**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!