Problem Analysis #7

[THOUGHT] First computation: define A and compute eigenvects to get eigenvalues and eigenvectors. [/THOUGHT] [PYTHON] from sympy import * A = Matrix([[0,1,2],[1,0,2],[2,2,3]]) ev = A.eigenvects() pprint(ev) [/PYTHON]

[THOUGHT] The eigenvalues are $\lambda_1 = -1$ and $\lambda_2 = 5$. The eigenspace $W(\lambda_1)$ has dimension $d_1 = 2$ and is spanned by $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$. The eigenspace $W(\lambda_2)$ is spanned by $\begin{pmatrix} 1/2 \\ 1/2 \\ 1 \end{pmatrix}$. An orthonormal eigenbasis can be obtained by Gram-Schmidt on the first eigenspace and normalizing the second eigenvector. The normalized $b_3 = \frac{1}{\sqrt{3/2}} \begin{pmatrix} 1/2 \\ 1/2 \\ 1 \end{pmatrix}$, so $\beta = |(b_3)_1|^2 = 1/6$. Then $S_{10} = 26041667$, $S = 1875000024$, and $\mathcal{K} = 3512109375089910575431$. [/THOUGHT] $\boxed{3512109375089910575431}$

Explanation

The provided AI-generated answer derives from the spectral decomposition of the symmetric matrix $A$ and targeted computations for part VI. The process begins with the eigenvalues and eigenvectors from the SymPy output: $\lambda_1 = -1$ (algebraic multiplicity 2, geometric multiplicity 2, spanned by $[-1, 1, 0]^T$ and $[-2, 0, 1]^T$) and $\lambda_2 = 5$ (multiplicity 1, spanned by $[1/2, 1/2, 1]^T$).

For part VI, compute $\beta = |(b_3)_1|^2$, where $b_3$ is the unit eigenvector for $\lambda_2$. The vector $u_3 = [1/2, 1/2, 1]^T$ has norm $\|u_3\| = \sqrt{(1/2)^2 + (1/2)^2 + 1^2} = \sqrt{3/2}$. Thus, $b_3 = u_3 / \sqrt{3/2}$, and the first component is $(1/2) / \sqrt{3/2} = \sqrt{2/12} = 1/\sqrt{6}$, so $\beta = (1/\sqrt{6})^2 = 1/6$.

Next, $S_{10} = \sum_{i,j} (A^{10})_{ij} = \mathbf{1}^T A^{10} \mathbf{1}$, where $\mathbf{1} = [1, 1, 1]^T$. Using spectral decomposition, $A^{10} = (-1)^{10} P_1 + 5^{10} P_2 = P_1 + 5^{10} P_2$, so $S_{10} = \mathbf{1}^T P_1 \mathbf{1} + 5^{10} \mathbf{1}^T P_2 \mathbf{1}$. Here, $\mathbf{1}^T P_2 \mathbf{1} = (\mathbf{1}^T b_3)^2$. Compute $\mathbf{1}^T u_3 = 2$, so $\mathbf{1}^T b_3 = 2 / \sqrt{3/2} = 2 \sqrt{2/3}$, and $\mathbf{1}^T P_2 \mathbf{1} = [2 \sqrt{2/3}]^2 = 8/3$. Then $\mathbf{1}^T P_1 \mathbf{1} = \mathbf{1}^T \mathbf{1} - 8/3 = 3 - 8/3 = 1/3$. With $5^{10} = 9,765,625$, $S_{10} = 1/3 + (8/3) \cdot 9,765,625 = (1 + 78,125,000)/3 = 78,125,001 / 3 = 26,041,667$.

Now, $S = (\lambda_2 - \lambda_1) d_1 \beta^{-1} S_{10} = (5 - (-1)) \cdot 2 \cdot 6 \cdot 26,041,667 = 6 \cdot 2 \cdot 6 \cdot 26,041,667 = 72 \cdot 26,041,667 = 1,875,000,024$.

Finally, $\mathcal{K} = 999 S^2 + 7$. Compute $S^2 = (1,875,000,024)^2 = 3,515,625,000,090,000,576$. Then $999 S^2 = 1,000 S^2 - S^2 = 3,515,625,000,090,000,576,000 - 3,515,625,000,090,000,576 = 3,515,625,000,090,000,576,000 - 3,515,625,000,090,000,576$ (via digit-wise subtraction with borrows), yielding $3,512,109,375,089,910,575,424$. Adding 7 gives $\mathcal{K} = 3,512,109,375,089,910,575,431$.