91Ó°ÊÓ

To find the eigenvalues of a matrix, one must first find its characteristic equation. The characteristic equation is derived from the determinant of the matrix A minus lambda times the identity matrix (\text{det}(A - \lambda I)). This step involves setting up the polynomial equation where \lambda represents eigenvalues.
The characteristic equation tells us where the matrix minus \lambda times the identity matrix results in a zero determinant. This is crucial in finding eigenvalues because they are solutions to this polynomial equation. For example, if we start with a matrix \begin{bmatrix} a & b \c & d \end{bmatrix}, we subtract \lambda I (where I is the identity matrix), resulting in the matrix \[ A - \lambda I = \begin{bmatrix} a - \lambda & b \ c & d - \lambda \end{bmatrix} \]. We then find the determinant of this new matrix to form the characteristic equation.

The determinant is a scalar value that helps us understand certain properties of a matrix, such as whether it is invertible. For our current problem, computing the determinant is key in forming the characteristic equation.
Given a 2x2 matrix \[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix}, \] the determinant is calculated as \text{det}(A) = ad - bc.
When we create the matrix (A - \lambda I) for the characteristic equation, we also need to find the determinant of this new matrix. This involves substituting all instances of A’s values appropriately, like so: \[ \text{det}(A - \lambda I) = \begin{vmatrix} a - \lambda & b \ c & d - \lambda \end{vmatrix} = (a - \lambda)(d - \lambda) - bc. \] Solving this determinant and setting it to zero gives a polynomial equation in terms of \lambda.

After writing down the characteristic equation as a polynomial, our goal is to solve for the eigenvalues (\text{\lambda}). If the polynomial is of degree 2, we have a quadratic equation. A standard quadratic equation is in the form \ ax^2 + bx + c = 0.
To solve for \lambda here, we use the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting our specific coefficients (a, b, and c) into this formula will give us the eigenvalues.
For instance, in the problem, we have the characteristic equation as \lambda^2 - 10\lambda + 3 = 0. Here, a = 1, b = -10, and c = 3. Plugging into the quadratic formula, we find our eigenvalues as \[ \lambda_1 = 9 \quad \text{and} \quad \lambda_2 = 1 \] It means that \lambda_1 and \lambda_2 are the values at which the matrix becomes singular.

Linear algebra is a branch of mathematics focusing on vector spaces and linear mappings between them. It is fundamental in understanding eigenvalues and eigenvectors.
An eigenvector of a matrix A is a non-zero vector v such that multiplying A by v results in a vector that is a scalar multiple of v. This scalar is known as the eigenvalue (\lambda).
In matrix form, this is written as: \[ A\mathbf{v} = \lambda\mathbf{v} \] Finding an eigenvector involves solving the matrix equation (A - \lambda I)v = 0.
For eigenvalues \lambda_1=9 and \lambda_2=1 derived from our example, we substitute these values back into (A - \lambda I) and solve for v. Different eigenvalues may have different eigenvectors, capturing how the matrix stretches or squashes vectors in its vector space.
This step-by-step understanding of eigenvalues and eigenvectors shows not just computational skills but also conceptual insights into the transformations involved in linear algebra.