We are given a matrix $A = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$ and need to determine which vector from the provided options is an eigenvector of this matrix.

Steps to determine the eigenvector:

1. To be an eigenvector, a vector $v$ must satisfy the equation $A \cdot v = \lambda \cdot v$, where $\lambda$ is the eigenvalue, and $v$ is the eigenvector.

2. First, find the eigenvalues $\lambda$ by solving the characteristic equation: $\text{det}(A - \lambda I) = 0$ where $I$ is the identity matrix.

   We compute the determinant: $\text{det} \begin{bmatrix} -\lambda & 3 \\ -3 & -\lambda \end{bmatrix} = \lambda^2 + 9 = 0$ Solving $\lambda^2 + 9 = 0$ gives $\lambda = \pm 3i$.

3. Now, find the eigenvectors corresponding to $\lambda = 3i$ and $\lambda = -3i$.

Eigenvectors for $\lambda = 3i$:

We solve $(A - 3iI)v = 0$. This leads to a system of linear equations, and solving it yields the eigenvector corresponding to $\lambda = 3i$, which involves complex numbers like $1 + i$ or $1 - i$.

Conclusion:

From the list of options, the vector $\begin{bmatrix} 1 + i \\ i \end{bmatrix}$ matches an eigenvector associated with the eigenvalue $3i$. Therefore, the correct answer is the second option: $\begin{bmatrix} 1 + i \\ i \end{bmatrix}$.

Would you like more details on any step? Here are related questions to deepen your understanding:

1. How do you compute the determinant of a 2x2 matrix?
2. What are the properties of complex eigenvalues and their eigenvectors?
3. Can a matrix have both real and complex eigenvalues?
4. How do you solve for eigenvectors once eigenvalues are found?
5. Why does the characteristic polynomial have complex roots in this case?

Tip: When dealing with complex eigenvalues, remember that their corresponding eigenvectors will also involve complex components.