Problem 4 - A Book of Abstract Algebra

Question

The conjugate of \(\alpha\) is $$ \bar{\alpha}=\left(\begin{array}{lr} a-b i & -c-d i \\ c-d i & a+b i \end{array}\right) $$ The norm of \(\alpha\) is \(a^{2}+b^{2}+c^{2}+d^{2}\), and is written \(\|\alpha\| .\) Show directly (by matrix multiplication) that $$ \bar{\alpha} \alpha=\alpha \bar{\alpha}=\left(\begin{array}{cc} t & 0 \\ 0 & t \end{array}\right) \quad \text { where } t=\|\alpha\| $$ Conclude that the multiplicative inverse of \(\alpha\) is \((1 / t) \vec{\alpha}\).

Step-by-Step Solution

Answer

The inverse of \(\alpha\) is \((1/t)\bar{\alpha}\) where \(t = a^2 + b^2 + c^2 + d^2\).

Step 1: Understand the Problem

We are given a matrix \(\alpha\) and its conjugate \(\bar{\alpha}\). We are asked to prove that the product of \(\bar{\alpha}\) and \(\alpha\) (and vice versa) results in a diagonal matrix with equal entries \(t\), where \(t = \|\alpha\|\). Then, determine the inverse of \(\alpha\) based on this result.

Step 2: Determine Matrix Multiplication

First, we need to perform the matrix multiplication \(\bar{\alpha}\alpha\) and \(\alpha\bar{\alpha}\). Recall that the product of two matrices \(A\) and \(B\), where \(A = \left(\begin{array}{cc}a_{11} & a_{12} \ a_{21} & a_{22}\end{array}\right)\) and \(B = \left(\begin{array}{cc}b_{11} & b_{12} \ b_{21} & b_{22}\end{array}\right)\), is given by:\[AB = \left(\begin{array}{cc}a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22}\end{array}\right)\]Compute this specifically for our matrices.

Step 3: Compute \(\bar{\alpha}\alpha\)

Multiply \(\bar{\alpha} = \left(\begin{array}{cc} a-b i & -c-d i \ c-d i & a+b i\end{array}\right)\) by \(\alpha = \left(\begin{array}{cc} a+b i & c+d i \ -c+d i & a-b i\end{array}\right)\):\[\bar{\alpha}\alpha = \left(\begin{array}{cc} (a-bi)(a+bi) + (-c-di)(-c+di) & (a-bi)(c+di) + (-c-di)(a-bi) \ (c-di)(a+bi) + (a+bi)(-c+di) & (c-di)(c+di) + (a+bi)(a-bi)\end{array}\right)\]Simplify this to determine the diagonal form.

Step 4: Simplify Diagonal Elements

Calculate each element:-Diagonal element \((1,1)\):\[(a^2+b^2) + (c^2+d^2) = a^2+b^2+c^2+d^2 = t\]-Diagonal element \((2,2)\):\[(c^2+d^2) + (a^2+b^2) = a^2+b^2+c^2+d^2 = t\]Both diagonal elements are \(t\) as expected.

Step 5: Verify Non-Diagonal Elements

Compute non-diagonal elements to confirm they are zero:-Element \((1,2)\):\[(a-bi)(c+di) + (-c-di)(a-bi) = ac+bdi-aci-bc-ad-bdi+aci-bci = 0\]-Element \((2,1)\):\[(c-di)(a+bi) + (a+bi)(-c+di) = ca-cdi+abi-bdi-ca+adi+bci-badi = 0\]Thus, all non-diagonal elements are zero.

Step 6: Conclude Matrix Representation

From Steps 4 and 5, we conclude that \(\bar{\alpha}\alpha = \alpha\bar{\alpha} = \left(\begin{array}{cc} t & 0 \ 0 & t \end{array}\right)\), confirming the problem's assertion.

Step 7: Determining the Inverse

Given \(\bar{\alpha}\alpha = \left(\begin{array}{cc} t & 0 \ 0 & t \end{array}\right)\), we see that the matrix is \(tI\) where \(I\) is the identity matrix. Therefore, the inverse of \(\alpha\) can be expressed as:\[(1/t) \bar{\alpha}\]

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental concept in linear algebra, crucial for solving problems involving matrices. It involves taking two matrices and producing another matrix. The key rule is that the number of columns in the first matrix must equal the number of rows in the second matrix. Consider two matrices, say, Matrix A and Matrix B.

• Matrix A is of size m x n (m rows and n columns).
• Matrix B is of size n x p (n rows and p columns).

The resulting matrix from multiplying A and B, denoted as AB, will have dimensions m x p. The element of AB in row i and column j is obtained by taking the dot product of the i-th row of A and the j-th column of B. This involves multiplying corresponding elements and summing these products.
Understanding this rule is essential for calculating matrix products effectively. In our problem, we see matrix multiplication in action to confirm results that show special properties when complex numbers are involved.

Norm of a Matrix

The norm of a matrix provides a measure of its size or length. It gives us an idea about the matrix's magnitude. For matrices, several norms exist, but here we focus on a particular norm derived from complex numbers within the matrix. In this specific exercise, we are given that the norm of the matrix \( \alpha \) is \( a^2 + b^2 + c^2 + d^2 \).
This norm is akin to calculating the magnitude of a vector stemming from the matrix entries rather than its geometric interpretation. To compute this, treat each entry of the matrix squared, like raising scalars to the power of two, and then summing these values.

• Each component is squared individually: \( a^2, b^2, c^2, \) and \( d^2 \).
• All squared components are summed: \(a^2 + b^2 + c^2 + d^2 \).
• This total gives the norm \( \|\alpha\| \).

Understanding this provides insight into the internal magnitude of any matrix defined numerically.

Inverse of a Matrix

An inverse of a matrix is another matrix that, when multiplied with the original matrix, results in the identity matrix. The identity matrix acts as the number 1 in matrix algebra. For any square matrix \( A \), its inverse \( A^{-1} \) satisfies \( AA^{-1} = A^{-1}A = I \), where \( I \) is the identity matrix.
Finding the inverse involves ensuring the matrix is square and its determinant is non-zero. Calculating the inverse depends on several methods like Gaussian elimination, but in this exercise, we used a simpler approach:

• The product \( \bar{\alpha} \alpha \) equates to a scaled identity matrix \( tI \), where \( t = a^2 + b^2 + c^2 + d^2 \).
• The inverse thus becomes simply \( (1/t)\bar{\alpha} \), since multiplying by \( \bar{\alpha} \) scaled down by \( 1/t \) will yield \( I \).

This exercise highlights how understanding the special properties of matrix products helps us derive the inverse readily.

Complex Numbers in Matrices

Complex numbers are foundational in many aspects of mathematics, especially when they are integrated into matrices. Each element within a matrix could potentially be a complex number, expressed as \( a + bi \), where \( i \) is the imaginary unit.
Dealing with complex numbers in matrices becomes particularly interesting because it adds another dimension to computation and requires specific operations like finding the conjugate. The conjugate of a complex number negates the imaginary part, which has significant algebraic implications:

• For a matrix conjugate, each element in the matrix has its imaginary part negated.
• In our exercise, this operation helps in demonstrating specific properties when multiplying matrices involving complex numbers.
• The conjugate leads to the simplification in proving the matrix equivalences \( \bar{\alpha}\alpha = \alpha\bar{\alpha} \).

Understanding how complex numbers operate within matrices is essential for tackling problems across mathematics and engineering where non-real conditions are considered.