29.8. Example Problems

Let:

$A = \begin{bmatrix} 2 & 1 \\\ 1 & 2 \\ \end{bmatrix}$

Step 1: Find characteristic equation:

$\text{det}(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 1 \\\ 1 & 2 - \lambda \end{vmatrix} = (2 - \lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0 \Rightarrow \lambda = 1, 3$

Step 2: Find eigenvectors:

For $\lambda = 1$:

$(A - I)x = 0 \Rightarrow \begin{bmatrix} 1 & 1 \\\ 1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\\ x_2 \end{bmatrix} = 0 \Rightarrow x = \begin{bmatrix} 1 \\\ 1 \end{bmatrix} \text{ or any scalar multiple}$

For $\lambda = 3$:

$(A - 3I)x = 0 \Rightarrow \begin{bmatrix} -1 & 1 \\\ 1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\\ x_2 \end{bmatrix} = 0 \Rightarrow x = \begin{bmatrix} 1 \\\ -1 \end{bmatrix} \text{ or any scalar multiple}$

In this example, we are tasked with finding the eigenvalues and eigenvectors of a 2x2 matrix A. We start with the matrix $A = \begin{bmatrix} 2 & 1 \\\ 1 & 2 \end{bmatrix}$. We first need to find the characteristic polynomial by calculating the determinant of $(A - \lambda I)$ where $I$ is the identity matrix. This gives us the equation $(2 - \lambda)^2 - 1 = 0$, which simplifies to $\lambda^2 - 4\lambda + 3 = 0$. Factoring this polynomial results in the eigenvalues $\lambda = 1$ and $\lambda = 3$. Then, to find the corresponding eigenvectors, we substitute each eigenvalue back into the equation $(A - \lambda I)x = 0$. For $\lambda = 1$, solving gives us the eigenvector $\begin{bmatrix} 1 \\\ 1 \end{bmatrix}$, and for $\lambda = 3$, we find the eigenvector $\begin{bmatrix} 1 \\\ -1 \end{bmatrix}$.

Think of eigenvalues and eigenvectors like a rubber band and the way it stretches. When you pull and stretch the rubber band in certain directions (which represents the eigenvectors), it will stretch more or less depending on how strong you pull (which represents the eigenvalues). The eigenvalues tell you how much stretching occurs in those specific directions.

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