Can a complex non-skew Hermitian matrix have purely imaginary eigenvalues? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:59:16Z http://mathoverflow.net/feeds/question/59069 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59069/can-a-complex-non-skew-hermitian-matrix-have-purely-imaginary-eigenvalues Can a complex non-skew Hermitian matrix have purely imaginary eigenvalues? dan 2011-03-21T15:30:48Z 2011-03-29T13:12:26Z <p>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.</p> <p>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 $(j\omega I + L)_{(kl)}$. </p> <p>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.</p> <p>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 $\omega \in R$.</p> <p>If for example, $L_{(kl)}$ does contain a purely imaginary eigenvalue, then there exists an $\omega$ that makes the matrix singular.</p> http://mathoverflow.net/questions/59069/can-a-complex-non-skew-hermitian-matrix-have-purely-imaginary-eigenvalues/59110#59110 Answer by Peter Shor for Can a complex non-skew Hermitian matrix have purely imaginary eigenvalues? Peter Shor 2011-03-21T21:46:58Z 2011-03-29T13:12:26Z <p>Take $$L = \left( \begin{array} {rrrr} 7&amp;-2&amp;-2&amp;-3\\ -2 &amp; 4 &amp; 0 &amp; -2 \\ -2 &amp; 0 &amp;4 &amp; -2 \\ -3 &amp; -2 &amp; -2 &amp; 7 \end{array} \right).$$ Remove the last row and first column. The remaining matrix has two purely imaginary eigenvalues. Does this answer your question?</p> <p>UPDATE:</p> <p>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.</p>