@MISC{Robinson_testfor, author = {R. Clark Robinson}, title = {TEST FOR POSITIVE AND NEGATIVE DEFINITENESS}, year = {} }

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Abstract

We want a computationally simple test for a symmetric matrix to induce a positive definite quadratic form. We first treat the case of 2 × 2 matrices where the result is simple. Then, we present the conditions for n × n symmetric matrices to be positive definite. Finally, we state the corresponding condition for the symmetric matrix to be negative definite or neither. Before starting all these cases, we recall the relationship between the eigenvalues and the determinant and trace of a matrix. For a matrix A, the determinant and trace are the product and sum of the eigenvalues: det(A) = λ1 · · · λn, tr(A) = λ1 + · · · + λn, where λj are the n eigenvalues of A. (Here we list an eigenvalue twice if it has multiplicity two, etc.) 1. Two by two matrices a b Let A = be a general 2 × 2 symmetric matrix. We will see in general that the quadratic b c form for A is positive definite if and only if all the eigenvalues are positive. Since, det(A) = λ1λ2, it is necessary that the determinant of A be positive. On the other hand if the determinant is positive, then either (i) both eigenvalues are positive, or (ii) both eigenvalues are negative. Since tr(A) = λ1 + λ2, if det(A)> 0 and tr(A)> 0 then both eigenvalues must be positive. We want to give this in a slightly different form that is more like what we get in the n × n case. If det(A) = ac − b 2> 0, then ac> b 2 ≥ 0, and a and c must have the same sign. Thus det(A)> 0 and tr(A)> 0 is equivalent to the condition that det(A)> 0 and a> 0. Therefore, a necessary and sufficient condition for the quadratic form of a symmetric 2 × 2 matrix to be positive definite is for det(A)> 0 and a> 0. We want to see the connection between the condition on A to be positive definite and completion of the squares. Q(x, y) = (x, y)A x