91Ó°ÊÓ

To determine if a matrix is diagonalizable, we begin by finding its characteristic polynomial. This polynomial is fundamental in identifying eigenvalues of the matrix. The characteristic polynomial of a matrix \( A \) is calculated using the determinant expression \( \det(A - \lambda I) \), where \( \lambda \) are the eigenvalues we seek, and \( I \) is the identity matrix of the same size as \( A \). For the given matrix \( A = \begin{pmatrix} k & 1 \ 1 & 0 \end{pmatrix} \), we subtract \( \lambda I \) from \( A \), which results in a new matrix. The determinant of this new matrix \( \begin{pmatrix} k - \lambda & 1 \ 1 & -\lambda \end{pmatrix} \) gives us the characteristic polynomial:

• Calculate: \((k-\lambda)(-\lambda) - 1 \cdot 1 = \lambda^2 - k\lambda - 1\)

Thus, the characteristic polynomial is \( \lambda^2 - k\lambda - 1 \). This polynomial is pivotal in finding the eigenvalues next.

Eigenvalues are the special set of scalars associated with a matrix that provide insight into its properties and transformation capabilities. To find the eigenvalues of the matrix \( A \), we solve its characteristic polynomial. In our example, we need to solve the equation \( \lambda^2 - k\lambda - 1 = 0 \) for \( \lambda \). Solving this quadratic equation yields the eigenvalues of the matrix. The roots \( \lambda \) represent the values where the transformation equates to a simple scaling represented by the matrix. By solving: \[ \lambda = \frac{k \pm \sqrt{k^2 + 4}}{2} \]We find the eigenvalues. These eigenvalues are distinct due to the nature of the discriminant \( k^2 + 4 \), which is always positive, ensuring that the matrix \( A \) is diagonalizable for every real value of \( k \).

The quadratic formula is a cornerstone in solving quadratic equations, like the characteristic polynomial in our case. This formula provides the easiest mechanism for finding roots of a quadratic equation of the form \( ax^2 + bx + c = 0 \). It is expressed as:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Applying the quadratic formula to the characteristic polynomial \( \lambda^2 - k\lambda - 1 = 0 \), where \( a = 1 \), \( b = -k \), and \( c = -1 \), we determine the eigenvalues \( \lambda \):

• \[ \lambda = \frac{k \pm \sqrt{k^2 + 4}}{2} \]

This succinctly solves for the values of \( \lambda \) that satisfy the polynomial equation, illustrating how the quadratic formula is utilized in matrix analysis to identify eigenvalues.

The determinant of a matrix is a scalar value that provides important insights into the matrix characteristics, such as invertibility and the behavior of the linear transformation it represents. When determining diagonalizability via the characteristic polynomial, we rely on the determinant. In this exercise, we calculate the determinant of the modified matrix \( A - \lambda I \):

• \[ \det \begin{pmatrix} k - \lambda & 1 \ 1 & -\lambda \end{pmatrix} = (k-\lambda)(-\lambda) - 1\]

This determinant results in the characteristic polynomial \( \lambda^2 - k\lambda - 1 \). The value of the determinant is crucial as it directly affects the nature and calculation of the eigenvalues. A zero determinant implies a singular matrix, but in the context of the characteristic equation, it specifies the change from which eigenvalues can be derived. This understanding aids in determining whether a matrix can be diagonalized through examination of its eigenvalues and their algebraic multiplicities.