The sum of the eigenvalues of an $n \times n$ matrix $A$ equals
A$\det A$, the determinant
B$\text{trace}(A)$, diagonal sum
C$n$, the matrix size only
D$0$, always equals zero
Answer & Solution
Correct answer: B. $\text{trace}(A)$, diagonal sum
1. Express the characteristic polynomial: $\det(A - \lambda I) = (-1)^n \lambda^n + (-1)^{n-1} \text{tr}(A)\, \lambda^{n-1} + \ldots + \det A$.
2. By Vieta's: sum of roots = (negation of next-to-leading coefficient over leading coefficient).
3. Result: $\lambda_1 + \lambda_2 + \ldots + \lambda_n = \text{trace}(A)$.
4. PRODUCT of eigenvalues = $\det A$ (option A's case for PRODUCT not SUM).
5. So trace = sum, det = product. Useful identities for quick eigenvalue computation.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.5 + §4.6 (Eigenvalue invariants)._
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