I'm attempting to solve problem B5 from Putnam 1999:
For an integer $n \geq 3$, let $\theta = \frac{2 \pi}{n}$. Evaluate the determinant of the $n \times n$ matrix $I + A$, where $I$ is the $n \times n$ identity matrix and $A = (a_{jk})$ has entries $a_{jk} = \cos(j \theta + k \theta)$ for all $j,k$.
I don't want to look at the official solution as I want to go through the problem myself. I calculated the determinants for $A_3$ through $A_6$, and they were $-1.25, -3, -5.25, -8$. The matrices are also all symmetrical, which is obvious from the definition of $A$. Here is $A_3$, for example:
\begin{bmatrix} -0.5 & 1 & -0.5\\ 1 & -0.5 & -0.5\\ -0.5 & -0.5 & 1 \end{bmatrix}
I found the pattern $\det(I_n+A_n)=1-\frac{n^2}{4}$, and as most mathematical competitions go, this is surely the solution, however, I now have to prove it. My teacher suggests using a proof by induction, however, I lack a general formula for the determinant of an $n \times n$ matrix.
In looking for one, I have come across the fact that the product of the eigenvalues of a matrix is equal to its determinant. I have also found out that symmetrical matrices have exactly $n$ eigenvalues and there exists a set of eigenvectors that are mutually orthogonal.
How do I calculate the eigenvalues and eigenvectors of an $n \times n$ matrix in this form?
To remove this question from the "unanswered" tab, here is a solution.
Let $v=(e^{i\theta},e^{2i\theta},\ldots,e^{in\theta})^\top$. Then $v^\top v=0$ and $A=\frac12\pmatrix{v&\bar{v}}\pmatrix{v^\top\\ \bar{v}^\top}$. Using the fact that $XY$ and $YX$ have identical nonzero eigenvalues, we see that the only two nonzero eigenvalues of $A$ are those of $\frac12\pmatrix{v^\top\\ \bar{v}^\top}\pmatrix{v&\bar{v}}=\frac12\pmatrix{0&n\\ n&0}$, i.e. $\pm\frac n2$. Hence $\det(I+A)=(1-\frac n2)(1+\frac n2)=1-\frac{n^2}4$.