Which one of the following statements is $\textbf{TRUE}$ about every $n \times n$ matrix with only real eigenvalues?

Let the real eigenvalues of the $n \times n$ matrix be $\lambda_1, \lambda_2, \dots, \lambda_n$.

• The determinant of the matrix is the product of its eigenvalues: $\det(A) = \prod_{i=1}^n \lambda_i$.

• The trace of the matrix is the sum of its eigenvalues: $\text{tr}(A) = \sum_{i=1}^n \lambda_i$.

Let us evaluate each option:

• Option A: If $\det(A) < 0$, then $\prod_{i=1}^n \lambda_i < 0$. For a product of real numbers to be negative, there must be an odd number of negative factors, which means there is at least one negative eigenvalue. Thus, statement (A) is always TRUE.

• Option B: Counterexample: $\lambda_1 = 10, \lambda_2 = -1$. Trace is $9 > 0$, but not all eigenvalues are positive.

• Option C: Counterexample: $\lambda_1 = -2, \lambda_2 = -2$. Determinant is $4 > 0$, but the eigenvalues are negative.

• Option D: Counterexample: for a $2 \times 2$ matrix with $\lambda_1 = -2, \lambda_2 = -4$, $\text{tr}(A) = -6$ and $\det(A) = 8$. Their product is $(-6)(8) = -48$. If we take a $3 \times 3$ matrix with $\lambda_1 = -2, \lambda_2 = -2, \lambda_3 = 5$, $\text{tr}(A) = 1 > 0$ and $\det(A) = 20 > 0$, product is positive, but not all eigenvalues are positive.

Therefore, (A) is the only universally correct statement.