GATE CS 2014 Set 3 Q4 — Eigenvalues and Eigenvectors, LU Decomposition

Let the eigenvalues of the $n \times n$ matrix be $\lambda_1, \lambda_2, \dots, \lambda_n$. Since the matrix has only real eigenvalues, each $\lambda_i \in \mathbb{R}$.

Recall the fundamental properties of eigenvalues:

• Trace: $\text{Tr}(A) = \sum_{i=1}^n \lambda_i$
• Determinant: $\text{Det}(A) = \prod_{i=1}^n \lambda_i$

Let's analyze each option:

• Option A: We are given $\text{Tr}(A) > 0$ and $\text{Det}(A) < 0$.
  Suppose none of the eigenvalues are negative (i.e., $\lambda_i \ge 0$ for all $i$). Then, their product must be non-negative: $$\text{Det}(A) = \prod_{i=1}^n \lambda_i \ge 0$$ This directly contradicts the given condition that $\text{Det}(A) < 0$. Thus, at least one eigenvalue must be negative. Statement A is TRUE.
• Option B: If $\text{Tr}(A) > 0$, all eigenvalues need not be positive.
  Counterexample: For a $2 \times 2$ matrix with eigenvalues $\{5, -1\}$, the trace is $4 > 0$, but one eigenvalue is negative.
• Option C: If $\text{Det}(A) > 0$, all eigenvalues need not be positive.
  Counterexample: For a $2 \times 2$ matrix with eigenvalues $\{-2, -3\}$, the determinant is $6 > 0$, but both eigenvalues are negative.
• Option D: If $\text{Tr}(A) \cdot \text{Det}(A) > 0$, all eigenvalues need not be positive.
  Counterexample: For a $3 \times 3$ matrix with eigenvalues $\{-1, -2, 6\}$, we have: $$\text{Tr}(A) = -1 - 2 + 6 = 3 > 0$$ $$\text{Det}(A) = (-1)(-2)(6) = 12 > 0$$ The product $\text{Tr}(A) \cdot \text{Det}(A) = 36 > 0$, yet there are negative eigenvalues.