91Ó°ÊÓ

Eigenvalues are fundamental in the study of linear transformations represented by matrices. They provide insight into the matrix's behavior and characteristics. An eigenvalue, denoted as \( \lambda \), is a scalar associated with a square matrix \( A \) and a non-zero vector \( \mathbf{v} \). When \( A \) is applied to \( \mathbf{v} \), the vector is stretched by the factor \( \lambda \) without changing direction.

• To find eigenvalues of a matrix, compute the determinant of \( A - \lambda I \), where \( I \) is the identity matrix.
• The result of this determinant being zero gives us the characteristic equation.
• The solutions of this equation are the eigenvalues.

In our example, we calculated the characteristic equation of \( A = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix} \) to be \( \lambda^2 - 4\lambda + 3 = 0 \). Solving it provides us with the eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = 1 \). These are the values for which the transformation scales the associated eigenvector without changing its direction.

Eigenvectors are vectors that remain in the same span when transformed by matrix \( A \), only scaled by their corresponding eigenvalues. Each eigenvector is tied to an eigenvalue.

• For a given eigenvalue \( \lambda \), the eigenvector \( \mathbf{v} \) satisfies the equation \( (A - \lambda I)\mathbf{v} = 0 \).
• A non-zero solution to this equation gives us the eigenvector associated with \( \lambda \).
• The process typically involves solving a system of linear equations.

In our practical problem, for \( \lambda_1 = 3 \), solving the equation gives the eigenvector \( \mathbf{v_1} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \). For \( \lambda_2 = 1 \), we find \( \mathbf{v_2} = \begin{pmatrix} 1 \ -1 \end{pmatrix} \). These vectors are critical in forming the matrix \( S \) used in the diagonalization of \( A \). They provide the direction which our transformations respect.

The inverse of a matrix is a crucial component in diagonalization, allowing transformations to be reversed. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse can be found provided the determinant \( ad-bc eq 0 \).

• The formula for the inverse is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
• Matrix inverses are essential in solving systems of linear equations and are the key in transforming systems back to their original state.
• In our exercise, finding the inverse of \( S \) was necessary to demonstrate the formula for \( A^k \).

For the matrix \( S = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \), its inverse \( S^{-1} \) was computed as \( \frac{1}{2} \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \). This transformation is essential in verifying the relationship with \( A^k = S \Lambda^k S^{-1} \). It makes the reversal from \( \Lambda^k \) back to \( A^k \) possible.

The characteristic equation is the core tool used to find eigenvalues. Formulated as \( \det(A - \lambda I) = 0 \), it accounts for scalar multiples that maintain the vector's span direction during the matrix transformation.

• The equation is derived from setting the determinant of \( A - \lambda I \) to zero.
• This approach reveals the eigenvalues as it results in a polynomial in \( \lambda \), called the characteristic polynomial.
• Solving this polynomial provides the eigenvalues of the matrix.

In our example with matrix \( A \), setting up \( \det \begin{pmatrix} 2 - \lambda & 1 \ 1 & 2 - \lambda \end{pmatrix} = \lambda^2 - 4\lambda + 3 = 0 \) led us to the solutions \( \lambda_1 = 3 \) and \( \lambda_2 = 1 \). Understanding and solving the characteristic equation is indispensable for tasks involving matrix diagonalization because it sets the stage for finding eigenvalues and consequently the corresponding eigenvectors.