# Why coefficient of states for non-positive Hamiltonian matrix are all non-negative?

For a Hamiltonian $$H$$, if the all elements of matrix is non-positive under a set of basis $$\{|\phi\rangle\}$$:$$\langle\phi|H|\phi'\rangle\leq0$$ then the ground state of $$H$$ will be the linear combination of basis with non-negative coefficient:$$|G\rangle=\sum_{\{\phi\}}\chi_\phi|\phi\rangle\\\chi_\phi\geq0$$ I am confused this is why?

Well... you can take $$-\vert G\rangle$$ and then all your coefficients would be negative. The point is: the signs of the coefficients in your eigenvector are not related to the signs of the eigenvalues or the signs of the entries in the matrix. For instance, consider the matrix $$M(\lambda)=\lambda \left(\begin{array}{cc} 1&0\\ 0&1\end{array}\right)+ \left(\begin{array}{cc} 0&-1\\ -1&0\end{array}\right)$$ You can make $$M(\lambda)$$ non-positive by choosing any $$\lambda<0$$, but you can also make $$M(\lambda)$$ to be neither non-positive or non-negative by choosing $$\lambda>0$$. Whichever you choose,
the eigenvectors fact do not depend on $$\lambda$$ as $$\lambda$$ multiplies the unit matrix: the eigenvectors are always $$(\pm 1,1)$$. In fact you can use a generalization of this to add to any negative matrix an arbitrary multiple of the unit: this will shift the eigenvalues but will not affect the eigenvectors.
More generally, the results on non-negative matrices are due to Perron and Frobenius. The eigenvector $$\vert\Psi\rangle$$ for the largest eigenvalue can be chosen to have positive only components. Since $$-\vert\Psi\rangle$$ is also an eigenvector corresponding to this eigenvalue, the components of $$-\vert\Psi\rangle$$ would all be negative.