CAYLEY-HAMILTON THEOREM states that for any $n \times n$ matrix $A$ with characteristic polynomial $p$
A$p(\lambda) = \det(A)$
B$p(A) = 0$
C$p$ has degree 1
D$A$ has $n$ distinct eigenvalues
Answer & Solution
Correct answer: B. $p(A) = 0$
1. CAYLEY-HAMILTON: every square matrix satisfies its own characteristic equation.
2. Formally: if $p(\lambda) = \det(A - \lambda I)$, then $p(A) = 0$ (where 0 is the zero matrix).
3. EXAMPLE: $A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$. $p(\lambda) = (\lambda-2)(\lambda-3) = \lambda^2 - 5\lambda + 6$. Then $A^2 - 5A + 6I = \begin{pmatrix}4&0\\0&9\end{pmatrix} - \begin{pmatrix}10&0\\0&15\end{pmatrix} + \begin{pmatrix}6&0\\0&6\end{pmatrix} = 0$. ✓
4. APPLICATIONS: compute inverses, simplify high powers of $A$.
5. Other options misstate Cayley-Hamilton.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.9 (Cayley-Hamilton)._
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