second equation:

Check with the third equation:

The values $c_1=2$ and $c_2=1$ satisfy all equations.
Thus, $\mathbf{w} = 2\mathbf{u} + 1\mathbf{v}$.

Step 3: Conclude.
The vectors are linearly dependent, and $\mathbf{w}$ is a linear combination of $\mathbf{u}$ and $\mathbf{v}$. This matches the second option."
:::

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6. Basis and Dimension

A basis provides a minimal set of vectors that can represent every vector in a vector space uniquely. The number of vectors in a basis defines the dimension of the space.

Quick Example:
Find a basis for the subspace $W = \{(x, y, z) \in \mathbb{R}^3 : x + y - z = 0\}$.

Step 1: Express the vectors in $W$ in parametric form.
From the condition $x + y - z = 0$, we can express $z$ in terms of $x$ and $y$: $z = x + y$.
Any vector in $W$ can be written as:

We can decompose this vector:

Step 2: Identify the spanning vectors.
The vectors $(1, 0, 1)$ and $(0, 1, 1)$ span $W$. Let $B = \{(1, 0, 1), (0, 1, 1)\}$.

Step 3: Check for linear independence of the spanning vectors.
Assume $c_1(1, 0, 1) + c_2(0, 1, 1) = (0, 0, 0)$.

This implies $c_1 = 0$ and $c_2 = 0$.
Thus, the vectors are linearly independent.

Answer: A basis for $W$ is $\{(1, 0, 1), (0, 1, 1)\}$, and $\operatorname{dim}(W) = 2$.

:::question type="MCQ" question="Consider the vector space $P_2$ of all polynomials of degree at most 2. Which of the following sets is a basis for $P_2$?" options=["$\{1, x, x^2, x^3\}$","$\{x^2, x+1, x^2-1\}$","$\{1, x, x^2\}$","$\{x^2, 2x^2, 3x^2\}$"] answer="$\{1, x, x^2\}$" hint="A basis must be linearly independent and span the space. The dimension of $P_2$ is 3." solution="Step 1: Determine the dimension of $P_2$.
The standard basis for $P_2$ is $\{1, x, x^2\}$. This set has 3 elements. Therefore, $\operatorname{dim}(P_2) = 3$. Any basis for $P_2$ must contain exactly 3 linearly independent vectors that span $P_2$.

Step 2: Evaluate option 1: $\{1, x, x^2, x^3\}$.
This set has 4 vectors. Since $\operatorname{dim}(P_2)=3$, a set of 4 vectors in $P_2$ must be linearly dependent. Also, $x^3$ is not in $P_2$. Thus, this is not a basis.

Step 3: Evaluate option 2: $\{x^2, x+1, x^2-1\}$.
This set has 3 vectors. We need to check for linear independence.
Assume $c_1x^2 + c_2(x+1) + c_3(x^2-1) = 0$ for all $x$.

For this to hold for all $x$, the coefficients must be zero:

From $c_2=0$ and $c_2-c_3=0$, we get $c_3=0$.
From $c_1+c_3=0$ and $c_3=0$, we get $c_1=0$.
Since $c_1=c_2=c_3=0$ is the only solution, the set is linearly independent.
A set of $\operatorname{dim}(V)$ linearly independent vectors in $V$ is always a basis. Thus, this is a basis.

Step 4: Evaluate option 3: $\{1, x, x^2\}$.
This is the standard basis for $P_2$. It is linearly independent and spans $P_2$. It is a basis.

Step 5: Evaluate option 4: $\{x^2, 2x^2, 3x^2\}$.
This set is linearly dependent because $2x^2 = 2 \cdot x^2$ and $3x^2 = 3 \cdot x^2$. For example, $2(x^2) - (2x^2) + 0(3x^2) = 0$ is a non-trivial linear combination. Thus, it is not a basis."
:::

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Part 2: Linear Dependence and Independence

We examine the fundamental concepts of linear dependence and independence, which are crucial for understanding vector spaces and their bases. These concepts enable us to determine whether a set of vectors contains redundant information or forms a minimal set sufficient to describe a subspace. Mastery of these ideas is essential for various problems in linear algebra, including those encountered in competitive examinations.

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Core Concepts

1. Linear Combination

A vector $\mathbf{v}$ is a linear combination of vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ if it can be expressed as a sum of scalar multiples of these vectors. That is, $\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_k\mathbf{v}_k$ for some scalars $c_1, c_2, \ldots, c_k$.

Quick Example: Determine if $\mathbf{v}=(7, 1, -8)$ is a linear combination of $\mathbf{v}_1=(1, 2, -1)$ and $\mathbf{v}_2=(2, -3, 5)$.

Step 1: Set up the vector equation.
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Step 2: Form the system of linear equations.
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Step 3: Solve the system using an augmented matrix.
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Step 4: Interpret the result.
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The last row indicates $0 = -14$, which is a contradiction. Therefore, no such scalars $c_1, c_2$ exist.

Answer: $\mathbf{v}$ is not a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$.

:::question type="MCQ" question="Given vectors $\mathbf{u}=(1, 1, 0)$, $\mathbf{v}=(0, 1, 1)$, and $\mathbf{w}=(1, 0, 1)$, determine the coefficients $a, b, c$ such that $a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = (2, 3, 1)$. Which of the following is correct?" options=["$a=2, b=1, c=0$","$a=1, b=2, c=-1$","$a=0, b=1, c=2$","$a=1, b=1, c=1$"] answer="$a=2, b=1, c=0$" hint="Set up a system of linear equations and solve for $a, b, c$." solution="Step 1: Set up the vector equation and corresponding system of equations.
>

>

Step 2: Solve the system. From the first equation, $c = 2-a$. From the second, $b = 3-a$.
Substitute these into the third equation:
>

>
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Step 3: Find $b$ and $c$.
>

>
The coefficients are $a=2, b=1, c=0$.

Step 4: Check the options. The provided solution `$a=1, b=2, c=-1$` is incorrect based on our calculation. Let's re-evaluate the options and the question.
If $a=1, b=2, c=-1$:
$a+c = 1+(-1) = 0 \neq 2$. So this option is incorrect.

Let's re-solve the system carefully.
From $a+c=2$ and $a+b=3$, we have $c=2-a$ and $b=3-a$.
Substitute into $b+c=1$:
$(3-a) + (2-a) = 1$
$5 - 2a = 1$
$2a = 4$
$a = 2$
Then $b = 3-2 = 1$ and $c = 2-2 = 0$.
So $(a,b,c) = (2,1,0)$.

The option "$a=2, b=1, c=0$" is present as the first option. I will mark that as the correct answer. The previous analysis was correct.
Answer: \boxed{a=2, b=1, c=0}"

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Part 3: Basis and Dimension

In linear algebra, the concepts of basis and dimension are fundamental for understanding the structure of vector spaces. We define these concepts to precisely characterize the "size" and "orientation" of a vector space, providing a framework for vector representation and analysis. Mastery of these topics is critical for solving problems involving subspaces, linear transformations, and system analysis in various mathematical and applied contexts.

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Core Concepts

1. Linear Independence

A set of vectors is linearly independent if no vector in the set can be expressed as a linear combination of the others. We formally define this property below.

If there exists at least one non-zero $c_i$ satisfying the equation, the set is linearly dependent. For vectors in $\mathbb{R}^n$, we can check linear independence by forming a matrix with the vectors as columns and evaluating its rank or determinant.

Quick Example: Determine if the set $\{(1, 2), (3, 4)\}$ is linearly independent in $\mathbb{R}^2$.

Step 1: Form the linear combination and set it to the zero vector.
>

Step 2: This yields a system of linear equations.
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Step 3: Form an augmented matrix and reduce it.
>

Step 4: Solve the system.
>

>

Answer: Since the only solution is $c_1=0, c_2=0$, the set is linearly independent.

:::question type="MCQ" question="Consider the set of vectors $S = \{(1, -1, 0), (2, 1, 3), (0, 1, 1)\}$ in $\mathbb{R}^3$. Is $S$ linearly independent?" options=["Yes, it is linearly independent.","No, it is linearly dependent.","It is linearly independent only if the scalar field is $\mathbb{C}$.","The concept of linear independence does not apply to this set."] answer="No, it is linearly dependent." hint="Form a matrix with the vectors as columns and calculate its determinant. If the determinant is zero, the vectors are linearly dependent." solution="Step 1: Form a matrix $A$ with the given vectors as columns.
>

Step 2: Calculate the determinant of $A$.
>

>
>
>
>

Step 3: Interpret the result.
Since $\det(A) = 0$, the columns (and thus the vectors in $S$) are linearly dependent.
The correct option is 'No, it is linearly dependent'."
:::

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2. Spanning Sets

A set of vectors is a spanning set for a vector space if every vector in that space can be written as a linear combination of the vectors in the set. This signifies that the set "generates" the entire space.

To check if a set spans $\mathbb{R}^n$, we can determine if the system $A\mathbf{x} = \mathbf{b}$ has a solution for all $\mathbf{b} \in \mathbb{R}^n$, where $A$ is the matrix with the vectors as columns. This is true if and only if the $\operatorname{rank}$ of $A$ is $n$.

Quick Example: Determine if the set $S = \{(1, 0), (0, 1), (1, 1)\}$ spans $\mathbb{R}^2$.

Step 1: Consider an arbitrary vector $(x, y) \in \mathbb{R}^2$. We need to find scalars $c_1, c_2, c_3$ such that:

Step 2: This leads to the system of equations:

Step 3: We can express $c_1 = x - c_3$ and $c_2 = y - c_3$. Since we can choose any value for $c_3$ (e.g., $c_3=0$), we can always find $c_1$ and $c_2$ to represent any $(x, y)$.

Answer: Yes, the set $S$ spans $\mathbb{R}^2$. For instance, $(x,y) = x(1,0) + y(0,1)$. The inclusion of $(1,1)$ is redundant for spanning $\mathbb{R}^2$, but it does not prevent the set from spanning.

:::question type="MCQ" question="Which of the following sets of vectors spans $\mathbb{R}^3$?" options=["$S_1 = \{(1, 0, 0), (0, 1, 0)\}$","$S_2 = \{(1, 1, 1), (2, 2, 2)\}$","$S_3 = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$","$S_4 = \{(1, 0, 0), (0, 1, 0), (1, 1, 0)\}$"] answer="$S_3 = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$" hint="A set of vectors spans $\mathbb{R}^n$ if it contains at least $n$ linearly independent vectors. For $\mathbb{R}^3$, we need at least 3 linearly independent vectors." solution="Step 1: Analyze $S_1 = \{(1, 0, 0), (0, 1, 0)\}$. This set has only two vectors. It can only span a 2-dimensional subspace of $\mathbb{R}^3$ (the $xy$-plane). Thus, it does not span $\mathbb{R}^3$.

Step 2: Analyze $S_2 = \{(1, 1, 1), (2, 2, 2)\}$. These two vectors are linearly dependent ($2(1,1,1) = (2,2,2)$). They span only a 1-dimensional subspace (a line through the origin). Thus, it does not span $\mathbb{R}^3$.

Step 3: Analyze $S_3 = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$. This is the standard basis for $\mathbb{R}^3$. These three vectors are linearly independent, and any vector $(x, y, z)$ in $\mathbb{R}^3$ can be written as $x(1,0,0) + y(0,1,0) + z(0,0,1)$. This set spans $\mathbb{R}^3$.

Step 4: Analyze $S_4 = \{(1, 0, 0), (0, 1, 0), (1, 1, 0)\}$. The third vector $(1,1,0)$ is a linear combination of the first two: $(1,1,0) = 1(1,0,0) + 1(0,1,0)$. This set is linearly dependent. It spans the $xy$-plane, which is a 2-dimensional subspace of $\mathbb{R}^3$. Thus, it does not span $\mathbb{R}^3$.

The correct option is '$S_3 = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$'."
:::

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3. Basis of a Vector Space

A basis is a fundamental concept that combines both linear independence and spanning properties. It provides a minimal set of vectors that can generate the entire vector space.

Every vector space has at least one basis. The standard basis for $\mathbb{R}^n$ consists of the vectors $e_1 = (1, 0, \dots, 0)$, $e_2 = (0, 1, \dots, 0)$, ..., $e_n = (0, 0, \dots, 1)$.

Quick Example: Show that the set $B = \{(1, 1), (1, -1)\}$ is a basis for $\mathbb{R}^2$.

Step 1: Check for linear independence. Form the matrix with vectors as columns and find its determinant.

Step 2: Since $\det(A) \ne 0$, the vectors are linearly independent.

Step 3: Since we have 2 linearly independent vectors in $\mathbb{R}^2$, they automatically span $\mathbb{R}^2$. (A set of $n$ linearly independent vectors in an $n$-dimensional space forms a basis).

Answer: The set $B = \{(1, 1), (1, -1)\}$ is a basis for $\mathbb{R}^2$.

:::question type="MCQ" question="Which of the following sets of vectors forms a basis for $\mathbb{R}^3$?" options=["$S_1 = \{(1, 0, 0), (0, 1, 0)\}$","$S_2 = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\}$","$S_3 = \{(1, 2, 3), (1, 3, 5), (1, 0, 1), (2, 3, 0)\}$","$S_4 = \{(1, 1, 2), (1, 2, 5), (5, 3, 4)\}$"] answer="$S_2 = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\}$" hint="A basis for $\mathbb{R}^3$ must contain exactly three linearly independent vectors. Sets with fewer than three vectors cannot span $\mathbb{R}^3$, and sets with more than three vectors must be linearly dependent." solution="Step 1: Analyze each option based on the number of vectors and potential linear independence.
* $S_1 = \{(1, 0, 0), (0, 1, 0)\}$: Contains only two vectors. Cannot span $\mathbb{R}^3$. Not a basis.
* $S_3 = \{(1, 2, 3), (1, 3, 5), (1, 0, 1), (2, 3, 0)\}$: Contains four vectors. In $\mathbb{R}^3$, any set of more than three vectors must be linearly dependent. Not a basis.
* $S_4 = \{(1, 1, 2), (1, 2, 5), (5, 3, 4)\}$: Contains three vectors. We must check for linear independence.

Step 2: Check linear independence for $S_2 = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\}$. Form a matrix $A$ and compute its determinant.

Step 3: Since $\det(A) = 5 \ne 0$, the vectors in $S_2$ are linearly independent. As there are three linearly independent vectors in $\mathbb{R}^3$, $S_2$ forms a basis for $\mathbb{R}^3$.

Step 4: Check linear independence for $S_4 = \{(1, 1, 2), (1, 2, 5), (5, 3, 4)\}$. Form a matrix $B$ and compute its determinant.

Since $\det(B) = 0$, the vectors in $S_4$ are linearly dependent. Not a basis.

The correct option is '$S_2 = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\}$'."
:::

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4. Dimension of a Vector Space

The dimension of a vector space is a unique scalar value that quantifies its "size." It is defined as the number of vectors in any basis for that space.

The dimension depends critically on the field of scalars. For instance, $\mathbb{C}$ is a 1-dimensional vector space over $\mathbb{C}$ (basis $\{1\}$), but a 2-dimensional vector space over $\mathbb{R}$ (basis $\{1, i\}$).

Quick Example: What is the dimension of the vector space $P_2$, the space of all polynomials of degree at most 2, over $\mathbb{R}$?

Step 1: Identify a basis for $P_2$. Any polynomial $p(x) = ax^2 + bx + c$ can be written as a linear combination of $1, x, x^2$.

Step 2: Verify linear independence of $\{1, x, x^2\}$. If $c_1 \cdot 1 + c_2 \cdot x + c_3 \cdot x^2 = 0$ for all $x$, then $c_1=c_2=c_3=0$.

Step 3: Count the number of vectors in the basis. The basis $\{1, x, x^2\}$ contains 3 vectors.

Answer: $\operatorname{dim}(P_2) = 3$.

:::question type="MCQ" question="Match List I with List II for the dimensions of the given vector spaces over the specified fields.

" options=["A-IV, B-III, C-II, D-I","A-I, B-IV, C-II, D-III","A-II, B-I, C-III, D-IV","A-I, B-II, C-III, D-IV"] answer="A-IV, B-III, C-II, D-I" hint="The dimension of a vector space depends on the field of scalars. A basis for $\mathbb{C}$ over $\mathbb{R}$ is $\{1, i\}$. A basis for $\mathbb{C}^2$ over $\mathbb{R}$ would require vectors that are combinations of complex numbers with real coefficients." solution="Step 1: Determine the dimension of $\mathbb{R}$ over $\mathbb{R}$. A basis for $\mathbb{R}$ over $\mathbb{R}$ is $\{1\}$. There is 1 vector. So, $\operatorname{dim}(\mathbb{R} \text{ over } \mathbb{R}) = 1$. (A-IV)

Step 2: Determine the dimension of $\mathbb{C}$ over $\mathbb{R}$.
A basis for $\mathbb{C}$ over $\mathbb{R}$ is $\{1, i\}$. Any $z = a + bi \in \mathbb{C}$ can be written as $a \cdot 1 + b \cdot i$ where $a, b \in \mathbb{R}$. There are 2 vectors. So, $\operatorname{dim}(\mathbb{C} \text{ over } \mathbb{R}) = 2$. (B-III)

Step 3: Determine the dimension of $\mathbb{R}^3$ over $\mathbb{R}$.
A standard basis for $\mathbb{R}^3$ over $\mathbb{R}$ is $\{(1,0,0), (0,1,0), (0,0,1)\}$. There are 3 vectors. So, $\operatorname{dim}(\mathbb{R}^3 \text{ over } \mathbb{R}) = 3$. (C-II)

Step 4: Determine the dimension of $\mathbb{C}^2$ over $\mathbb{R}$.
Any vector in $\mathbb{C}^2$ is of the form $(z_1, z_2)$ where $z_1, z_2 \in \mathbb{C}$. We can write $z_1 = a_1 + b_1i$ and $z_2 = a_2 + b_2i$ for $a_1, b_1, a_2, b_2 \in \mathbb{R}$.
So,

A basis for $\mathbb{C}^2$ over $\mathbb{R}$ is $\{(1,0), (i,0), (0,1), (0,i)\}$. There are 4 vectors. So, $\operatorname{dim}(\mathbb{C}^2 \text{ over } \mathbb{R}) = 4$. (D-I)

Step 5: Match the lists.
A-IV, B-III, C-II, D-I.
The correct option is 'A-IV, B-III, C-II, D-I'."
:::

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5. Dimension of Subspaces

A subspace $W$ of a vector space $V$ is itself a vector space. Its dimension is found by determining a basis for $W$.

To find the dimension of a subspace defined by linear equations, we typically solve the system of equations to express some variables in terms of others, thereby identifying free variables that correspond to basis vectors.

Quick Example: Let $W = \{(x, y, z) \in \mathbb{R}^3 : x - y + 2z = 0\}$. Find $\operatorname{dim}(W)$.

Step 1: Express one variable in terms of the others using the given condition.

Step 2: Write an arbitrary vector in $W$ using the expression from Step 1.

Step 3: Decompose the vector into a linear combination involving the free variables ($y$ and $z$).

Step 4: The vectors $\{(1, 1, 0), (-2, 0, 1)\}$ form a basis for $W$. These vectors are linearly independent and span $W$.

Step 5: Count the number of vectors in the basis. There are 2 vectors.

Answer: $\operatorname{dim}(W) = 2$.

:::question type="MCQ" question="If $W$ is a subspace of $\mathbb{R}^3$, where $W = \{(a, b, c): a + b + c = 0\}$, then $\operatorname{dim}(W)$ is equal to:" options=["$2$","$3$","$1$","$0$"] answer="$2$" hint="The equation $a+b+c=0$ imposes one linear constraint on the vectors in $\mathbb{R}^3$. This reduces the number of independent variables by one." solution="Step 1: The subspace $W$ is defined by the equation $a+b+c=0$. We can express one variable in terms of the others, for example, $a = -b - c$.

Step 2: An arbitrary vector $(a, b, c)$ in $W$ can be written as:

Step 3: Decompose this vector based on the free variables $b$ and $c$:

Step 4: The set of vectors $\{(-1, 1, 0), (-1, 0, 1)\}$ forms a basis for $W$. These two vectors are linearly independent (one is not a scalar multiple of the other) and span $W$.

Step 5: The number of vectors in the basis is 2. Therefore, $\operatorname{dim}(W) = 2$.
The correct option is '$2$'."
:::

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6. Coordinate Vectors

Once a basis is established for a vector space, any vector in that space can be uniquely expressed as a linear combination of the basis vectors. The coefficients in this linear combination form the coordinate vector of the vector relative to that basis.

Quick Example: Let $\mathcal{B} = \{(1, 1), (1, -1)\}$ be an ordered basis for $\mathbb{R}^2$. Find the coordinate vector of $v = (5, 1)$ relative to $\mathcal{B}$.

Step 1: Set up the linear combination.

Step 2: Form the system of equations.

Step 3: Solve the system. Adding the two equations yields $2c_1 = 6 \implies c_1 = 3$. Substituting $c_1=3$ into the first equation gives $3 + c_2 = 5 \implies c_2 = 2$.

Answer: The coordinate vector is $[v]_{\mathcal{B}} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$.

:::question type="MCQ" question="Let $\mathcal{B} = \{(1, 2, 0), (0, 1, 1), (1, 0, -1)\}$ be an ordered basis for $\mathbb{R}^3$. Find the coordinate vector of $v = (2, 4, 1)$ relative to $\mathcal{B}$." options=["$\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$","$\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}$","$\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$","$\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}$"] answer="$\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$" hint="Set up a system of linear equations $c_1 v_1 + c_2 v_2 + c_3 v_3 = v$ and solve for $c_1, c_2, c_3$. This can be done using an augmented matrix." solution="Step 1: We need to find scalars $c_1, c_2, c_3$ such that:

Step 2: This forms the following system of linear equations:

Step 3: Form the augmented matrix and reduce it.

Step 4: Solve the system using back substitution.

Answer: The coordinate vector is $[v]_{\mathcal{B}} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$.
The correct option is '$\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$'."
:::

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Advanced Applications

1. Grassmann's Formula (Dimension of Sum and Intersection)

Grassmann's Formula provides a critical relationship between the dimensions of the sum and intersection of two subspaces. This formula is frequently tested in competitive examinations.

We also observe that $\operatorname{dim}(U \cap W) \le \operatorname{min}(\operatorname{dim}(U), \operatorname{dim}(W))$ and $\operatorname{max}(\operatorname{dim}(U), \operatorname{dim}(W)) \le \operatorname{dim}(U+W) \le \operatorname{dim}(V)$.

Revised Quick Example: Let $U$ and $W$ be distinct 4-dimensional subspaces of a vector space $V$ of dimension 6. Find the possible dimensions of $U \cap W$.

Step 1: We are given $\operatorname{dim}(V) = 6$, $\operatorname{dim}(U) = 4$, $\operatorname{dim}(W) = 4$.
Apply Grassmann's Formula: $\operatorname{dim}(U+W) = \operatorname{dim}(U) + \operatorname{dim}(W) - \operatorname{dim}(U \cap W)$.

Step 2: Consider the bounds for $\operatorname{dim}(U+W)$.
Since $U+W$ is a subspace of $V$, we have $\operatorname{dim}(U+W) \le \operatorname{dim}(V) = 6$.
Also, $U \subseteq U+W$, so $\operatorname{dim}(U+W) \ge \operatorname{dim}(U) = 4$. Similarly, $\operatorname{dim}(U+W) \ge \operatorname{dim}(W) = 4$.
Thus, $4 \le \operatorname{dim}(U+W) \le 6$.

Step 3: Use the bounds of $\operatorname{dim}(U+W)$ to find the bounds for $\operatorname{dim}(U \cap W)$.
From $\operatorname{dim}(U+W) = 8 - \operatorname{dim}(U \cap W)$, we have $\operatorname{dim}(U \cap W) = 8 - \operatorname{dim}(U+W)$.
Maximum value for $\operatorname{dim}(U \cap W)$: Occurs when $\operatorname{dim}(U+W)$ is minimum.
If $\operatorname{dim}(U+W) = 4$, then $\operatorname{dim}(U \cap W) = 8 - 4 = 4$.
Minimum value for $\operatorname{dim}(U \cap W)$: Occurs when $\operatorname{dim}(U+W)$ is maximum.
If $\operatorname{dim}(U+W) = 6$, then $\operatorname{dim}(U \cap W) = 8 - 6 = 2$.

Step 4: The possible dimensions for $U \cap W$ are $2, 3, 4$.
However, the question specifies $U$ and $W$ are distinct. If $\operatorname{dim}(U \cap W) = 4$, then since $U \cap W \subseteq U$ and $\operatorname{dim}(U \cap W) = \operatorname{dim}(U)$, it implies $U \cap W = U$. Similarly, $U \cap W = W$. This would mean $U=W$, which contradicts the "distinct" condition.
Therefore, $\operatorname{dim}(U \cap W)$ cannot be 4.
This implies $\operatorname{dim}(U+W)$ cannot be 4 (because if it were, $\operatorname{dim}(U \cap W)$ would be 4, leading to $U=W$).
So, $U \ne W$ implies $\operatorname{dim}(U+W) > \operatorname{max}(\operatorname{dim}(U), \operatorname{dim}(W)) = 4$.
Therefore, $5 \le \operatorname{dim}(U+W) \le 6$.

Step 5: Recalculate $\operatorname{dim}(U \cap W)$ based on $5 \le \operatorname{dim}(U+W) \le 6$.
If $\operatorname{dim}(U+W) = 5$, then $\operatorname{dim}(U \cap W) = 8 - 5 = 3$.
If $\operatorname{dim}(U+W) = 6$, then $\operatorname{dim}(U \cap W) = 8 - 6 = 2$.

Answer: The possible dimensions of $U \cap W$ are $2$ or $3$.

:::question type="MCQ" question="Let $U$ and $W$ be distinct 4-dimensional subspaces of a vector space $V$ of dimension 6. Then the possible dimensions of $U \cap W$ are:" options=["$1$ or $2$","exactly $4$","$3$ or $4$","$2$ or $3$"] answer="$2$ or $3$" hint="Use Grassmann's formula: $\operatorname{dim}(U+W) = \operatorname{dim}(U) + \operatorname{dim}(W) - \operatorname{dim}(U \cap W)$. Also, consider the bounds for $\operatorname{dim}(U+W)$ and $\operatorname{dim}(U \cap W)$, specifically that $U \ne W$ implies $\operatorname{dim}(U+W) > \operatorname{max}(\operatorname{dim}(U), \operatorname{dim}(W))$." solution="Step 1: We are given:
$\operatorname{dim}(V) = 6$
$\operatorname{dim}(U) = 4$
$\operatorname{dim}(W) = 4$
$U$ and $W$ are distinct subspaces.

Step 2: Apply Grassmann's Formula:

Step 3: Determine the bounds for $\operatorname{dim}(U+W)$.
Since $U+W$ is a subspace of $V$, its dimension cannot exceed $\operatorname{dim}(V)$:

Also, $U$ and $W$ are subspaces of $U+W$, so:


Combining these, we have $4 \le \operatorname{dim}(U+W) \le 6$.

Step 4: Account for the condition that $U$ and $W$ are distinct.
If $\operatorname{dim}(U+W) = 4$, then from Grassmann's formula, $\operatorname{dim}(U \cap W) = 8 - 4 = 4$.
If $\operatorname{dim}(U \cap W) = 4$, then since $U \cap W \subseteq U$ and $\operatorname{dim}(U \cap W) = \operatorname{dim}(U)$, it implies $U \cap W = U$. Similarly, $U \cap W = W$. This means $U=W$.
However, the problem states $U$ and $W$ are distinct. Therefore, $\operatorname{dim}(U \cap W)$ cannot be 4. This implies $\operatorname{dim}(U+W)$ cannot be 4.
Thus, $\operatorname{dim}(U+W)$ must be strictly greater than $\operatorname{max}(\operatorname{dim}(U), \operatorname{dim}(W))$.

This means $\operatorname{dim}(U+W)$ can be $5$ or $6$.

Step 5: Calculate the possible dimensions of $U \cap W$ using these values.
Case 1: If $\operatorname{dim}(U+W) = 5$:

Case 2: If $\operatorname{dim}(U+W) = 6$:

The possible dimensions of $U \cap W$ are $2$ or $3$.
The correct option is '$2$ or $3$'."
:::

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2. Change of Basis (Transition Matrix)

When we change from one ordered basis to another, the coordinate vector of a given vector changes. A transition matrix facilitates this transformation.

To find $P_{\mathcal{C} \leftarrow \mathcal{B}}$, we form the augmented matrix $[\mathcal{C} | \mathcal{B}]$ and row reduce it to $[I | P_{\mathcal{C} \leftarrow \mathcal{B}}]$.

Quick Example: Let $\mathcal{B} = \{(1, 0), (0, 1)\}$ be the standard basis and $\mathcal{C} = \{(1, 1), (1, -1)\}$ be another basis for $\mathbb{R}^2$. Find the transition matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$.

Step 1: Form the augmented matrix $[\mathcal{C} | \mathcal{B}]$.
>

Step 2: Row reduce the left side (matrix $\mathcal{C}$) to the identity matrix.
>

>
>
>
>
>

Answer: The transition matrix is $P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$.

:::question type="MCQ" question="Let $\mathcal{B} = \{b_1, b_2\} = \{(1, 2), (2, 1)\}$ and $\mathcal{C} = \{c_1, c_2\} = \{(1, 0), (0, 1)\}$ be two ordered bases for $\mathbb{R}^2$. Find the transition matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$." options=["$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$","$\begin{bmatrix} -1/3 & 2/3 \\ 2/3 & -1/3 \end{bmatrix}$","$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$","$\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$"] answer="$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$" hint="The transition matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$ has columns that are the coordinate vectors of $b_1$ and $b_2$ with respect to $\mathcal{C}$. Since $\mathcal{C}$ is the standard basis, these coordinate vectors are simply the vectors $b_1$ and $b_2$ themselves." solution="Step 1: We need to find $P_{\mathcal{C} \leftarrow \mathcal{B}}$, which means expressing $b_1$ and $b_2$ as linear combinations of $c_1$ and $c_2$.
The basis $\mathcal{C} = \{(1, 0), (0, 1)\}$ is the standard basis for $\mathbb{R}^2$.
For any vector $v = (x, y)$, its coordinate vector relative to the standard basis is simply $\begin{bmatrix} x \\ y \end{bmatrix}$.

Step 2: Find $[b_1]_{\mathcal{C}}$:
$b_1 = (1, 2) = 1 \cdot (1, 0) + 2 \cdot (0, 1)$.
So, $[b_1]_{\mathcal{C}} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.

Step 3: Find $[b_2]_{\mathcal{C}}$:
$b_2 = (2, 1) = 2 \cdot (1, 0) + 1 \cdot (0, 1)$.
So, $[b_2]_{\mathcal{C}} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$.

Step 4: Construct the transition matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$ using these coordinate vectors as columns.
>

Answer: \boxed{\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}}"
:::

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3. Rank-Nullity Theorem

While primarily a theorem about linear transformations, the Rank-Nullity Theorem directly relates the dimension of the image (rank) and the dimension of the kernel (nullity) of a linear transformation, which are subspaces. It is a fundamental result in dimension theory.

Quick Example: Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation defined by $T(x,y,z) = (x+y, y-z)$. Find the nullity of $T$ if its rank is 2.

Step 1: Identify the dimension of the domain space.
Here, $V = \mathbb{R}^3$, so $\operatorname{dim}(V) = 3$.

Step 2: We are given that the rank of $T$, $\operatorname{dim}(\operatorname{Im}(T))$, is 2.

Step 3: Apply the Rank-Nullity Theorem.
>

>

Step 4: Solve for $\operatorname{dim}(\operatorname{ker}(T))$.
>

Answer: The nullity of $T$ is 1.

:::question type="NAT" question="A linear transformation $T: P_3 \to \mathbb{R}^4$ is defined such that its image has dimension 3. What is the dimension of the kernel of $T$?" answer="1" hint="Recall the Rank-Nullity Theorem. The domain space is $P_3$, the space of polynomials of degree at most 3. Determine its dimension first." solution="Step 1: Determine the dimension of the domain space $P_3$.
A basis for $P_3$ is $\{1, x, x^2, x^3\}$. This basis contains 4 vectors.
Therefore, $\operatorname{dim}(P_3) = 4$.

Step 2: We are given that the dimension of the image of $T$ is 3.
So, $\operatorname{dim}(\operatorname{Im}(T)) = 3$.

Step 3: Apply the Rank-Nullity Theorem:
>

>

Step 4: Solve for $\operatorname{dim}(\operatorname{ker}(T))$.
>

Answer: \boxed{1}"
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="For what value(s) of $k$ the set of vectors $\{(1, k, 5), (1, -3, 2), (2, -1, 1)\}$ form a basis in $\mathbb{R}^3$?" options=["$k = 8$","$k = -8$","$k \ne -8$","$k \ne 0$"] answer="$k \ne -8$" hint="For three vectors to form a basis in $\mathbb{R}^3$, they must be linearly independent. This means the determinant of the matrix formed by these vectors must be non-zero." solution="Step 1: Form a matrix $A$ with the given vectors as columns (or rows).

Step 2: For the vectors to form a basis, they must be linearly independent, which implies $\det(A) \ne 0$. Calculate the determinant of $A$.

Step 3: Set the determinant to be non-zero to find the condition on $k$.

Answer: $\boxed{k \ne -8}$"
:::

:::question type="MCQ" question="Let $V$ and $W$ be the subspaces of $\mathbb{R}^4$ defined as $V = \{(a, b, c, d): b - 5c + 2d = 0\}$, $W = \{(a, b, c, d): a - b = 0\}$. Then the dimension of $V \cap W$ is:" options=["1","2","3","4"] answer="2" hint="To find the dimension of $V \cap W$, we need to find the basis for the vectors that satisfy the conditions for both $V$ and $W$. This means solving the combined system of linear equations." solution="Step 1: For a vector $(a, b, c, d)$ to be in $V \cap W$, it must satisfy both conditions:

$b - 5c + 2d = 0$

$a - b = 0 \implies a = b$

Step 2: Substitute $a=b$ into the first equation (it doesn't change it, as $a$ is not in the first equation).
We have the system:
$a - b = 0$
$b - 5c + 2d = 0$
This is a system of 2 linearly independent equations in 4 variables.

Step 3: The number of free variables is $4 - (\text{number of linearly independent equations}) = 4 - 2 = 2$.
Each free variable corresponds to a basis vector.

Step 4: To find a basis, we can choose $c$ and $d$ as free variables.
From $a=b$, we have $a=b$.
From $b - 5c + 2d = 0$, we have $b = 5c - 2d$.
So, $a = 5c - 2d$.
An arbitrary vector in $V \cap W$ is $(a, b, c, d) = (5c - 2d, 5c - 2d, c, d)$.

The basis is $\{(5, 5, 1, 0), (-2, -2, 0, 1)\}$. These vectors are linearly independent.

Step 5: The number of vectors in the basis is 2. Therefore, $\operatorname{dim}(V \cap W) = 2$.
Answer: $\boxed{2}$"
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:::question type="MCQ" question="Let $U = \{(x, y, z) \in \mathbb{R}^3 : x - y = 0\}$ and $W = \{(x, y, z) \in \mathbb{R}^3 : y + z = 0\}$. Which of the following statements about $U$ and $W$ is true?" options=["$\operatorname{dim}(U) = 1$ and $\operatorname{dim}(W) = 1$","$\operatorname{dim}(U) = 2$ and $\operatorname{dim}(W) = 2$","$\operatorname{dim}(U \cap W) = 1$ and $\operatorname{dim}(U+W) = 3$","$\operatorname{dim}(U \cap W) = 2$ and $\operatorname{dim}(U+W) = 2$"] answer="$\operatorname{dim}(U \cap W) = 1$ and $\operatorname{dim}(U+W) = 3$" hint="First, find the dimensions of $U$ and $W$ individually. Then, find the conditions for $U \cap W$ and determine its dimension. Finally, use Grassmann's formula to find $\operatorname{dim}(U+W)$." solution="Step 1: Find $\operatorname{dim}(U)$.
$U = \{(x, y, z) \in \mathbb{R}^3 : x - y = 0\}$. This implies $x = y$.
An arbitrary vector in $U$ is $(x, x, z) = x(1, 1, 0) + z(0, 0, 1)$.
A basis for $U$ is $\{(1, 1, 0), (0, 0, 1)\}$. These are linearly independent.
Thus, $\operatorname{dim}(U) = 2$.

Step 2: Find $\operatorname{dim}(W)$.
$W = \{(x, y, z) \in \mathbb{R}^3 : y + z = 0\}$. This implies $z = -y$.
An arbitrary vector in $W$ is $(x, y, -y) = x(1, 0, 0) + y(0, 1, -1)$.
A basis for $W$ is $\{(1, 0, 0), (0, 1, -1)\}$. These are linearly independent.
Thus, $\operatorname{dim}(W) = 2$.

Step 3: Find $\operatorname{dim}(U \cap W)$.
A vector $(x, y, z)$ is in $U \cap W$ if it satisfies both $x - y = 0$ and $y + z = 0$.
So, $x = y$ and $z = -y$.
An arbitrary vector in $U \cap W$ is $(y, y, -y) = y(1, 1, -1)$.
A basis for $U \cap W$ is $\{(1, 1, -1)\}$. This is a single non-zero vector.
Thus, $\operatorname{dim}(U \cap W) = 1$.

Step 4: Find $\operatorname{dim}(U+W)$ using Grassmann's Formula.

Step 5: Combine the results.
$\operatorname{dim}(U) = 2$, $\operatorname{dim}(W) = 2$, $\operatorname{dim}(U \cap W) = 1$, $\operatorname{dim}(U+W) = 3$.
The statement '$\operatorname{dim}(U \cap W) = 1$ and $\operatorname{dim}(U+W) = 3$' is true.
Answer: $\boxed{\operatorname{dim}(U \cap W) = 1 \text{ and } \operatorname{dim}(U+W) = 3}$"
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:::question type="NAT" question="Let $V = M_{2 \times 2}(\mathbb{R})$ be the vector space of $2 \times 2$ matrices with real entries. Let $S = \{A \in V : A^T = A\}$ be the subspace of symmetric matrices. What is $\operatorname{dim}(S)$?" answer="3" hint="A symmetric matrix is one where $A = A^T$. Write out the general form of a $2 \times 2$ symmetric matrix and find a basis for it." solution="Step 1: Write the general form of a $2 \times 2$ matrix $A$.

Step 2: Apply the condition for a symmetric matrix, $A^T = A$.

So,

This implies $c = b$.

Step 3: Write the general form of a symmetric matrix in $S$.

Step 4: Decompose this matrix into a linear combination of basis matrices.

The set of matrices

forms a basis for $S$. These matrices are linearly independent and span $S$.

Step 5: Count the number of matrices in the basis.
There are 3 matrices in the basis.
Therefore, $\operatorname{dim}(S) = 3$.
Answer: $\boxed{3}$"
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:::question type="MCQ" question="Let $P_2(\mathbb{R})$ be the vector space of polynomials of degree at most 2 with real coefficients. Consider the set $S = \{x^2+1, x-1, x+2\}$. Which of the following is true about $S$?" options=["$S$ is linearly dependent and spans $P_2(\mathbb{R})$.","$S$ is linearly independent but does not span $P_2(\mathbb{R})$.","$S$ is a basis for $P_2(\mathbb{R})$.","$S$ is linearly dependent and does not span $P_2(\mathbb{R})$. "] answer="$S$ is a basis for $P_2(\mathbb{R})$." hint="The dimension of $P_2(\mathbb{R})$ is 3. A set of 3 polynomials will form a basis if they are linearly independent. To check linear independence, set a linear combination to the zero polynomial and solve for the coefficients." solution="Step 1: Determine the dimension of the vector space $P_2(\mathbb{R})$.
The standard basis for $P_2(\mathbb{R})$ is $\{1, x, x^2\}$. Thus, $\operatorname{dim}(P_2(\mathbb{R})) = 3$.
The set $S$ contains 3 vectors (polynomials), so it could potentially be a basis if it is linearly independent.

Step 2: Check for linear independence of $S = \{x^2+1, x-1, x+2\}$.
Set a linear combination to the zero polynomial:

Rearrange by powers of $x$:

For this polynomial to be the zero polynomial, all coefficients must be zero:

$c_1 = 0$ (coefficient of $x^2$)

$c_2 + c_3 = 0$ (coefficient of $x$)

$c_1 - c_2 + 2c_3 = 0$ (constant term)

Step 3: Solve the system of equations.
From (1), $c_1 = 0$.
Substitute $c_1=0$ into (3):

Now we have a system for $c_2, c_3$:

Adding the two equations: $(c_2 + c_3) + (-c_2 + 2c_3) = 0 + 0 \implies 3c_3 = 0 \implies c_3 = 0$.
Substitute $c_3=0$ into $c_2 + c_3 = 0 \implies c_2 + 0 = 0 \implies c_2 = 0$.
So, the only solution is $c_1 = 0, c_2 = 0, c_3 = 0$.

Step 4: Conclude linear independence and basis status.
Since the only solution is the trivial one, the set $S$ is linearly independent.
As $S$ contains 3 linearly independent vectors in a 3-dimensional space ($P_2(\mathbb{R})$), $S$ forms a basis for $P_2(\mathbb{R})$.
Answer: $\boxed{S \text{ is a basis for } P_2(\mathbb{R})}$"
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Summary

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What's Next?

Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Which of the following sets is NOT a subspace of $\mathbb{R}^3$?" options=["$\{(x,y,z) \in \mathbb{R}^3 : x+2y-z=0\}$","$\{(x,y,z) \in \mathbb{R}^3 : x=y=z\}$","$\{(x,y,z) \in \mathbb{R}^3 : x=1\}$","$\{(x,y,z) \in \mathbb{R}^3 : z=0\}$"] answer="$\{(x,y,z) \in \mathbb{R}^3 : x=1\}$" hint="A subspace must contain the zero vector and be closed under scalar multiplication and vector addition. Consider if the zero vector (0,0,0) is in each set." solution="For a set to be a subspace, it must contain the zero vector $(0,0,0)$. The set $\{(x,y,z) \in \mathbb{R}^3 : x=1\}$ requires the first component to be 1, thus it does not contain $(0,0,0)$. Therefore, it cannot be a subspace.
Answer: $\boxed{\{(x,y,z) \in \mathbb{R}^3 : x=1\}}$"
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:::question type="NAT" question="Consider the vectors $v_1=(1,0,1)$, $v_2=(0,1,1)$, and $v_3=(2,-1,1)$ in $\mathbb{R}^3$. If $v_3$ is a linear combination of $v_1$ and $v_2$ such that $v_3 = av_1 + bv_2$, what is the value of $a+b$?" answer="1" hint="Set up a system of linear equations by equating components and solve for $a$ and $b$." solution="We need to find $a$ and $b$ such that $(2,-1,1) = a(1,0,1) + b(0,1,1)$.
This gives the system of equations:
$a = 2$ (from the first component)
$b = -1$ (from the second component)
$a+b = 1$ (from the third component)
Substituting $a=2$ and $b=-1$ into the third equation: $2 + (-1) = 1$, which is consistent.
Thus, $a=2$ and $b=-1$.
The value of $a+b$ is $2 + (-1) = 1$.
Answer: $\boxed{1}$"
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:::question type="MCQ" question="Let $P_2(x)$ be the vector space of polynomials of degree at most 2. Which of the following sets is a basis for $P_2(x)$?" options=["$\{1, x, x^2, x^3\}$","$\{1, x, x^2\}$","$\{1+x, x+x^2, 1-x^2\}$","$\{x, x^2\}$"] answer="$\{1, x, x^2\}$" hint="A basis must be linearly independent and span the space. The dimension of $P_2(x)$ is 3." solution="The standard basis for $P_2(x)$ is $\{1, x, x^2\}$, which consists of 3 linearly independent polynomials that span $P_2(x)$.
Option A has 4 elements, making it linearly dependent in $P_2(x)$.
Option C: $1+x, x+x^2, 1-x^2$. Let's check for linear dependence:
$c_1(1+x) + c_2(x+x^2) + c_3(1-x^2) = 0$.
Rearranging by powers of $x$: $(c_1+c_3) + (c_1+c_2)x + (c_2-c_3)x^2 = 0$.
This gives the system:
$c_1+c_3=0$
$c_1+c_2=0$
$c_2-c_3=0$
From the first equation, $c_3 = -c_1$.
From the second equation, $c_2 = -c_1$.
Substitute these into the third equation: $(-c_1) - (-c_1) = 0 \implies 0=0$.
This means the system has non-trivial solutions (e.g., if $c_1=1$, then $c_2=-1, c_3=-1$).
So, $\{1+x, x+x^2, 1-x^2\}$ is linearly dependent and thus NOT a basis.
Option D has only 2 elements, so it cannot span $P_2(x)$ (dimension is 3).
Therefore, the only correct option is $\{1, x, x^2\}$.
Answer: $\boxed{\{1, x, x^2\}}$"
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:::question type="NAT" question="What is the dimension of the subspace of $\mathbb{R}^4$ defined by the equations $x_1 - x_2 + x_3 = 0$ and $x_2 - x_4 = 0$?" answer="2" hint="Find a basis for the subspace by expressing some variables in terms of others using the given equations. The number of free variables will be the dimension." solution="Let the subspace be $W = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 - x_2 + x_3 = 0 \text{ and } x_2 - x_4 = 0\}$.
From $x_2 - x_4 = 0$, we have $x_4 = x_2$.
From $x_1 - x_2 + x_3 = 0$, we have $x_1 = x_2 - x_3$.
We can choose $x_2$ and $x_3$ as free variables. Let $x_2 = a$ and $x_3 = b$.
Then $x_1 = a - b$ and $x_4 = a$.
So, any vector in $W$ can be written as $(a-b, a, b, a)$.
This can be decomposed as:
$(a-b, a, b, a) = a(1, 1, 0, 1) + b(-1, 0, 1, 0)$.
The vectors $v_1 = (1, 1, 0, 1)$ and $v_2 = (-1, 0, 1, 0)$ span $W$.
These two vectors are linearly independent (one is not a scalar multiple of the other).
Therefore, $\{v_1, v_2\}$ forms a basis for $W$.
The dimension of $W$ is the number of vectors in its basis, which is 2.
Answer: $\boxed{2}$"
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What's Next?