91Ó°ÊÓ

Eigenvalues are numbers that help us understand the behavior of matrices, especially in linear transformations. When you have a matrix, such as matrix \( A \) in the provided problem, eigenvalues are the special numbers \( \lambda \) that make the equation \( A\mathbf{v} = \lambda\mathbf{v} \) true for a non-zero vector \( \mathbf{v} \), known as the eigenvector. Simplifying, this means applying the matrix \( A \) to \( \mathbf{v} \) amounts to just scaling \( \mathbf{v} \) by \( \lambda \). In our exercise, the eigenvalues given are \( \lambda = 1 \) and \( \lambda = 5 \). These were verified by solving the equation \( \det(A - \lambda I) = 0 \), ensuring they satisfy the conditions for eigenvalues in the matrix.

A basis is a set of vectors that span a space and are linearly independent. For eigenspaces (a subspace formed by eigenvectors corresponding to a given eigenvalue), finding a basis means determining a set of vectors that can combine to form any vector in that eigenspace without redundancy. In this context, after finding the eigenvectors for the given eigenvalues, we find specific vectors to form a basis. For example, for \( \lambda = 1 \), after solving \((A-\lambda I)\mathbf{v} = \mathbf{0}\), we found a vector \( \begin{bmatrix} -2 \ 3 \end{bmatrix} \) that can represent all solutions, forming a basis of the eigenspace.

Matrix operations include many useful computations like addition, subtraction, multiplication, and finding determinants. When dealing with eigenvalues and eigenspaces, a common operation is matrix subtraction and multiplication, particularly when finding \( A - \lambda I \). Here \( I \) is the identity matrix. This operation has to be performed to evaluate the matrix for solving \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). Matrix operations allow transitions between different forms to simplify equations and eventually find the solutions we need, like the eigenvectors.

The characteristic equation is a polynomial equation derived from a square matrix, and its roots are the eigenvalues of that matrix. To find the characteristic equation, you subtract \( \lambda \) times the identity matrix from your matrix \( A \) to form \( A - \lambda I \), and then find the determinant: \( \det(A - \lambda I) = 0 \). Solving this equation gives you the eigenvalues. For instance, in our problem, after computing the determinant with \( A - \lambda I \), we checked it was zero for \( \lambda = 1 \) and \( \lambda = 5 \), confirming they are indeed eigenvalues of matrix \( A \). This equation is essential because it provides a systematic method to verify potential eigenvalues.