GATE CS 2014 Set 3 Q.14 (CS Q.4): Linear Algebra

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

If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative.

If the trace of the matrix is positive, all its eigenvalues are positive.

If the determinant of the matrix is positive, all its eigenvalues are positive.

If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.

Official answer and solution

Solution

Relate the trace and determinant of a matrix to the sum and product of its real eigenvalues.

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

By linear algebra identities:
- $\text{Trace} = \sum_{i=1}^n \lambda_i$ .
- $\text{Determinant} = \prod_{i=1}^n \lambda_i$ .
If the determinant is negative ( $\prod_{i=1}^n \lambda_i < 0$ ):
The product of real numbers can only be negative if there is an odd number of negative eigenvalues (at least one negative eigenvalue).
Therefore, at least one eigenvalue must be negative.
This directly confirms that option A is always TRUE.
Counterexamples for other options:
- For B: $\lambda = (3, -1) \implies \text{Trace} = 2 > 0$ , but not all eigenvalues are positive.
- For C: $\lambda = (-2, -3) \implies \text{Det} = 6 > 0$ , but no eigenvalue is positive.
- For D: $\lambda = (-2, -3, 6) \implies \text{Trace} = 1 > 0, \text{Det} = 36 > 0$ , yet two eigenvalues are negative.

Hence, statement A is correct.

Final answer

Common trap

Assuming that a positive trace or positive determinant guarantees that all eigenvalues are positive.

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