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.
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2 
- $\begingroup$ Have you tried computing the characteristic polynomial? $\endgroup$Rodrigo de Azevedo– Rodrigo de Azevedo2019-11-24 08:48:07 +00:00Commented Nov 24, 2019 at 8:48
- $\begingroup$ Hint: split the eigenvectors $\nu$ of $M$ into $[u,v]^T$ and write down the eigenvalue equations in terms of $u$ and $v$ $\endgroup$user619894– user6198942019-11-24 08:54:11 +00:00Commented Nov 24, 2019 at 8:54
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1 Answer
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1 Hint: Eigenvalues are $\pm \sqrt{-2+2\cos(\theta_j)}, \quad \theta_j=j\pi/(N+1), \quad j=1,\ldots,N$.
- $\begingroup$ 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. $\endgroup$user532956– user5329562019-11-26 11:32:29 +00:00Commented Nov 26, 2019 at 11:32