eigenvector {{1, 2}, {2, 1}}, given eigenvalue lambda = (1 + 2^(1/2))

To find the eigenvector corresponding to the eigenvalue \(\lambda = 1 + \sqrt{2}\) for the matrix 

\[
A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix},
\]

we need to solve the equation:

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

where \(I\) is the identity matrix and \(\mathbf{v}\) is the eigenvector corresponding to \(\lambda\).

### Step-by-Step Solution

1. **Form the matrix \(A - \lambda I\):**

\[
A - \lambda I = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} - (1 + \sqrt{2}) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.
\]

2. **Subtract \(\lambda I\) from \(A\):**

\[
A - \lambda I = \begin{bmatrix} 1 - (1 + \sqrt{2}) & 2 \\ 2 & 1 - (1 + \sqrt{2}) \end{bmatrix}.
\]

Simplifying further:

\[
A - \lambda I = \begin{bmatrix} -\sqrt{2} & 2 \\ 2 & -\sqrt{2} \end{bmatrix}.
\]

3. **Set up the system of linear equations \((A - \lambda I) \mathbf{v} = 0\):**

Let \(\mathbf{v} = \begin{bmatrix} x \\ y \end{bmatrix}\). Then:

\[
\begin{bmatrix} -\sqrt{2} & 2 \\ 2 & -\sqrt{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.
\]

This results in the following two equations:

\[
-\sqrt{2} x + 2y = 0,
\]

\[
2x - \sqrt{2} y = 0.
\]

4. **Solve for \(x\) and \(y\):**

From the first equation:

\[
-\sqrt{2} x + 2y = 0 \implies y = \frac{\sqrt{2}}{2} x.
\]

From the second equation:

\[
2x - \sqrt{2} y = 0 \implies 2x = \sqrt{2} y.
\]

Substitute \(y = \frac{\sqrt{2}}{2} x\) into the second equation:

\[
2x = \sqrt{2} \left( \frac{\sqrt{2}}{2} x \right),
\]

\[
2x = x.
\]

Since \(x \neq 0\), the above simplifies to a tautology. Thus, any scalar multiple of \(\mathbf{v} = \begin{bmatrix} 1 \\ \frac{\sqrt{2}}{2} \end{bmatrix}\) is an eigenvector. To get a simpler form, we multiply by 2:

\[
\mathbf{v} = \begin{bmatrix} 2 \\ \sqrt{2} \end{bmatrix}.
\]

### Conclusion

The eigenvector corresponding to the eigenvalue \(\lambda = 1 + \sqrt{2}\) for the matrix 

\[
A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}
\]

is \(\mathbf{v} = \begin{bmatrix} 2 \\ \sqrt{2} \end{bmatrix}\).

Would you like further details on this topic or have any questions?

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### Related Questions:

1. What is an eigenvalue, and how is it different from an eigenvector?
2. How do you find the eigenvalues of a matrix?
3. What is the geometric interpretation of eigenvectors and eigenvalues?
4. Can there be multiple eigenvectors corresponding to the same eigenvalue?
5. How does the determinant of a matrix relate to its eigenvalues?

### Tip:

When solving for eigenvectors, always remember that eigenvectors are determined up to a scalar multiple, meaning any scalar multiple of an eigenvector is also an eigenvector for the same eigenvalue.

Learn how to find the eigenvector corresponding to the eigenvalue λ = 1 + √2 for the matrix A = [[1, 2], [2, 1]]. Detailed step-by-step solution involving linear algebra concepts, matrix operations, and eigenvalue equations.

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