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I am trying to produce an example of a (necessarily non-normal) matrix that has only eigenvalues with positive real part, but whose numerical range contains elements with strictly negative real part. Of course, I could take the standard counterexample $$ B=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} $$ and simply change its sign.

However, for some reason I need a matrix whose columns (but not rows!) sum up to 0 - indeed, I am looking for a Kirchhoff matrix of a graph, in the jargon of Tutte - and I am having some difficulty finding a matrix with this property. Is there any theorem that forbids such a behaviour? I am aware of results on the numerical range of doubly stochastic matrices, but nothing about simply stochastic matrices or matrices of the form I am considering.

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Let $$A=\pmatrix{1&-1/2\cr-1&1/2\cr}$$ The column sums are zero and the eigenvalues are zero and $3/2$. $$A+A^t=\pmatrix{2&-3/2\cr-3/2&1\cr}$$ which has determinant less than zero, so is not positive semidefinite. Wikipedia says the numerical range of $A$ is a closed subset of the right-half plane if and only if $A+A^t$ is positive semidefinite.

As it seems to be desired in adition that all entries be taken from $\{-1,0,1\}$, let $$A=\pmatrix{1&-1&-1\cr-1&1&0\cr0&0&1\cr}$$ All entries are appropriate, eigenvalues are $0,1,2$. $$A+A^t=\pmatrix{2&-2&-1\cr-2&2&0\cr-1&0&2\cr}$$ which has the negative eigenvalue $2-\sqrt5$.

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  • $\begingroup$ I have tried so far to modify your counterexample to get a matrix whose entries are only 0/1/-1, but so far without success. But your counterexample is already useful, thanks. $\endgroup$ – Delio Mugnolo Jul 30 '13 at 5:01
  • $\begingroup$ update: thanks for your new version. both your matrices are perfect. additionally, i was able to reproduce the same phenomenon with the matrix $A=\begin{pmatrix}2 & 0 & -1\\ -2 & 1 & 0\\ 0 & -1 & 1\end{pmatrix}$: eigenvalues $0,2\pm i$ but negative value of $(Ax|x)$ for $x=(4,6,5)$. this matrix does not have a 0,1,-1 pattern, but it has the advantage that it is the kirchhoff matrix of a simple graph. but it is just a matter of taste. $\endgroup$ – Delio Mugnolo Jul 30 '13 at 19:35

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