Eigenvalues and Eigenvectors
Definition and Significance
- An eigenvector of a square matrix $A$ is a non-zero vector $\mathbf{v}$ that, when multiplied by $A$, results in a scalar multiple of itself.
- This scalar is called the eigenvalue corresponding to that eigenvector.
Mathematically, this is expressed as:
$$A\mathbf{v} = \lambda\mathbf{v}$$
where $\lambda$ is the eigenvalue.
NoteEigenvalues and eigenvectors provide crucial information about the behavior of linear transformations represented by matrices.
Characteristic Polynomial
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:
$$\lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}$$
ExampleConsider 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: \[\lambda_1 = 4, \quad\lambda_2 = 2 \]
Finding Eigenvectors
Once we have the eigenvalues, we can find the corresponding eigenvectors by solving the equation $(A - \lambda I)\mathbf{v} = \mathbf{0}$ for each eigenvalue.
ExampleBased on the previous example:
For the eigenvector corresponding to \(\lambda_1 = 4 \): \[ (A - 4I) \mathbf{v} = 0 \]\[\begin{bmatrix} 3 - 4 & 1 \\ 1 & 3 - 4 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]\[\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] From the first equation: \[ - v_1 + v_2 = 0 \Rightarrow v_1 = v_2 \] Thus, an eigenvector for \(\lambda_1 = 4 \) is: \[\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] Similarly, for \(\lambda_2 = 2 \): \[ (A - 2I) \mathbf{v} = 0 \]\[\begin{bmatrix} 3 - 2 & 1 \\ 1 & 3 - 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]\[\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] From the first equation: \[ v_1 + v_2 = 0 \Rightarrow v_1 = -v_2 \] Thus, an eigenvector for \(\lambda_2 = 2 \) is: \[\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \]
Diagonalization of 2x2 Matrices
- Diagonalization is a process of transforming a matrix into a diagonal matrix.
- For 2x2 matrices with distinct real eigenvalues, this process is always possible.
Process of Diagonalization
- Find the eigenvalues of the matrix.
- Find the corresponding eigenvectors.
- Form the matrix $P$ with eigenvectors as columns.
