Around The 100-dimensional Unit Sphere

The desired maximum value of $S$ is half the largest eigenvalue of the symmetric $100 \times 100$ matrix $B[100]$ , where $B[100]_{ij} \; = \; \left\{ \begin{array}{lll} 1 & \qquad & |i-j| = 1\;, \\ 0 & \qquad & \mbox{otherwise}\;, \end{array} \right.$ for $1 \le i,j \le 100$ .

The vector $\mathbf{x}(\alpha)$ , where $\mathbf{x}(\alpha)_j \; = \; \sin j\alpha \;, \qquad \qquad 1 \le j \le 100 \;,$ satisfies the identity $B[100] \mathbf{x}(\alpha) \; = \; 2\cos\alpha \mathbf{x}(\alpha) + \left(\begin{array}{c} 0 \\ 0 \\ \vdots \\ 0 \\ -\sin101\alpha\end{array}\right) \;,$ and hence $\mathbf{x}(\alpha)$ is an eigenvector of $B[100]$ , with eigenvalue $2\cos\alpha$ , provided that $\mathbf{x}(\alpha) \neq \mathbf{0}$ and $\sin101\alpha = 0$ .

Thus the eigenvalues of $B[100]$ are $2\cos \tfrac{k\pi}{101}$ for $1 \le k \le 100$ , and so the largest eigenvalue of $B[100]$ is $2\cos\tfrac{\pi}{101}$ . The answer to the question is thus $\lfloor 10^5 \cos\tfrac{\pi}{101} \rfloor \,=\, \boxed{99951}$ .

If you prefer, we can show that the characteristic polynomial $\chi_{B[100]}(\lambda)$ of the matrix $B[100]$ is $U_{100}(\tfrac12\lambda)$ , where $U_{100}$ is a Chebyshev polynomial of the second kind. This is because, if $B[m]$ is the analogously-defined $m \times m$ matrix, then elementary determinant calculations give $\chi_{B[m+1]}(\lambda) + \chi_{B[m-1]}(\lambda) \; = \; \lambda \chi_{B[m]}(\lambda)$ for $m \ge 2$ , while $\chi_{B[1]}(\lambda) \; = \; \lambda \qquad \qquad \chi_{B[2]}(\lambda) \; =\; \lambda^2 - 1 \;.$ It is now clear that $\chi_{B[m]}(2\lambda) \,=\, U_m(\lambda)$ , as required. Since $U_m(\cos\theta) \; = \; \frac{\sin(m+1)\theta}{\sin\theta} \;,$ we deduce that the eigenvalues of $B[100]$ are $2\cos\tfrac{k\pi}{100}$ for $1 \le k \le 100$ , as before.