Reordering the rows of a matrix and positive definiteness I have two questions:
1) Does reordering the rows of a matrix affect its positive definiteness property? 
2) If all elements of a non-symmetric matrix are positive, does this automatically imply positive definiteness?
Thank you.  
 A: The answers to your questions are 1) Yes and 2) No.
A matrix $A \in \mathbb{R}^{m \times m} $ is positive definite if the eigenvalues of the symmetric matrix $\frac{1}{2}(A+A^T)$ are strictly positive. If $A$ is symmetric and positive definite we say that $A$ is symmetric positive definite.
Consider the matrix $$A = \begin{bmatrix} 2 & 1 \\ 1 & 8 \end{bmatrix}.$$ We have $A = \frac{1}{2}(A+A^T)$ because $A$ is symmetric. The eigenvalues are real. The product of the eigenvalues equals the determinant which is $15$. The sum of the eigenvalues equals the trace which is $10$. The eigenvalues are strictly positive. This matrix is symmetric positive definite. 
Now swap the rows of $A$ to form the matrix 
$$ B = \begin{bmatrix} 1 & 8 \\ 2 & 1 \end{bmatrix}.$$
This matrix is not positive definite. To see this we examine
$$\frac{1}{2}(B + B^T) = \begin{bmatrix} 1 & 5 \\ 5 & 1 \end{bmatrix}$$
The eigenvalues are real. The determinant is $-24$. We conclude that there are two nonzero eigenvalues with different signs.

It is true that many texts assume symmetry before discussing definiteness. However, in the field of numerical linear algebra, symmetry is not a requirement. A prominent example is the GMRES algorithm for solving linear systems of the form $Ax=b$. Here one of the principal theorems asserts the convergence when $A$ is positive definite. This link goes to the relevant portion of Yousef Saad's textbook: "Iterative methods for sparse linear systems" on his personal website. Section 6.11.4 contains the relevant definition and the quoted result.
