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3 fixed typo

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.

Post Undeleted by Peter Shor
2 fixed matrix

Take $$L = \left( \begin{array} {rrrr} 6&-2&-1&-3\\ 7&-2&-2&-3\\ -2 & 2 4 & 0 & -1 2 \\ -1 2 & 0 & 2 &4 & -1 2 \\ -3 & -1 2 & -1 2 & 6 7 \end{array} \right).$$ Remove the last row and first column. The remaining matrix has two purely imaginary eigenvalues. Does this answer your question?

Post Deleted by Peter Shor
1