[THOUGHT] I will define the matrix $A$ using SymPy to ensure exact arithmetic. Then I will calculate the eigenvalues and eigenvectors. I will identify $\lambda_1$ and $\lambda_2$ such that $\lambda_1 < \lambda_2$. I will determine the dimension $d_1$ of the eigenspace $W(\lambda_1)$. I will find the eigenvector $b_3$ associated with $\lambda_2$, normalize it, and find its first component to compute $\beta$. I will compute $A^{10}$ and the sum of its entries $S_{10}$. Finally, I will compute $S$ and $\mathcal{K}$.

[/THOUGHT] [PYTHON] import sympy as sp

Define the matrix A

A = sp.Matrix([ [0, 1, 2], [1, 0, 2], [2, 2, 3] ])

1. Find eigenvalues and eigenvectors

eigen_info = A.eigenvects()

eigen_info is a list of tuples: (eigenvalue, multiplicity, [eigenvectors])

print("Eigenvalues and Eigenvectors:") for val, mult, vecs in eigen_info: print(f"Lambda: {val}, Multiplicity: {mult}, Vectors: {vecs}")

Extract eigenvalues

eigenvalues = sorted([val for val, _, _ in eigen_info]) print(f"Sorted unique eigenvalues: {eigenvalues}")

Assign lambda_1 and lambda_2

The problem implies there are distinct eigenvalues lambda_1 < ... < lambda_r

lambda_1 = eigenvalues[0] lambda_2 = eigenvalues[1] # Assuming there are at least 2. If there are 3, I need to check the problem statement carefully.

The problem mentions lambda_1 < lambda_2 and b_3 in W(lambda_2).

If there are 3 distinct eigenvalues, the notation lambda_1 < lambda_2 usually refers to the sorted list.

However, the problem later says "assume b_3 in W(lambda_2)". If there were 3, b_3 would be the 3rd basis vector.

Let's check the output first.

2. Calculate d_1

d_1 is dim(W(lambda_1))

d_1 = [mult for val, mult, _ in eigen_info if val == lambda_1][0] print(f"d_1 (dim W(lambda_1)): {d_1}")

3. Find b_3 and beta

b_3 is in W(lambda_2).

We need an orthonormal basis.

Find eigenvectors for lambda_2

vecs_lambda2 = [vec for val, _, vecs in eigen_info if val == lambda_2 for vec in vecs]

If dim(W(lambda_2)) == 1, b_3 is just the normalized eigenvector.

If dim > 1, there's ambiguity, but let's see the multiplicity first.

v3 = vecs_lambda2[0] norm_v3 = v3.norm() b3 = v3 / norm_v3 print(f"b_3 (unsimplified): {b3}")

beta = |(b_3)_1|^2

b3_1 = b3[0] beta = sp.Abs(b3_1)**2 print(f"beta: {beta}")

4. Calculate A^10 and S_10

A_10 = A**10 S_10 = sum(A_10) print(f"S_10: {S_10}")

5. Calculate S and K

S = (lambda_2 - lambda_1) * d_1 * beta^(-1) * S_10

K = 999 * S^2 + 7

S = (lambda_2 - lambda_1) * d_1 * (1/beta) * S_10 print(f"S (symbolic): {S}") K = 999 * S**2 + 7 print(f"K (symbolic): {K}") print(f"K (numeric): {float(K)}") [/PYTHON]

The eigenvalues of $A$ are $\lambda_1 = -1$ (multiplicity 2) and $\lambda_2 = 5$ (multiplicity 1). The dimension of the eigenspace $W(\lambda_1)$ is $d_1 = 2$. The eigenvector corresponding to $\lambda_2 = 5$ is $v_3 = (1, 1, 2)^T$. Normalizing this vector gives $b_3 = \frac{1}{\sqrt{6}}(1, 1, 2)^T$. The first component of $b_3$ is $(b_3)_1 = \frac{1}{\sqrt{6}}$, so $\beta = |(b_3)_1|^2 = \frac{1}{6}$. The sum of entries of $A^{10}$ is $S_{10} = 26041667$. The value $S$ is calculated as: [ S = (\lambda_2 - \lambda_1) d_1 \beta^{-1} S_{10} = (5 - (-1)) \cdot 2 \cdot 6 \cdot 26041667 = 6 \cdot 12 \cdot 26041667 = 72 \cdot 26041667 = 1875000024. ] Finally, $\mathcal{K} = 999 S^2 + 7$. Using Python for the large integer arithmetic: [ \mathcal{K} = 999 \times (1875000024)^2 + 7 = 3512109464910000575431. ]

\boxed{3512109464910000575431}