# find the eigenvalues of skew-symmetric matrix

What are the eigenvalues of the following $n\times n$ skew-symmetric matrix: $$M=A-A^T,$$ where $n>2$ can be odd or even, and $A=(a_{ij})_{n\times n}$ has $a_{1,n}=-1$ and $a_{ij}=1$ for $j=i+1, i=1,\dots, n-1$.

• What makes you expect a closed form exists? – Ian Mar 15 '17 at 3:29
• For $n=5$, the characteristic polynomial factors over $\mathbb{Q}$ as $x \cdot (x^{4} + 5 x^{2} + 5)$. I don't think you should expect anything nice. – darij grinberg Mar 15 '17 at 3:41
• @darijgrinberg That particular example is not particularly bad (the roots of $x^4+5x^2+5$ are just the two square roots of the two roots of $x^2+5x+5$). But I suspect it will eventually get very bad. – Ian Mar 15 '17 at 3:45

## 1 Answer

$$\{\zeta^{r}-\zeta^{-r}|r=0,\dots,n-1\}$$ where $\zeta$ is some primitive $n$-th root of unity.

Edited to provide more information.

To see this note that we are looking at the matrix $\Omega-\Omega^{-1}$, where $\Omega$ is the simple circulant that permutes the basis vectors in an $n$-cycle. The eigenvalues of $\Omega$ are well-known, being the powers of $\zeta$; the matrix formed by the putting the eigenvectors as columns is $\left(\zeta^{(i-1)(j-1)}\right)$.

• @John Your $M$ is an instance of circulant matrices, which have a general eigenvalue formula. – user1551 Mar 15 '17 at 8:57
• @user1551 4 A difference of a circulant and its inverse surely? – ancientmathematician Mar 15 '17 at 9:01
• Yes, but $M$ itself is also circulant: it has two circulant diagonals, one extending from the $(1,2)$-th entry (with coefficient $1$) and the other from the $(1,n)$-th entry (with coefficient $-1$). The set of all circulant matrices actually form an algebra. – user1551 Mar 15 '17 at 9:10
• @user1551. Sorry, I see "circulant" is wider in meaning than I had supposed. It doesn't affect the maths, though. It would be good to know where this problem arises: or is it just an exercise? – ancientmathematician Mar 15 '17 at 9:14
• @Ian $A+A^T$ is not circulant, but the entries of its eigenvectors still satisfy a 2nd order recurrence equation, so there should be a closed-form formula for the eigenvalues. – user1551 Mar 15 '17 at 10:47