Question

Which one of the following statements is true for every square matrix with only real eigenvalues?

• A. If the trace of the matrix is positive and the determinant of the matrix is negative, then at least one of its eigenvalues is negative.
• B. If the trace of the matrix is positive, then all its eigenvalues are positive.
• C. If the determinant of the matrix is positive, then all its eigenvalues are positive.
• D. If the product of the trace and determinant of the matrix is positive, then all its eigenvalues are positive.

Hint:

Use det = product of eigenvalues; a negative product of real numbers always has an odd count of negative factors.

Answer 1

The Correct Option is A

Step 1: For a square matrix with real eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_n\), the trace equals the sum of the eigenvalues and the determinant equals the product of the eigenvalues:

\[ \text{trace}(A) = \sum_{i=1}^{n} \lambda_i, \qquad \det(A) = \prod_{i=1}^{n} \lambda_i \]

Step 2: Check option (A). If \(\det(A) < 0\), the product of the real eigenvalues is negative. A product of real numbers is negative only when an odd number of them are negative, so at least one eigenvalue must be negative. This conclusion does not depend on the sign of the trace, so (A) always holds.

Step 3: Check option (B) with a counterexample. Take \(\text{diag}(3, -1)\). Its trace is \(3 + (-1) = 2 > 0\), but the eigenvalue \(-1\) is negative. So (B) is false.

Step 4: Check option (C) with a counterexample. Take \(\text{diag}(-1, -1)\). Its determinant is \((-1)(-1) = 1 > 0\), but both eigenvalues are negative. So (C) is false.

Step 5: Check option (D) with a counterexample. Take \(\text{diag}(3, -1, -1)\). The trace is \(3 - 1 - 1 = 1 > 0\) and the determinant is \(3 \times (-1) \times (-1) = 3 > 0\), so their product is \(3 > 0\), yet not all eigenvalues are positive. So (D) is false.

Only option (A) holds for every such matrix.

\[\boxed{\text{Option (A)}}\]