# Prove the determinants of these related matrices are zero.

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?

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What does «extending it to the 3×3 case» mean? Also, when you say «taking the zeroes» you probably mean «evaluating at zero». –  Mariano Suárez-Alvarez Dec 23 '11 at 22:38
How are you extending the matrices to higher $n$? I don't follow. –  anon Dec 23 '11 at 23:08
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? –  user4536 Dec 23 '11 at 23:11
Which antiderivative? A function has many... –  Ｊ. Ｍ. Dec 23 '11 at 23:14
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... –  Ｊ. Ｍ. Dec 23 '11 at 23:18

I should also point out that symmetric matrices need not have determinant zero; you may be misremembering the fact that $\textit{anti}$-symmetric $n \times n$ matrices with $n$ odd (i.e. those $A$ with $A^t = -A$) have zero determinant.
Excelent. So the determinant is zero for any n greater than two. Two questions: would this result extend to the general case e.g. the matrices of trig functions? And, can this proof be expressed in terms of $S_n$? –  user4536 Dec 23 '11 at 23:42
With regards to your first question - yes! The third row of your general matrix is always the negative of the first. And as far as your second goes, I'm not quite sure what you mean when you ask if it could be "expressed in terms of $S_n$". Do you mean whether or not you could prove it by using the formula for the determinant as a sum over permutations? I would think that the answer to that would be "yes, but why on earth would you want to?". The permutation formula is nice insofar as it allows us to actually compute determinants, but (at least in my opinion) it's a real mess. (continued...) –  NKS Dec 23 '11 at 23:53