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.



numerical range of a column-zero-sum matrix
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](http://en.wikipedia.org/wiki/Numerical_range) 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 $2-\sqrt5$. 


