To find the eigenvectors of the matrix $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$, we need to follow these steps:

1. Find the Eigenvalues:

The eigenvalues $\lambda$ are found by solving the characteristic equation:

$\det(A - \lambda I) = 0$

Where $I$ is the identity matrix:

$A - \lambda I = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1-\lambda & 2 \\ 2 & 3-\lambda \end{bmatrix}$

The determinant of this matrix is:

$\det(A - \lambda I) = (1-\lambda)(3-\lambda) - (2)(2)$

$= \lambda^2 - 4\lambda - 1 - 4 = \lambda^2 - 4\lambda - 4$

Now, solve the characteristic equation:

$\lambda^2 - 4\lambda - 3 = 0$

Use the quadratic formula:

$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 1$, $b = -4$, and $c = -3$:

$\lambda = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$

So the eigenvalues are:

$\lambda_1 = 2 + \sqrt{7}, \quad \lambda_2 = 2 - \sqrt{7}$

2. Find the Eigenvectors:

For each eigenvalue $\lambda$, solve the equation $(A - \lambda I)\mathbf{v} = 0$ to find the corresponding eigenvector $\mathbf{v}$.

For $\lambda_1 = 2 + \sqrt{7}$:

$(A - \lambda_1 I) = \begin{bmatrix} 1-(2+\sqrt{7}) & 2 \\ 2 & 3-(2+\sqrt{7}) \end{bmatrix} = \begin{bmatrix} -1-\sqrt{7} & 2 \\ 2 & 1-\sqrt{7} \end{bmatrix}$

Set up the system $(A - \lambda_1 I)\mathbf{v} = 0$:

$\begin{bmatrix} -1-\sqrt{7} & 2 \\ 2 & 1-\sqrt{7} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

This simplifies to:

(-1-\sqrt{7})x_1 + 2x_2 = 0 \\ 2x_1 + (1-\sqrt{7})x_2 = 0 \end{cases}$$ We can solve for $$x_2$$ in terms of $$x_1$$ from the first equation: $$x_2 = \frac{1+\sqrt{7}}{2} x_1$$ So, the eigenvector corresponding to $$\lambda_1 = 2 + \sqrt{7}$$ is: $$\mathbf{v}_1 = \begin{bmatrix} 1 \\ \frac{1+\sqrt{7}}{2} \end{bmatrix}$$ #### For $$\lambda_2 = 2 - \sqrt{7}$$: $$(A - \lambda_2 I) = \begin{bmatrix} 1-(2-\sqrt{7}) & 2 \\ 2 & 3-(2-\sqrt{7}) \end{bmatrix} = \begin{bmatrix} -1+\sqrt{7} & 2 \\ 2 & 1+\sqrt{7} \end{bmatrix}$$ Set up the system $$(A - \lambda_2 I)\mathbf{v} = 0$$: $$\begin{cases} (-1+\sqrt{7})x_1 + 2x_2 = 0 \\ 2x_1 + (1+\sqrt{7})x_2 = 0 \end{cases}$$ We can solve for $$x_2$$ in terms of $$x_1$$ from the first equation: $$x_2 = \frac{1-\sqrt{7}}{2} x_1$$ So, the eigenvector corresponding to $$\lambda_2 = 2 - \sqrt{7}$$ is: $$\mathbf{v}_2 = \begin{bmatrix} 1 \\ \frac{1-\sqrt{7}}{2} \end{bmatrix}$$ ### Final Answer: The eigenvectors of the matrix \( A = \