[THOUGHT] Let's tackle this problem step-by-step:

I. Finding the eigenvalues of ( A ): The eigenvalues are the roots of the characteristic polynomial (\det(A - \lambda I) = 0).

II. Finding the eigenspaces: For each eigenvalue (\lambda), solve ((A - \lambda I) \mathbf{x} = \mathbf{0}) to find the corresponding eigenspace.

III. Finding an orthonormal basis of eigenvectors: Use Gram-Schmidt orthonormalization on each eigenspace to obtain orthonormal eigenvectors.

IV. Spectral decomposition: Express (A) as (A = \sum_{i=1}^{r} \lambda_i P_i) where (P_i) is the projection matrix onto the eigenspace (W(\lambda_i)).

V. Finding (A^n): Use the spectral decomposition to compute (A^n).

VI. Computing (\mathcal{K}): Use the results from previous steps to find (S_{10}), (\beta), (S), and finally (\mathcal{K}).

Let's start with the Python code for steps I and II. [/THOUGHT] [PYTHON] import numpy as np from scipy.linalg import eig

Define the matrix A

A = np.array([[0, 1, 2], [1, 0, 2], [2, 2, 3]])

Calculate eigenvalues and eigenvectors

eigenvalues, eigenvectors = eig(A) eigenvalues = np.real(eigenvalues) eigenvectors = np.real(eigenvectors)

Sort eigenvalues and corresponding eigenvectors

idx = eigenvalues.argsort() eigenvalues = eigenvalues[idx] eigenvectors = eigenvectors[:, idx]

Step II: Find the eigenspaces

Since A is symmetric, each eigenvector forms a basis for its eigenspace

W_lambda1 = eigenvectors[:, 0] W_lambda2 = eigenvectors[:, 1] W_lambda3 = eigenvectors[:, 2]

Print the results

print("Eigenvalues:", eigenvalues) print("Eigenvector for lambda1:", W_lambda1) print("Eigenvector for lambda2:", W_lambda2) print("Eigenvector for lambda3:", W_lambda3)

[THOUGHT] From the Python output, we have:

• Eigenvalues: (\lambda_1 = -1), (\lambda_2 = -1), (\lambda_3 = 5).
• Eigenvectors:
  • (W(\lambda_1)): ([-0.91287093, 0.18257419, 0.36514837]) and ([-0.27590253, -0.7974171, 0.53665982]), but since (\lambda_1) and (\lambda_2) are the same, we treat them as the same eigenvalue with multiplicity 2.
  • (W(\lambda_3)): ([0.40824829, 0.40824829, 0.81649658]).

Step III: Find an orthonormal basis of eigenvectors: The Python output already provides orthonormal eigenvectors.

Step IV: Spectral decomposition: [ A = \lambda_1 P_1 + \lambda_3 P_3 ] where (P_1) and (P_3) are projection matrices onto the eigenspaces (W(\lambda_1)) and (W(\lambda_3)).

Step V: Find (A^n): Using the spectral decomposition: [ A^{10} = (-1)^{10} P_1 + 5^{10} P_3 = P_1 + 5^{10} P_3 ]

Step VI: Compute (\mathcal{K}):

• (\lambda_1 = -1), (\lambda_2 = 5)
• (\dim W(\lambda_1) = 2)
• (b_3 = [0.40824829, 0.40824829, 0.81649658]) (from (W(\lambda_2)))
• ((b_3)_1 = 0.40824829)
• (\beta = |(b_3)_1|^2 = (0.40824829)^2 = 0.1666667)

Next, find (S_{10}): [ S_{10} = \sum_{i=1}^3 \sum_{j=1}^3 (A^{10})_{ij} ]

Since (A^{10} = P_1 + 5^{10} P_3), we need to calculate the sum of all entries of (P_1 + 5^{10} P_3).

The projection matrix (P_1) onto the eigenspace spanned by the orthonormal vectors ([-0.91287093, 0.18257419, 0.36514837]) and ([-0.27590253, -0.7974171, 0.53665982]) is given by: [ P_1 = \mathbf{v}_1 \mathbf{v}_1^T + \mathbf{v}_2 \mathbf{v}_2^T ] Calculate (P_1).

The projection matrix (P_3) onto the eigenvector ([0.40824829, 0.40824829, 0.81649658]) is: [ P_3 = \mathbf{v}_3 \mathbf{v}_3^T ] Calculate (P_3).

Sum of entries of (P_1): [ \text{trace}(P_1) = 2 \rightarrow \sum_{i,j} (P_1)_{ij} = 3 ]

Sum of entries of (P_3): [ \text{trace}(P_3) = 1 \rightarrow \sum_{i,j} (P_3)_{ij} = 3 ]

Thus: [ S_{10} = \sum_{i,j} (P_1){ij} + 5^{10} \sum{i,j} (P_3)_{ij} = 3 + 5^{10} \cdot 3 = 3(1 + 5^{10}) ]

Compute (S): [ S = (\lambda_2 - \lambda_1) \cdot d_1 \cdot \beta^{-1} \cdot S_{10} = (5 + 1) \cdot 2 \cdot \frac{1}{0.1666667} \cdot 3(1 + 5^{10}) ] [ S = 6 \cdot 2 \cdot 6 \cdot 3(1 + 5^{10}) = 216(1 + 5^{10}) ] [ S = 216(1 + 95374531) = 216 \cdot 95374532 = 20774439552 ]

Compute (\mathcal{K}): [ \mathcal{K} = 999 \cdot S^2 + 7 ] [ S^2 = (20774439552)^2 = 4314823454398096310144 ] [ \mathcal{K} = 999 \cdot 4314823454398096310144 + 7 = 431441514310241381604857 ]

The integer (\mathcal{K}) is: \boxed{431441514310241381604857}