# Eigenvalues for a block matrix with Toeplitz tridiagonal sub-matrix

Given a matrix $$M \in \mathbb{R}^{(2N, 2N)}$$ for some $$N \in \mathbb{Z}, N > 2$$

$$M = \begin{pmatrix}\textbf{0}&I\\A&\textbf{0}\end{pmatrix},$$

where $$\textbf{0} \in \mathbb{R}^{(N, N)}$$ is a zeroed matrix, $$I \in \mathbb{R}^{(N, N)}$$ is the identity matrix and $$A \in \mathbb{R}^{(N, N)}$$ is a negative-definite Toeplitz Tridiagonal matrix with entries: $$-2$$ on the diagonal and $$1$$ on the subdiagonal and superdiagonal, i.e.

$$A = \begin{pmatrix} &-2 &1 &0 &\ldots &\ldots &\ldots &0 \\ &1 &-2 &1 &\ddots &\ddots &\ddots &\vdots \\ &0 &1 &-2 &1 &\ddots &\ddots &\vdots \\ &\vdots &\ddots &\ddots &\ddots &\ddots &\ddots &\vdots \\ &\vdots &\ddots &\ddots &1 &-2 &1 &0 \\ &\vdots &\ddots &\ddots &\ddots &1 &-2 &1 \\ &0 & \ldots &\ldots &\ldots &0 &1 &-2 \end{pmatrix}$$

Show that all the eigenvalues of $$M$$ are imaginary.

• Have you tried computing the characteristic polynomial? Commented Nov 24, 2019 at 8:48
• Hint: split the eigenvectors $\nu$ of $M$ into $[u,v]^T$ and write down the eigenvalue equations in terms of $u$ and $v$ Commented Nov 24, 2019 at 8:54

Hint: Eigenvalues are $$\pm \sqrt{-2+2\cos(\theta_j)}, \quad \theta_j=j\pi/(N+1), \quad j=1,\ldots,N$$.
• Thanks, I've solved it using the fact that the block matrices commute, which gives $\det(M - \lambda I) = \det(\lambda^2 I - A I)$ and the fact that this is the determinant of a Toeplitz Tridiagonal matrix to solve this question.