Let's find the eigenvectors corresponding to the eigenvalues \( \lambda_1 = 2 + 2i \) and \( \lambda_2 = 2 - 2i \).

### Step 1: Eigenvector for \( \lambda_1 = 2 + 2i \)

We need to solve the equation:

\[
(A - \lambda_1 I) \mathbf{v}_1 = 0
\]

Substituting \( \lambda_1 = 2 + 2i \) into the matrix:

\[
A - \lambda_1 I = \begin{pmatrix} 1-(2+2i) & -5 \\ 1 & 3-(2+2i) \end{pmatrix} = \begin{pmatrix} -1-2i & -5 \\ 1 & 1-2i \end{pmatrix}
\]

Now, we solve the system:

\[
\begin{pmatrix} -1-2i & -5 \\ 1 & 1-2i \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

This leads to the following two equations:

\[
\begin{cases}
(-1 - 2i)v_{11} - 5v_{12} = 0 \\
v_{11} + (1 - 2i)v_{12} = 0
\end{cases}
\]

From the first equation:

\[
v_{11} = \frac{-5v_{12}}{-1-2i} = \frac{5v_{12}}{1+2i}
\]

To simplify \( \frac{5}{1+2i} \), multiply numerator and denominator by the complex conjugate of the denominator:

\[
\frac{5}{1+2i} \cdot \frac{1-2i}{1-2i} = \frac{5(1-2i)}{(1+2i)(1-2i)} = \frac{5(1-2i)}{1 + 4} = \frac{5 - 10i}{5} = 1 - 2i
\]

So,

\[
v_{11} = (1 - 2i)v_{12}
\]

Thus, the eigenvector corresponding to \( \lambda_1 = 2 + 2i \) can be written as:

\[
\mathbf{v}_1 = v_{12} \begin{pmatrix} 1 - 2i \\ 1 \end{pmatrix}
\]

For simplicity, let \( v_{12} = 1 \). Then the eigenvector is:

\[
\mathbf{v}_1 = \begin{pmatrix} 1 - 2i \\ 1 \end{pmatrix}
\]

### Step 2: Eigenvector for \( \lambda_2 = 2 - 2i \)

Similarly, for \( \lambda_2 = 2 - 2i \), we solve:

\[
(A - \lambda_2 I) \mathbf{v}_2 = 0
\]

Substituting \( \lambda_2 = 2 - 2i \) into the matrix:

\[
A - \lambda_2 I = \begin{pmatrix} 1-(2-2i) & -5 \\ 1 & 3-(2-2i) \end{pmatrix} = \begin{pmatrix} -1+2i & -5 \\ 1 & 1+2i \end{pmatrix}
\]

Now, solve the system:

\[
\begin{pmatrix} -1+2i & -5 \\ 1 & 1+2i \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

From the first equation:

\[
v_{21} = \frac{-5v_{22}}{-1+2i} = \frac{5v_{22}}{1-2i}
\]

Simplifying \( \frac{5}{1-2i} \):

\[
\frac{5}{1-2i} \cdot \frac{1+2i}{1+2i} = \frac{5(1+2i)}{(1-2i)(1+2i)} = \frac{5(1+2i)}{1 + 4} = \frac{5 + 10i}{5} = 1 + 2i
\]

So,

\[
v_{21} = (1 + 2i)v_{22}
\]

Thus, the eigenvector corresponding to \( \lambda_2 = 2 - 2i \) is:

\[
\mathbf{v}_2 = v_{22} \begin{pmatrix} 1 + 2i \\ 1 \end{pmatrix}
\]

For simplicity, let \( v_{22} = 1 \). Then the eigenvector is:

\[
\mathbf{v}_2 = \begin{pmatrix} 1 + 2i \\ 1 \end{pmatrix}
\]

### Summary

- The eigenvector corresponding to \( \lambda_1 = 2 + 2i \) is \( \mathbf{v}_1 = \begin{pmatrix} 1 - 2i \\ 1 \end{pmatrix} \).
- The eigenvector corresponding to \( \lambda_2 = 2 - 2i \) is \( \mathbf{v}_2 = \begin{pmatrix} 1 + 2i \\ 1 \end{pmatrix} \).

Do you have any more questions or need further clarification?

Here are 5 related questions:
1. How do eigenvectors help in solving systems of differential equations?
2. What is the significance of complex eigenvectors in the solution?
3. How do you interpret the real and imaginary parts of the eigenvectors?
4. Can you have multiple linearly independent eigenvectors for a single eigenvalue?
5. How would the solution look if the eigenvalues were purely real?

**Tip:** When dealing with complex eigenvalues, it is often helpful to express the solution in terms of real-valued functions using Euler's formula.

Explore how to find eigenvectors and eigenvalues for complex numbers in matrices. This detailed explanation covers the process of solving eigenvector equations, including examples with complex eigenvalues. Suitable for advanced undergraduate students and beyond.

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