Problem 1 - A Book of Abstract Algebra

Question

Prove that \(\\{(a, b, c): 2 a-3 b+c=0\\}\) is a subspace of \(\mathbb{R}^{3}\).

Step-by-Step Solution

Answer

{(a, b, c): 2a - 3b + c = 0} is a subspace of \(\mathbb{R}^{3}\) because it meets all three subspace criteria: contains zero vector, closed under addition, and closed under scalar multiplication.

Step 1: Definition of a Subspace

A subset of a vector space is a subspace if it satisfies three properties: (1) It contains the zero vector, (2) It is closed under addition, and (3) It is closed under scalar multiplication.

Step 2: Show Zero Vector is in the Subset

The zero vector in \(\mathbb{R}^3\) is \((0, 0, 0)\). Substitute these into the equation: \[2(0) - 3(0) + 0 = 0.\]The equation holds true, so the zero vector is in the subset.

Step 3: Verify Closure under Addition

Assume \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) are in the subset, meaning \[2a_1 - 3b_1 + c_1 = 0 \quad\text{and}\quad 2a_2 - 3b_2 + c_2 = 0.\]Add these equations:\[2(a_1+a_2) - 3(b_1+b_2) + (c_1+c_2) = (2a_1-3b_1+c_1) + (2a_2-3b_2+c_2) = 0 + 0 = 0.\]Therefore, \((a_1+a_2, b_1+b_2, c_1+c_2)\) is in the subset, showing closure under addition.

Step 4: Verify Closure under Scalar Multiplication

Let \((a, b, c)\) be in the subset satisfying the equation \[2a - 3b + c = 0.\]For any scalar \(k\), multiply each component by \(k\):\[2(ka) - 3(kb) + (kc) = k(2a - 3b + c) = k \cdot 0 = 0.\]Thus, \((ka, kb, kc)\) is in the subset, showing closure under scalar multiplication.

Key Concepts

Vector Space

A vector space is a fundamental concept in linear algebra that consists of vectors and operations defined on them. In simpler terms, a vector space forms a big family of vectors that work together under certain rules. These rules ensure that when you add vectors or multiply them by numbers (scalars), they still belong to the same family.

A vector space must satisfy certain properties:

• It includes an operation for vector addition.
• There's scalar multiplication (multiplying a vector by a real number).
• It must include a zero vector.
• It is closed under vector addition and scalar multiplication.

Understanding vector spaces is crucial because it helps us grasp the structures and systems in linear algebra, like solving equations and understanding geometrical concepts.

Zero Vector

A zero vector is a special type of vector in a vector space. Think of the zero vector as the starting point or the origin in a space like (0,0,0) for egin{math}\mathbb{R}^{3} egin{math} . It's essentially a vector with all components equal to zero.

The zero vector has some unique roles and properties:

• It is the only vector that doesn’t change other vectors when added to them. So, for any vector \((a, b, c)\), adding the zero vector \((0, 0, 0)\) leaves it unchanged: \((a, b, c) + (0, 0, 0) = (a, b, c)\).
• It must be part of every vector space, including any subspaces. This requirement is crucial for validating if a given subset is indeed a subspace.

Confirming that a subset includes the zero vector is often the first step in proving that something is a subspace.

Closure under Addition

Closure under addition is an important property for vector spaces and their possible subspaces. It means that if you take any two vectors within a set and add them together, the sum is also in the set. This property ensures that the group of vectors you are considering remains intact even after adding elements within it.

For instance, consider vectors \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) that satisfy the condition of a subspace, like \(2a - 3b + c = 0\) in egin{math}\mathbb{R}^{3}egin{math}. After adding these vectors, \((a_1 + a_2, b_1 + b_2, c_1 + c_2)\), the result must still meet the original condition to remain in the subspace.

• Mathematically, this is shown as: \[2(a_1 + a_2) - 3(b_1 + b_2) + (c_1 + c_2) = 0 + 0 = 0\]

Closure under addition keeps vector spaces and their subspaces organized and predictable.

Closure under Scalar Multiplication

Closure under scalar multiplication is another vital feature for identifying vector spaces and subspaces. Here, if you take any vector from the set and multiply it by a number, which is called a scalar, the resulting vector should still belong to the set.

This principle means that multiplying doesn’t cause vectors to step outside the defined space. For a subspace defined by a condition like \(2a - 3b + c = 0\), any vector \((a, b, c)\) within this subspace, when scaled by a scalar \(k\), would transform into \((ka, kb, kc)\). For the subset to be closed under scalar multiplication:

• Mathematically, it must hold true that: \[2(ka) - 3(kb) + (kc) = k(2a - 3b + c) = 0\]

Ensuring closure under this operation means the set remains stable, fostering an environment where linear combinations consistently belong to the subspace.