Putnam 2009, Problem A-3

Let $d_n$ be the determinant of the $n \times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos 1, \cos 2, \ldots, \cos n^2$. (For example,
\[
d_3 = \begin{vmatrix}
\cos 1 & \cos 2 & \cos 3 \\
\cos 4 & \cos 5 & \cos 6 \\
\cos 7 & \cos 8 & \cos 9
\end{vmatrix}.
\]
The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n \to \infty} d_n$.

Solution:
The limit is \( 0 \); we will show this by checking that \( d_n = 0 \) for all \( n \geq 3 \). Starting from the given matrix, add the third column to the first column; this does not change the determinant. However, thanks to the identity \( \cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2} \), the resulting matrix has the form
\[
\begin{pmatrix}
2\cos 2\cos 1 & \cos 2 & \cdots \\
2\cos(n+2)\cos 1 & \cos(n+2) & \cdots \\
2\cos(2n+2)\cos 1 & 2\cos(2n+2) & \cdots \\
\vdots & \vdots & \ddots
\end{pmatrix}
\]
with the first column being a multiple of the second. Hence \( d_n = 0 \).

**Remark.** Another way to draw the same conclusion is to observe that the given matrix is the sum of the two rank 1 matrices \( A_{jk} = \cos(j-1)n\cos k \) and \( B_{jk} = -\sin(j-1)n\sin k \), and so has rank at most 2. One can also use the matrices \( A_{jk} = e^{i(j-1)n+k} \), \( B_{jk} = e^{-i(j-1)n+k} \).