2x2 Eigenvalues

Implement a solution to compute the eigenvalues of a 2×2 matrix using the characteristic equation. This task involves finding the scalar values, λ, that represent how much change occurs in a linear transformation.

The concept of eigenvalues and eigenvectors is crucial in linear algebra, as it helps describe the properties of linear transformations. For a 2×2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the eigenvalues satisfy the equation $\det(A - \lambda I) = 0$ , where $I$ is the identity matrix.

Start with the matrix $A$ and the identity matrix $I$ .
Construct the matrix $A - \lambda I$ .
Compute the determinant of $A - \lambda I$ and set it equal to zero.

\lambda^2 - (a+d)\lambda + (ad-bc) = 0

This technique is widely used in computer vision for image processing and analysis.

Example:

Input:

A = [[4, 1], [2, 3]]

Output:

eigenvalues = [5.0, 2.0]

Reasoning:

Finding eigenvalues of a 2×2 matrix using the characteristic equation:

For matrix A = [[a, b], [c, d]], eigenvalues satisfy: $\det(A - \lambda I) = 0$

Set up the characteristic equation: For A = [[4, 1], [2, 3]]: $\det\begin{pmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{pmatrix} = 0$
Expand the determinant: $(4-\lambda)(3-\lambda) - (1)(2) = 0$ $12 - 4\lambda - 3\lambda + \lambda^2 - 2 = 0$ $\lambda^2 - 7\lambda + 10 = 0$
Apply the quadratic formula: $\lambda = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm 3}{2}$
Solve for both eigenvalues: $\lambda_1 = \frac{7 + 3}{2} = 5$ $\lambda_2 = \frac{7 - 3}{2} = 2$
Result: eigenvalues = [5.0, 2.0]

The trace (sum of diagonal) equals $4 + 3 = 7 = 5 + 2$ , and determinant equals $12 - 2 = 10 = 5 \times 2$ . ✓

Constraints:

A is a 2×2 real matrix
Return eigenvalues sorted in descending order
Round to 4 decimal places
Assume real eigenvalues (discriminant ≥ 0)

Example:

Constraints:

Test Results