Cayley-Hamilton
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Lemma: $$End_k(V) \otimes k[t] \cong End_{k[t]}(V \otimes k[t])$$ for $$V$$ a finite dimensional free $$k$$-module (see below)

Corollary [The Cayley-Hamilton Theorem]: For an endomorphism $$\phi : V \rightarrow V$$ of a finite dimensional vector space $$V$$ over a field $$k$$, $$p(\phi) = 0$$ where $$p \in End(V) \otimes k[t]$$ is the characteristic polynomial of $$\phi$$.

Proof: Consider how $$\text{det}(\Phi)$$ factors as $$\Phi \circ \text{adj}(\Phi) = \text{adj}(\Phi) \circ \Phi$$ in $$\text{End}(V)$$ for a finite dimensional vector space $$V$$ with endomorphism $$\Phi : V \rightarrow V$$. We want a factorization of $$p(t)$$ of $$\phi$$ into some polynomial analogous to the adjugate and a linear term $$t - \phi$$: $$p(t) = f(t)(t - \phi)$$ These two factorizations are analogous, and in fact, if we get the formality right, we can view these as corresponding factorizations in isomorphic rings $$\text{End}(V \otimes k[t]) \cong \text{End}(V) \otimes k[t]$$.

Construct a map

$$\text{End}_k (V) \otimes_k k[t] \rightarrow \text{Hom}_k(V, V \otimes_k k[t])$$

sending $$\phi \otimes t^n$$ to the map sending $$v$$ to $$\phi(v)t^n$$. This is injective, and surjective since $$V$$ is finitely generated. Composing these isomorphisms gives an isomorphism $$F : \text{End}_{k} (V)[t] \rightarrow \text{End}_{k[t]} (V \otimes_k k[t])$$.

View $$t - \phi$$ as a $$k[t]$$-linear endomorphism of $$V \otimes_k k[t]$$. Under the isomorphism $$F$$, $$\text{char}(\phi)$$ maps to $$\text{det} (t - \phi) 1_{V \otimes_k k[t]} )$$ and $$F ( t - \phi ) = t - \phi$$. $$t - \phi$$ divides $$\text{det}(t - \phi) 1_{V \otimes_k k[t]}$$ in $$\text{End}_{k[t]} (V \otimes_k k[t])$$, since $$\text{det} (t - \phi) 1_{V \otimes_k k[t]} = \text{adj}(t - \phi) (t - \phi)$$, where $$\text{adj}(t - \phi)$$ is the adjugate matrix. Therefore, $$t - \phi$$ divides $$\text{char}(\phi)$$ in $$\text{End}_{k}(V)[t]$$. So $$\text{char}(\phi)$$ has $$\phi$$ as a root in $$\text{End}_k(V)$$, so that the evaluation homomorphism $$\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$$ sends the characteristic polynomial $$\text{char}(\phi)$$ to $$0$$.

The characteristic polynomial $$p(t)$$ of $$\phi \in \text{End}_k (V)$$ naturally lives in $$\text{End}_k(V)[t]$$ from the natural map $$\text{End}_k(V) \rightarrow \text{End}_k(V) \otimes_k k[t]$$. View $$t \text{Id}_V - \phi$$ as having endomorphisms as coefficients, and then take the determinant, which is then in $$\text{End}_k (V)[t]$$.

In the isomorphism $$\text{End}_k ( V \otimes_k k[t]) \cong \text{End}_k (V)[t]$$ We have corresponding elements $$\Phi \leftrightarrow t - \phi$$ and $$\text{det}(\Phi) \leftrightarrow p(t)$$ Therefore, the factorization $$\text{det}(\Phi) 1_{V \otimes_k k[t]} = \text{adj}(\Phi) \Phi$$ corresponds to a factorization $$p(t) = f(t)(t-\phi)$$ in $$\text{End}_k (V) [t]$$. And that's the whole idea!

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