1
$\begingroup$

Suppose you are given this $2 \times 2$ matrix of trig functions:

\begin{vmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{vmatrix}

The zeros of which give the identity matrix:

\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}

Noticing the original matrix is an example of a wronskian, extending it to the $3 \times 3$ case:

\begin{vmatrix} \cos\theta & \sin\theta & -\cos\theta \\ -\sin\theta & \cos\theta & \sin\theta \\ -\cos\theta & -\sin\theta & \cos\theta \end{vmatrix}

Evaluating at zero yields:

\begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}

which is symmetric about the diagonal, leading me to believe that all the odd powers have determinants equal to zero. In the next even case e.g. $4 \times 4$:

\begin{vmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1 \end{vmatrix}

which is also symmetric about the diagonal, leading me to believe the even powers greater than two equal zero.

Originally when I attempted the case where n equals five, I wrote the determinant as the sum of products in terms of trig functions without evaluating at zero. I noticed that they grow as a function of n at the same rate as the elements of the symmetric group.

So, I have two questions:

  1. was I right? is the determinant equal to 0 for n greater than 2 based on the symmetry of the matrix about the diagonal?

  2. since there is an isomorphism between the sum of products representation of the determinant and $S_n$, is there a group theoretic proof to be had?

  • 5
    What does «extending it to the 3×3 case» mean? Also, when you say «taking the zeroes» you probably mean «evaluating at zero».2011-12-23
  • 0
    How are you extending the matrices to higher $n$? I don't follow.2011-12-23
  • 0
    You iterate down the column by taking the derivative of the previous element. You iterate across the row by taking the anti-derivative of the previous element. Is that clearer?2011-12-23
  • 0
    Which antiderivative? A function has many...2011-12-23
  • 0
    Note that in your $3\times 3$ case, the third row's the negative of the first; in the $4\times 4$ case, the first and third rows are negatives of each other, as are the second and fourth ones. *Of course* your determinants are then zero...2011-12-23
  • 0
    for the purposes of this question, the elements of the matrix are cosine, whose derivative is -sine, whose derivative again is -cosine, whose derivative is sine, whose derivative is cosine again. This forms a ring. The "anti-derivative" is equivalent to incrementing to the next element on this ring, but in the opposite direction of the derivative. Now, is THAT clearer?2011-12-23

1 Answers 1