Reordering the rows of a matrix and positive definiteness

Question

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.

Answer 1

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.

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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.