Which one of the following statements is TRUE about every n x n matrix with only real eigenvalues? A. If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative. B. If the trace of the matrix is positive, all its eigenvalues are positive. C. If the determinant of the matrix is positive, all its eigenvalues are positive. D. If the product of the trace and determinant is positive, all its eigenvalues are positive.

GATE 2014 · Engineering Mathematics · Eigen Value · medium

Answer: Statement A is the only one that is always TRUE.

1. Analyze option A: If trace > 0 (sum positive) and det < 0 (product negative), then the product of all eigenvalues is negative. A product of real numbers is negative iff there is an odd count of negative factors — so at least one eigenvalue must be negative. Option A is TRUE.
2. Disprove options B, C, D with counterexamples: B: eigenvalues 3, -1 => trace = 2 > 0 but one eigenvalue is -1. Disproved. C: eigenvalues -2, -3 => det = 6 > 0 but both negative. Disproved. D: eigenvalues -2, -3 => trace = -5, det = 6, product of trace and det = -30 < 0 (disproves D for that sign); but eigenvalues 5, -2 => trace = 3, det = -10, product = -30 < 0 (not positive), so D's premise doesn't help. For eigenvalues -1, -2, trace = -3, det = 2, trace*det = -6 < 0. With trace*det > 0: try 3, 2 => fine, but 3, -4, -1 => trace=-2, det=12, product=-24 < 0. Finding trace*det > 0 with a negative eigenvalue: eigenvalues 5, -2, -3 => trace=0 (not >0). Try 6, -2, -1 => trace=3>0, det=12>0, product=36>0 BUT both B and C already fail these. So D could fail: eigenvalues -2, -3, 7 => trace=2>0, det=42>0, product=84>0, yet two eigenvalues are negative. Disproved.