A matrix

Suppose that $N \ge 3$ and consider the $N \times N$ matrix $A$ with components $a_{jk} \; = \; \cos\left(\tfrac{2\pi(j+k)}{N}\right) \hspace{2cm} 1 \le j,k \le N$ If we define $X_r \; = \; \cos\left(\tfrac{2\pi(r+2)}{N} \right) \hspace{2cm} r \in \mathbb{Z}$ then $X_{r+N} = X_r$ for all $r$ , and we have $a_{jk} \; = \; X_{j+k-2} \hspace{2cm} 1 \le j,k \le N$ Let us define $\zeta = e^{\frac{2\pi i}{N}}$ and define the vector $\mathbf{v}(k) \in \mathbb{C}^N$ by $\mathbf{v}(k)_j \; = \; \zeta^{k(j-1)} \hspace{2cm} 1 \le j \le N$ It is well-known that $\big\{\mathbf{v}(k) \,:\, 1 \le k \le N\big\}$ is an orthogonal basis for $\mathbf{C}^N$ . If we define $\Lambda_j \; = \; \sum_{k=1}^N X_k \zeta^{kj} \hspace{2cm} 1 \le j \le N$ then we can show that $A\mathbf{v}(k) \; = \; \left\{ \begin{array}{lcl} \Lambda_k\mathbf{v}(N-k) & \hspace{1cm} & 1 \le k \le N-1 \\ \Lambda_N\mathbf{v}(N) & & k = N \end{array} \right.$ Thus $\mathbf{C}^N$ can be split into $A$ -invariant subspaces:

• one-dimensional subspaces spanned by $\mathbf{v}(N)$ and (if $N$ is even) $\mathbf{v}(\tfrac12N)$ ,
• two-dimensional subspaces spanned by $\{\mathbf{v}(k),\mathbf{v}(N-k)\}$ for any $1 \le k < \tfrac12N$ . Since the $X_k$ are real, we see that $\Lambda_{N-k}$ is the complex conjugate of $\Lambda_k$ , and so either $\Lambda_k$ and $\Lambda_{N-k}$ are both nonzero or both zero, and so this vector space has a basis of eigenvectors if $A$ ,

and hence there exists a basis for eigenvectors for $A$ , and hence for $I+A$ . Thus the eigenvalues of $I+A$ are

• $1 + \Lambda_N$ and (if $N$ is even) $1 + \Lambda_{\frac12N}$ ,

• $1 + \sqrt{\Lambda_k\Lambda_{N-k}}$ and $1 - \sqrt{\Lambda_k\Lambda_{N-k}}$ for all $1 \le k < \tfrac12N$ .

Thus we deduce that $\mathrm{det}\,(I+A) \; = \; \left\{ \begin{array}{lcl} \displaystyle(1 + \Lambda_N)\prod_{1 \le k < \frac12N}\big(1 - \Lambda_k\Lambda_{N-k}\big) & \hspace{1.5cm} & N\; \mathrm{odd} \\ \displaystyle (1 + \Lambda_N)(1 + \Lambda_{\frac12N})\prod_{1 \le k < \frac12N}\big(1 - \Lambda_k\Lambda_{N-k}\big) & & N\;\mathrm{even} \end{array}\right.$ We now calculate that $\begin{aligned} \Lambda_k & = \; \sum_{j=1}^N X_j \zeta^{kj} \; = \; \tfrac12\sum_{j=1}^N \big(\zeta^{j+2} + \zeta^{-j-2}\big)\zeta^{kj} \; = \; \tfrac12\zeta^2\sum_{j=1}^N \zeta^{(k+1)j} + \tfrac12\zeta^{-2}\sum_{j=1}^N \zeta^{(k-1)j} \\ & = \; \left\{ \begin{array}{lcl} \tfrac12N\zeta^{-2} & \hspace{1cm} & k = 1 \\ \tfrac12N\zeta^2 & & k = N-1 \\ 0 & & \mathrm{o.w.} \end{array}\right. \end{aligned}$ from which we deduce that $\mathrm{det}(I+A) \; = \; 1 - \tfrac14N^2 \hspace{2cm} N \ge 3$ irrespective of whether $N$ is even or odd. With $N = 1996$ we obtain the answer is $\boxed{-996003}$ .