# Is a matrix with non-negative entries which is positive definite over the reals alsopositive definite over the complex field?

in general, if a matrix is symmetric and positive definite over the real field, it is positive definite over the complex field. That if we drop the symmetric condition, this is no longer true is shown here https://en.wikipedia.org/wiki/Positive-definite_matrix#Consistency_between_real_and_complex_definitions .

Now, I was asking myself, if we instead of symmetry, assume the matrix having no negative entries, does it hold that it is positive definite over the real field?

## 1 Answer

This does not hold: Consider $M = \begin{pmatrix}1 & 3\\ 1 & 5\end{pmatrix}$. Then for every $x = (a,b)^T\in \mathbb R^n$ we have $$x^TMx = (a,b)\cdot \begin{pmatrix}a+3b\\ a+5b\end{pmatrix} = a^2 + 3ab + ab + 5b^2 = a^2 + 4ab + 5b^2 = (a+2b)^2 + b^2 \ge0.$$ But for $x = (1,i)^T\in \mathbb C^2$ we have $$x^*Mx = (1,-i)\cdot \begin{pmatrix}1 + 3i\\ 1 + 5i\end{pmatrix} = 1+3i -i + 5 = 6+2i,$$ which is not a real number.