Find the eigenvalues and eigenvectors of the following matrices. Do some problems by hand to be sure you understand what the process means. Then check your results by computer. $$\left(\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right)$$

To find the **eigenvalues** of a matrix, we start with the characteristic polynomial. This polynomial is essential as it helps us to determine whether a scalar \( \lambda \) is an eigenvalue of the matrix. The characteristic polynomial is derived from the determinant of \( A - \lambda I \), where \( A \) is the given matrix and \( I \) is the identity matrix.
For the given matrix:
\[ A = \begin{pmatrix} 1 & 3 \ 2 & 2 \end{pmatrix} \] We form \( A - \lambda I \):
\[ \begin{pmatrix} 1-\lambda & 3 \ 2 & 2-\lambda \end{pmatrix} \] Next, we calculate the determinant of this matrix to get our characteristic polynomial:
\[ \det \begin{pmatrix} 1-\lambda & 3 \ 2 & 2-\lambda \end{pmatrix} \] Use the determinant formula for a 2x2 matrix:
\[ (1-\lambda)(2-\lambda) - (3)(2) = 0 \] Simplify to get:
\[ \lambda^2 - 3\lambda - 4 = 0 \] This equation is the characteristic polynomial of our matrix.

After finding the characteristic polynomial, our next step is to solve it for \( \lambda \), the eigenvalues, using the quadratic formula.
The characteristic polynomial we have is:
\[ \lambda^2 - 3\lambda - 4 = 0 \] Now, let's solve it using the quadratic formula:
\( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \, \ b = -3 \, \ c = -4 \):
Substitute the values:
\[ \lambda = \frac{3 \pm \sqrt{9 + 16}}{2} \] Simplify:
\[ \lambda = \frac{3 \pm 5}{2} \] This gives us two eigenvalues:
\[ \lambda_1 = 4 \] and
\[ \lambda_2 = -1 \]

Now that we have the eigenvalues, it's time to find the corresponding eigenvectors. An eigenvector is a vector whose direction remains unchanged when a linear transformation is applied to it. For the given eigenvalue \( \lambda \), we solve \( (A - \lambda I)v = 0 \).

**Finding Eigenvectors for \( \lambda_1 = 4 \):**
For \( \lambda_1 = 4 \), solve: \[ (A - 4I)v = 0 \] which translates to:
\[ \begin{pmatrix} 1-4 & 3 \ 2 & 2-4 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] Simplify to:
\[ \begin{pmatrix} -3 & 3 \ 2 & -2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] This gives us the equation:
\[ -3x + 3y = 0 \] Simplifying, we get:
\[ x = y \] So, an eigenvector is:
\[ \begin{pmatrix} 1 \ 1 \end{pmatrix} \]

**Finding Eigenvectors for \( \lambda_2 = -1 \):**
For \( \lambda_2 = -1 \), solve:
\[ (A + I)v = 0 \] which translates to:
\[ \begin{pmatrix} 1+1 & 3 \ 2 & 2+1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] Simplify to:
\[ \begin{pmatrix} 2 & 3 \ 2 & 3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] This gives us the equation:
\[ 2x + 3y = 0 \] Solving for \( x \) in terms of \( y \):
\[ 2x = -3y \] \[ x = -\frac{3}{2}y \] So, an eigenvector is:
\[ \begin{pmatrix} -3 \ 2 \end{pmatrix} \]