The CHARACTERISTIC POLYNOMIAL of an $n \times n$ matrix $A$ is
A$p(\lambda) = \det(A - \lambda I)$
B$p(\lambda) = A - \lambda$
C$p(\lambda) = \lambda^n$
D$p(\lambda) = \text{trace}(A) - \lambda$
Answer & Solution
Correct answer: A. $p(\lambda) = \det(A - \lambda I)$
1. CHARACTERISTIC POLYNOMIAL of $A$: $p(\lambda) = \det(A - \lambda I)$ (or sometimes $\det(\lambda I - A)$, same roots).
2. The ROOTS of $p(\lambda) = 0$ are exactly the EIGENVALUES of $A$.
3. $p$ is a polynomial of degree $n$ in $\lambda$, so it has $n$ roots (counting multiplicity) over $\mathbb{C}$.
4. CAYLEY-HAMILTON THEOREM: every matrix satisfies its own characteristic polynomial, i.e. $p(A) = 0$.
5. Other options are not the characteristic polynomial.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.5 (Characteristic polynomial)._
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