Are these square matrices always diagonalisable?

Question

When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:

Let $A_n\in M_n(\mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?

For example, $$A_3=\begin{bmatrix}0&1&0\\-1&0&1\\0&-1&0\end{bmatrix},\quad A_5=\begin{bmatrix}0&1&0&0&0\\-1&0&1&0&0\\0&-1&0&1&0\\0&0&-1&0&1\\0&0&0&-1&0\end{bmatrix}.$$

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Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $\chi_n(t)\in\mathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that $$\chi_n(t)=-t\chi_{n-1}(t)+\chi_{n-2}(t)\tag{1}$$ for all $n\geq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even, $$\det(A_{2n}-\lambda I)=\det(A_{2n}^t-\lambda I)=\det(-A_{2n}-\lambda I)=\det(A_{2n}+\lambda I).$$ This implies that whenever $\lambda$ is an eigenvalue of $A_{2n}$, so is $-\lambda$. In other words, $\chi_{2n}(t)$ is always of the form $(t^2-\lambda _1^2)(t^2-\lambda_2^2)\dotsm(t^2-\lambda_n^2)$ for some $\lambda_i$.

And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that $$\chi_{2n}'(t)=2t\sum_{k=1}^n\frac{\chi_{2n}(t)}{t^2-\lambda_k^2}$$ never shares a common zero with $\chi_{2n}$ (which would resolve the even case), though I don't see how to make this work.

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Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $S\subseteq\mathbb Z$ for which the conclusion is proven for $n\in S$, so any such approach is welcome too.

Thank you in advance!

Answer 1

All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $\mathbb C$, by the spectral theorem.

Answer 2

The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $\delta = 0$ and off-diagonal entries $\tau = 1$ and $\sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$\lambda_k = 2i\cos\left(\dfrac{k\pi}{n+1}\right),$$ for $k = 1,\ldots,n$, and the corresponding eigenvectors $v_1,\ldots,v_n$ have entries $$v_k[m] = i^m\sin\left(\dfrac{mk\pi}{n+1}\right).$$

Answer 3

Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.