📐 Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic polynomial: if $p(\\lambda) = \\det(\\lambda I - A)$, then $p(A) = 0$.

Proof: **Proof using Jordan form:** Over $\\mathbb{C}$, $A$ is similar to its Jordan form $J$. Since $p(A)$ and $p(J)$ are similar, it suffices to show $p(J) = 0$.

For a Jordan block $J_k(\\lambda)$, the characteristic polynomial is $(t - \\lambda)^k$.

We have $(J_k(\\lambda) - \\lambda I)^k = N^k$ wh...

From: Advanced Linear Algebra

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