91Ó°ÊÓ

Finding the characteristic polynomial is the key step to determine the eigenvalues of a matrix. For any square matrix, its eigenvalues are the roots of its characteristic polynomial.
Given a matrix \(A\), the characteristic polynomial is determined by the equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix of the same size as \(A\), and \(\lambda\) represents the eigenvalues.
To construct this polynomial, you first form the matrix \(A - \lambda I\). For each entry in \(A\), you subtract the corresponding entry in the identity matrix multiplied by \(\lambda\).

• If \(A\) is a \(2 \times 2\) matrix, the result is a new \(2 \times 2\) matrix.
• The determinant of this new matrix gives us the characteristic polynomial.

This polynomial is what you will then solve to find the eigenvalues of \(A\). The degree of the polynomial is the same as the dimension of the matrix, so for a \(2 \times 2\) matrix, it is a quadratic equation.

Modulo arithmetic, often called modular arithmetic, is a system of arithmetic for integers. When discussing eigenvalues over a field like \(\mathbb{Z}_p\), where \(p\) is a prime number, all calculations are done using modulo \(p\).
This form of arithmetic wraps around after reaching \(p\), the modulus. Essentially, after dividing by \(p\), we only consider the remainder.
Consider modulo 3 for example:

• While calculating \(4 - 4\lambda + \lambda^2 - 1\), the operations are simplified by reducing each number modulo 3.
• Thus, 4 becomes 1 because \(4 \equiv 1 \mod 3\), and any result over \(p\) gets reduced similarly.

This simplifies solving many problems since complex arithmetic with large numbers reduces to simple operations with small integers.

The determinant of a matrix gives key insights into its properties, including the solutions to its characteristic polynomial. For a \( 2 \times 2 \) matrix, the determinant is simple to compute.
Given a matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated as \(ad - bc\). This value provides the characteristic polynomial when substituted for \(A - \lambda I\).

• For example, the determinant of \(\begin{pmatrix} 2 - \lambda & 1 \ 1 & 2 - \lambda \end{pmatrix}\) is \((2-\lambda)(2-\lambda) - 1\).
• This determinant equals the characteristic polynomial, which you solve to find the matrix's eigenvalues.

Understanding the role of the matrix determinant simplifies finding eigenvalues, as one needs to compute and set the determinant equation to zero to solve for eigenvalues efficiently.