The task requires finding eigenvalues $\lambda_1$ and $\lambda_2$ (with $\lambda_1 < \lambda_2$) and their corresponding eigenvectors for the matrix

$A = \begin{bmatrix} 5 & 0 \\ 6 & -1 \end{bmatrix}.$

Step 1: Find the eigenvalues

The eigenvalues of a matrix $A$ are found by solving the characteristic equation: $\det(A - \lambda I) = 0,$ where $I$ is the identity matrix.

For $A = \begin{bmatrix} 5 & 0 \\ 6 & -1 \end{bmatrix}$, we calculate: $A - \lambda I = \begin{bmatrix} 5 - \lambda & 0 \\ 6 & -1 - \lambda \end{bmatrix}.$

The determinant is: $\det(A - \lambda I) = (5 - \lambda)(-1 - \lambda) - (6 \cdot 0).$

Simplify: $\det(A - \lambda I) = (5 - \lambda)(-1 - \lambda) = -5 - 5\lambda + \lambda + \lambda^2 = \lambda^2 - 4\lambda - 5.$

Set this equal to 0: $\lambda^2 - 4\lambda - 5 = 0.$

Solve the quadratic equation using the quadratic formula: $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = -4$, and $c = -5$. Substituting: $\lambda = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)} = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2}.$

$\lambda = \frac{4 \pm 6}{2}.$

Thus: $\lambda_1 = \frac{4 - 6}{2} = -1, \quad \lambda_2 = \frac{4 + 6}{2} = 5.$

Step 2: Find eigenvectors

For each eigenvalue $\lambda$, solve $(A - \lambda I)\vec{X} = 0$.

For $\lambda_1 = -1$:

$A - (-1)I = A + I = \begin{bmatrix} 5 + 1 & 0 \\ 6 & -1 + 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 6 & 0 \end{bmatrix}.$

The system $(A - (-1)I)\vec{X} = 0$ becomes: $\begin{bmatrix} 6 & 0 \\ 6 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$

From the first row: $6x_1 = 0 \implies x_1 = 0.$

Let $x_2 = t$, where $t \in \mathbb{R}$. The eigenvector is: $\vec{X}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$

For $\lambda_2 = 5$:

$A - 5I = \begin{bmatrix} 5 - 5 & 0 \\ 6 & -1 - 5 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 6 & -6 \end{bmatrix}.$

The system $(A - 5I)\vec{X} = 0$ becomes: $\begin{bmatrix} 0 & 0 \\ 6 & -6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$

From the second row: $6x_1 - 6x_2 = 0 \implies x_1 = x_2.$

Let $x_1 = t$. Then $x_2 = t$, and the eigenvector is: $\vec{X}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$

Final Answer:

• Eigenvalues: $\lambda_1 = -1, \lambda_2 = 5$,
• Corresponding eigenvectors: $\vec{X}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad \vec{X}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$

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Follow-Up Questions:

1. How are eigenvalues used in real-world applications?
2. Can eigenvalues for $3 \times 3$ matrices be calculated similarly?
3. What happens if the eigenvalues are complex numbers?
4. How do eigenvectors relate to transformations in linear algebra?
5. What are some special properties of symmetric matrices' eigenvalues?

Tip:

Always double-check the characteristic equation for calculation errors when solving for eigenvalues!