Problem Setup.

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 \(\cos1,\cos2,\dots,\cos{n^2}\). For example, \[d_3 = \begin{vmatrix} \cos1 & \cos2 & \cos3 \\ \cos4 & \cos5 & \cos6 \\ \cos7 & \cos8 & \cos9 \end{vmatrix}\] Note that the argument of \(\cos\) is always in radians, not degrees. Evaluate \(\lim_{n\rightarrow\infty}d_n\).

Solution.

To begin, let us compute \(d_3\) explicitly: \[\begin{aligned} d_3 &= \begin{vmatrix} \cos1 & \cos2 & \cos3 \\ \cos4 & \cos5 & \cos6 \\ \cos7 & \cos8 & \cos9 \end{vmatrix} \\[5pt] &= \cos7 \begin{vmatrix} \cos2 & \cos3 \\ \cos5 & \cos6 \\ \end{vmatrix} - \cos8 \begin{vmatrix} \cos1 & \cos3 \\ \cos4 & \cos6 \\ \end{vmatrix} + \cos9 \begin{vmatrix} \cos1 & \cos2 \\ \cos4 & \cos5 \\ \end{vmatrix} \\[2.5pt] &= \cos7(\cos2\cos6-\cos3\cos5)-\cos8(\cos1\cos6-\cos3\cos4)+\cos9(\cos1\cos5-\cos2\cos4) \end{aligned}\] Note that each \(2 \times 2\) determinant is of the form \((\cos a\cos b-\cos c \cos d)\), where \(a + b = c + d\). This pattern works nicely with the trigonometric identity \(\cos\alpha\cos\beta=\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\): \[\begin{aligned} d_3 = \quad \cos7\Big(\frac{\cos8+\cos4}{2}-\frac{\cos8+\cos2}{2}\Big) \\[2.5pt] -\cos8\Big( \frac{\cos7+\cos5}{2} - \frac{\cos7+\cos1}{2}\Big) \\[2.5pt] + \cos9\Big( \frac{\cos6+\cos4}{2}-\frac{\cos6+\cos2}{2} \Big) \end{aligned}\] We see that the \(\cos(\alpha+\beta)\) terms cancel out, \[\begin{aligned} d_3 = \frac{\cos7}{2}(\cos4 - \cos2)-\frac{\cos8}{2}(\cos5 - \cos1)+\frac{\cos9}{2}(\cos4 - \cos2) \end{aligned}\] Regrouping, \[\begin{aligned} d_3 = \Big( \frac{\cos1\cos8}{2} - \frac{\cos2\cos7}{2} \Big) + \Big( \frac{\cos4\cos7}{2}-\frac{\cos2\cos9}{2} \Big) + \Big( \frac{\cos4\cos9}{2}-\frac{\cos5\cos8}{2} \Big) \end{aligned}\] Applying the cosine identity again, \[\begin{aligned} d_3 &= \frac{1}{2}\Big(\frac{\cos3-\cos7}{2}\Big) + \frac{1}{2}\Big(\frac{\cos5-\cos3}{2} \Big) + \frac{1}{2}\Big(\frac{\cos7-\cos5}{2}\Big) \\[5pt] &= 0 \end{aligned}\] Our guess, then, is \(\lim_{n\rightarrow\infty}d_n=0\). One way to show this for the general case of \(d_n\), is to show that two columns are linearly dependent. The general form of \(d_n\) is \[d_n = \begin{vmatrix} \cos1 & \cos2 & \cos3 & ... & \cos n\\ \cos(n+1) & \cos(n+2) & \cos(n+3) & ... & \cos 2n \\ \cos(2n+1) & \cos(2n+2) & \cos(2n+3) & ... & \cos3n \\ \cos(3n+1) & \cos(3n+2) & \cos(3n+3) & ... & \cos4n \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \cos\Big((n-1)n + 1\Big) & \cos\Big((n-1)n + 2\Big) & \cos\Big((n-1)n + 3\Big) & \dots & \cos n^2 \end{vmatrix}\] We can add column three to column one without changing the determinant, \[\begin{aligned} d_n = \begin{vmatrix} \cos1 + \cos3 & \cos2 & \cos3 & ... & \cos n\\ \cos(n+1) + \cos(n+3) & \cos(n+2) & \cos(n+3) & ... & \cos 2n \\ \cos(2n+1) + \cos(2n+3) & \cos(2n+2) & \cos(2n+3) & ... & \cos3n \\ \cos(3n+1) + \cos(3n+3) & \cos(3n+2) & \cos(3n+3) & ... & \cos4n \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \cos\Big((n-1)n + 1\Big) + \cos\Big((n-1)n + 3\Big) & \cos\Big((n-1)n + 2\Big) & \cos\Big((n-1)n + 3\Big) & \dots & \cos n^2 \end{vmatrix} \end{aligned}\] Using another trig identity, \(\cos\alpha+\cos\beta = 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\), to simplify column one: \[\begin{aligned} d_n = \begin{vmatrix} 2\cos2\cos1 & \cos2 & \cos3 & ... & \cos n\\ 2\cos(n + 2)\cos1 & \cos(n+2) & \cos(n+3) & ... & \cos 2n \\ 2\cos(2n+2)\cos1 & \cos(2n+2) & \cos(2n+3) & ... & \cos3n \\ 2\cos(3n+2)\cos1 & \cos(3n+2) & \cos(3n+3) & ... & \cos4n \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 2\cos\Big((n-1)n + 2\Big)\cos1 & \cos\Big((n-1)n + 2\Big) & \cos\Big((n-1)n + 3\Big) & \dots & \cos n^2 \end{vmatrix} \end{aligned}\] Finally, we can note that column one is a multiple of column two and therefore the two columns are linearly dependent; the determinant of a matrix with dependent columns is \(0\). Therefore, \(\lim_{n\rightarrow\infty}d_n = 0\).