To find the eigenvectors of the matrix \(\mathbf{A} = \begin{pmatrix} 4 & 2 \\ -1 & 6 \end{pmatrix}\), we need to follow these steps:

### Step 1: Find the Eigenvalues
The eigenvalues \(\lambda\) of matrix \(\mathbf{A}\) are found by solving the characteristic equation:

\[
\det(\mathbf{A} - \lambda \mathbf{I}) = 0
\]

Where \(\mathbf{I}\) is the identity matrix. For our matrix:

\[
\mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} 4 & 2 \\ -1 & 6 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4-\lambda & 2 \\ -1 & 6-\lambda \end{pmatrix}
\]

The determinant is:

\[
\det(\mathbf{A} - \lambda \mathbf{I}) = (4-\lambda)(6-\lambda) - (2)(-1) = (\lambda^2 - 10\lambda + 24) + 2 = \lambda^2 - 10\lambda + 26
\]

### Step 2: Solve the Characteristic Equation
Solve the quadratic equation:

\[
\lambda^2 - 10\lambda + 26 = 0
\]

The roots (eigenvalues) \(\lambda\) are given by the quadratic formula:

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

For \(a = 1\), \(b = -10\), and \(c = 26\):

\[
\lambda = \frac{10 \pm \sqrt{100 - 104}}{2} = \frac{10 \pm \sqrt{-4}}{2} = \frac{10 \pm 2i}{2} = 5 \pm i
\]

So, the eigenvalues are \(\lambda_1 = 5 + i\) and \(\lambda_2 = 5 - i\).

### Step 3: Find the Eigenvectors
For each eigenvalue, solve the equation \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0\) to find the eigenvectors.

Let's do this for \(\lambda_1 = 5 + i\):

\[
\mathbf{A} - (5 + i)\mathbf{I} = \begin{pmatrix} 4-(5+i) & 2 \\ -1 & 6-(5+i) \end{pmatrix} = \begin{pmatrix} -1-i & 2 \\ -1 & 1-i \end{pmatrix}
\]

Solve:

\[
\begin{pmatrix} -1-i & 2 \\ -1 & 1-i \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

This leads to the system:

\[
\begin{cases}
(-1-i)x_1 + 2x_2 = 0 \\
-x_1 + (1-i)x_2 = 0
\end{cases}
\]

From the first equation:

\[
x_1 = \frac{2}{1+i}x_2
\]

Simplify \(\frac{2}{1+i}\) by multiplying by the conjugate:

\[
x_1 = \frac{2(1-i)}{(1+i)(1-i)}x_2 = (1-i)x_2
\]

So, the eigenvector corresponding to \(\lambda_1 = 5 + i\) is:

\[
\mathbf{v}_1 = x_2 \begin{pmatrix} 1-i \\ 1 \end{pmatrix}
\]

Similarly, for \(\lambda_2 = 5 - i\), the eigenvector would be:

\[
\mathbf{v}_2 = x_2 \begin{pmatrix} 1+i \\ 1 \end{pmatrix}
\]

### Final Answer
The eigenvectors corresponding to the eigenvalues \(\lambda_1 = 5 + i\) and \(\lambda_2 = 5 - i\) are:

\[
\mathbf{v}_1 = \begin{pmatrix} 1 - i \\ 1 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 1 + i \\ 1 \end{pmatrix}
\]

Would you like more details, or do you have any questions?

### Related Questions
1. How do you find eigenvectors when the eigenvalues are real numbers?
2. What is the significance of complex eigenvalues in a matrix?
3. Can a matrix have eigenvectors with zero entries?
4. How do eigenvectors and eigenvalues relate to diagonalization?
5. What is the geometric interpretation of eigenvectors and eigenvalues?

### Tip
When dealing with complex eigenvalues and eigenvectors, it is helpful to understand how to handle complex arithmetic and conjugates, as they often arise in such problems.

Discover how to find eigenvectors of the matrix [4, 2; -1, 6] using step-by-step methods. This problem involves calculating eigenvalues, solving characteristic equations, and handling complex numbers in linear algebra. Suitable for advanced undergraduate studies.

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