Problem 2 - A Book of Abstract Algebra

Question

Let \(\mathscr{S}\) be the following subset of \(\mathscr{M}_{2}(\mathbb{R})\) : $$ \mathscr{S}=\left\\{\left(\begin{array}{rr} a & b \\ -b & a \end{array}\right): a, b \in \mathbb{R}\right\\} $$ Prove that the function $$ f(a+b \mathbf{i})=\left(\begin{array}{rr} a & b \\ -b & a \end{array}\right) $$ is an isomorphism from \(\mathbb{C}\) to \(\mathscr{S}\). [REMARK: You must begin by checking that \(f\) is a well-defined function; that is, if \(a+b \mathbf{i}=c+d \mathbf{i}\), then \(f(a+b \mathbf{i})=f(c+d \mathbf{i})\). To do this, note that if \(a+b \mathbf{i}=c+d \mathbf{i}\) then \(a-c=(d-b) \mathbf{i} ;\) this last equation is impossible unless both sides are equal to zero, for otherwise it would assert that a given real number is equal to an imaginary number.].

Step-by-Step Solution

Answer

The function \( f \) is an isomorphism from \( \mathbb{C} \) to \( \mathscr{S} \).

Step 1: Check if the function is well-defined

To show that the function \( f \) is well-defined, assume \( a + b \mathbf{i} = c + d \mathbf{i} \). This implies \( a = c \) and \( b = d \). Consequently, \( \begin{pmatrix} a & b \ -b & a \end{pmatrix} = \begin{pmatrix} c & d \ -d & c \end{pmatrix} \), proving that \( f(a + b \mathbf{i}) = f(c + d \mathbf{i}) \). Therefore, \( f \) is well-defined.

Step 2: Prove that f is a homomorphism

For any \( z_1 = a_1 + b_1 \mathbf{i} \) and \( z_2 = a_2 + b_2 \mathbf{i} \) in \( \mathbb{C} \), we have \( z_1 z_2 = (a_1a_2 - b_1b_2) + (a_1b_2 + a_2b_1)\mathbf{i} \). The function \( f(z_1z_2) \) then equals \( \begin{pmatrix} a_1a_2 - b_1b_2 & a_1b_2 + a_2b_1 \ -a_1b_2 - a_2b_1 & a_1a_2 - b_1b_2 \end{pmatrix} \). Now, \( f(z_1) \cdot f(z_2) = \begin{pmatrix} a_1 & b_1 \ -b_1 & a_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \ -b_2 & a_2 \end{pmatrix} \), which also equals \( \begin{pmatrix} a_1a_2 - b_1b_2 & a_1b_2 + a_2b_1 \ -a_1b_2 - a_2b_1 & a_1a_2 - b_1b_2 \end{pmatrix} \). This shows \( f(z_1z_2) = f(z_1) \cdot f(z_2) \), establishing \( f \) as a homomorphism.

Step 3: Show f is bijective

To show \( f \) is injective, assume \( f(a_1+b_1\mathbf{i}) = f(a_2+b_2\mathbf{i}) \). This implies \( a_1=a_2 \) and \( b_1=b_2 \), which means \( a_1+b_1\mathbf{i} = a_2+b_2\mathbf{i} \). Therefore, \( f \) is injective. To prove \( f \) is surjective, any matrix \( \begin{pmatrix} a & b \ -b & a \end{pmatrix} \) in \( \mathscr{S} \) can be written as \( f(a+b\mathbf{i}) \), making \( f \) surjective. Hence, \( f \) is bijective.

Step 4: Conclude that f is an isomorphism

As we have shown \( f \) to be well-defined, a homomorphism, and bijective, it follows that \( f \) is an isomorphism from \( \mathbb{C} \) to \( \mathscr{S} \).

Key Concepts

Complex Numbers

Complex numbers are fundamental in mathematics. They include a real part and an imaginary part. The general form is \(a + b\mathbf{i}\), where \(a\) and \(b\) are real numbers, and \(\mathbf{i}\) is the imaginary unit with the property \(\mathbf{i}^2 = -1\). This combination allows complex numbers to handle scenarios that real numbers alone cannot, such as the square roots of negative numbers.

In this exercise, complex numbers take center stage by providing a framework to explore isomorphism between these numbers and a set of real matrices. Each complex number can be uniquely represented in a matrix form, bridging the gap between algebraic expressions and linear algebra settings. This association surfaces in our isomorphic function \(f\), which maps a complex number to a specific matrix.

The connection formed by this mapping underscores the versatility and depth of complex numbers, transforming abstract algebraic concepts into tangible matrix operations. This exercise exemplifies the notion that complex numbers, though abstract, provide profound insights and applications across mathematical fields.

Real Matrices

Real matrices are arrays of real numbers arranged in rows and columns. They form the backbone of linear algebra. In the context of this exercise, real matrices help represent complex numbers in a two-by-two matrix form.

Consider the general form of a matrix in our set \(\mathscr{S}\):

• Each entry is a real number, organized in a symmetric way to represent complex numbers.
• The diagonal terms are equal to \(a\), the real part.
• The off-diagonal terms, \(b\) and \(-b\), capture the imaginary part \(b\mathbf{i}\).

This structured form allows for operations like addition and multiplication, drawing parallels with operations on complex numbers. These matrices ensure that each complex number corresponds precisely to one matrix in \(\mathscr{S}\), making \(f\) a clear representation mapping in this exercise.
This matrix representation bridges complex numbers to linear algebra, highlighting how different mathematical domains overlap and enhance our understanding of functional mappings.

Linear Algebra

Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between them. It plays a vital role in simplifying complex problems through vector and matrix representations. In our discussion, linear algebra helps connect complex numbers and real matrices under the function \(f\).

Through the mapping, linear algebra shows:

• Properties of matrices, such as identity and inverses, find parallels in the addition and multiplication behavior in complex numbers.
• The set \(\mathscr{S}\), composed of matrices, serves as a vector space with addition and scalar multiplication resembling complex number operations.
• Operations on matrices reflect arithmetic on complex numbers, reinforcing the structure and symmetry present in both mathematical entities.

Understanding linear algebra equips you to grasp the structure of complex numbers further, especially when exploring transformations and representations in vector spaces. It shows us that abstract mathematical concepts can have concrete, visual representations, making them easier to work with and understand.

Function Mapping

Function mapping refers to the process of associating every element from one set with an element from another set, providing a method to re-express mathematical ideas. Our function \(f\) maps complex numbers to real matrices, explicitly demonstrating an isomorphic relationship.

This mapping defines:

• For every complex number \(a+b\mathbf{i}\), there is a unique corresponding matrix and vice versa.
• The operation preserves the structure – operations on complex numbers mirror those on their matrix counterparts.
• A well-defined, bijective function, as shown in our solution, ensures one-to-one correspondence between elements in \(\mathbb{C}\) and matrices in \(\mathscr{S}\).

This mapping is foundational in understanding how different algebraic structures can describe the same underlying concepts, a keystone in mathematical abstraction and representation. It exemplifies how seemingly different areas of math share deep connections, allowing for seamless transition and translation between contexts.