Mathematically, this is expressed as:
$$A\mathbf{v} = \lambda\mathbf{v}$$
where $\lambda$ is the eigenvalue.
Eigenvalues and eigenvectors provide crucial information about the behavior of linear transformations represented by matrices.
For a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the characteristic polynomial is given by:
$$p(\lambda) = \det(A - \lambda I) = \lambda^2 - (a+d)\lambda + (ad-bc)$$
where $I$ is the 2x2 identity matrix.
The roots of this polynomial are the eigenvalues of the matrix. For a 2x2 matrix, we can find these roots using the quadratic formula:
$$\lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}$$
Consider the matrix \( A \): \[ A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix} \] The characteristic polynomial is given by: \[ p(\lambda) = \det(A - \lambda I) = \begin{vmatrix} 3 - \lambda& 1 \\ 1 & 3 - \lambda\end{vmatrix} \] Expanding the determinant: \[ (3 - \lambda)(3 - \lambda) - (1)(1) = \lambda^2 - 6\lambda + 8 \] Solving the quadratic equation: \[\lambda^2 - 6\lambda + 8 = 0 \] Factoring: \[ (\lambda - 4)(\lambda - 2) = 0 \] Thus, the eigenvalues are: \[\lambda_1 = 4, \quad\lambda_2 = 2 \]
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