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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.

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The matrix $\left(\begin{matrix i & i \\ 0 & i\end{matrix}\right)$ hs $i$ as only eigenvalue, is not skew Hermitian (not even diagonalizable). So...? –  Stefan Waldmann Mar 21 '11 at 15:54
    
Yes, that provides a simple example. I've modified my question to further require that the matrix is irreducible, which I neglected to mention originally. –  dan Mar 21 '11 at 16:48
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If you multiply by $i$, then the question is: is a complex matrix having only real eigenvalues necessarily Hermitian. The answer is "no". For example $\left(\begin{array} {cc} 5&3\\2&4 \end{array}\right)$. –  Peter Shor Mar 21 '11 at 18:58
    
To prove the theorem you want (if it's true), I suspect you would have to use the fact that $L$ has all positive entries. –  Peter Shor Mar 21 '11 at 19:00
    
Is the $j$ in $j\omega I$ the imaginary unit or the column index as in $L_{(i,j)}$? –  Federico Poloni Mar 21 '11 at 21:01
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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.

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This is a counter example to the general question, thank-you. However, for the question related to the Laplacian, I am only considering unweighted graphs. –  dan Mar 22 '11 at 6:38
    
In fact, it's a counterexample to the Laplacian question for weighted graphs, which makes me suspect that there is probably a counterexample for unweighted graphs. –  Peter Shor Mar 22 '11 at 11:18
    
That would be very interesting. May I ask how you constructed the example for weighted graphs? As I mentioned, I have yet to find an example for unweighted graphs. –  dan Mar 22 '11 at 12:05
    
@dan: I stuck a lot of parameters in (with the symmetry you see) and did a small-scale computer search. I don't know how well that would work for unweighted graphs, especially if you've been testing them. –  Peter Shor Mar 22 '11 at 14:57
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