Question

I am trying to determine if a certain matrix can have purely imaginary eigenvalues. My question in its most general form is weather a complex matrix that is not skew-Hermitian and irreducible can contain eigenvalues on the imaginary axis.

My question, however, arises from a more particular instance. I am trying to determine if there can be eigenvalues on the imaginary axis of the matrix $(j\omega I + L)_{(kl)}$.

Here, $\omega$ is some real number, $A_{(kl)}$ denotes the sub-matrix of $A$ obtained by deleting the $k$-th row and $l$th column, and $L$ is the combinatorial Laplacian of a connected graph.

The Matrix-Tree theorem tells us that the determinant of any sub-matrix of $L$ is equal to the number of spanning trees in the graph. This, of course, implies that $L_{(k,l)}$ is invertible. I would like to know $(j\omega I + L)_{(kl)}$ inherits that property, i.e. it is invertible for any choice of $\omega \in R$.

If for example, $L_{(kl)}$ does contain a purely imaginary eigenvalue, then there exists an $\omega$ that makes the matrix singular.

Answer 1

Take $$ L = \left( \begin{array} {rrrr} 7&-2&-2&-3\\ -2 & 4 & 0 & -2 \\ -2 & 0 &4 & -2 \\ -3 & -2 & -2 & 7 \end{array} \right).$$ Remove the last row and first column. The remaining matrix has two purely imaginary eigenvalues. Does this answer your question?

UPDATE:

Can I point out that, even though this matrix has imaginary eigenvalues, there is no value of $\omega$ such that $(j\omega I + L)_{(k,\ell)}$ has determinant zero, where $j = \sqrt{-1}$. This is because the operations of adding the identity and removing row $k$ and column $\ell$ do not commute. You may want to rethink your question.