Consider the matrix \(M = \begin{bmatrix} 2 & -1 \\ 3 & 1 \\ \end{bmatrix} \).…

For a 2×2 matrix A, the eigenvalues are the two roots of the characteristic polynomial λ2 − (trace A)λ + det(A) = 0. The discriminant Δ = (trace A)2 − 4·det(A) fixes their nature: Δ > 0 gives two distinct real eigenvalues, Δ = 0 gives one repeated real eigenvalue, and Δ < 0 gives a complex-conjugate pair. Because det(A) always equals the product of the eigenvalues, det(A) = 0 forces at least one eigenvalue to be zero, and det(A) < 0 forces two real eigenvalues to have opposite sign.

1. Compute trace(M) = 2 + 1 = 3 and det(M) = 2·1 − (−1)·3 = 2 + 3 = 5.

2. Write the characteristic polynomial: λ2 − 3λ + 5 = 0.

3. Compute the discriminant: Δ = 32 − 4·5 = 9 − 20 = −11, which is negative.

4. By the rule above, Δ < 0 means M’s eigenvalues form a complex-conjugate pair; solving the quadratic gives λ = (3 ± √−11)/2 = (3 ± i√11)/2.

Cross-check: the sum of the two eigenvalues, (3 + i√11)/2 + (3 − i√11)/2 = 3, matches trace(M); their product, [32 − (i√11)2]/4 = (9 + 11)/4 = 5, matches det(M). Since det(M) = 5 ≠ 0, no eigenvalue equals zero, and since Δ < 0, neither eigenvalue is real — which rules out any statement claiming real eigenvalues, a zero eigenvalue, or one non-negative and one negative real eigenvalue.

So the only true statement is that the eigenvalues of M form a complex-conjugate pair.