Trig and Matrices

Geometry Level 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 $\cos 1, \cos 2, \dots, \cos n^2$ . (For example, $d_3 = \left| \begin{matrix} \cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{matrix} \right|.$ The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n\to\infty} d_n$ .

The answer is 0.

1 solution

Otto Bretscher
Dec 14, 2018

It turns out that $d_n=0$ for $n\geq 3$ , so that the limit is $\boxed{0}$ as well.

The reason, in a nutshell, is that the vectors $(\cos 1, \cos 2,..., \cos n)$ and $(\sin 1, \sin 2,..., \sin n)$ span the row space of $A_n$ , so that $rank A_n\leq 2$ and $A_n$ fails to be invertible for $n\geq 3$ .

We will illustrate this observation in the case of the second row of the given matrix $A_3$ ; it is a simple application of the addition theorem for the cosine:

$(\cos 4, \cos5, \cos6)= \left(\cos(3+1),\cos(3+2),\cos(3+3)\right)= \cos 3(\cos 1, \cos2, \cos3)- \sin 3(\sin 1, \sin 2,\sin3)$

PS: This works out the same way if were to measure angles in degrees, of course.

Nice solution, thank you for sharing it.

Hana Wehbi - 2 years, 5 months ago

Trig and Matrices

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