I'm currently studying the Cayley-Hamilton theorem for an exam, and I do not quite get the proof presented in the lecture. It was structured as follows: first we'll prove it over $\mathbb{C}$ using the fact that every matrix is trigonalizable over $\mathbb{C}$ and then generalize to an arbitrary field $\mathbb{K}$. The first part of the proof is very comprehensive, what I don't get is the final generalization:
We know that $\forall A \in \text{M}_{n\text{x}n}(\mathbb{K}): \text{det}(A) = \sum_{\sigma \in S_n}\text{sgn}(\sigma)\cdot A(1,\sigma 1) \cdot ... \cdot A(n, \sigma n)$
$\Longrightarrow \text{char}_{A}(x) = \text{det}(A-xI_n)\in \mathbb{K}[x, A(1,1), A(1,2), ..., A(n,n)]$
But from the formula from the determinant we get that the coefficients are in $\mathbb{R}$, in fact in {-1,1} $\Longrightarrow \text{char}_{A}(x)\in \mathbb{R}[x, A(1,1), A(1,2), ..., A(n,n)]$
Consider now the matrix $\text{char}_{A}(A)$ and write it as: $\text{char}_{A}(A) = (q_{i,j}(A(1,1), ..., A(n,n)))_{i,j}$ with $q_{i,j} \in \mathbb{R}[A(1,1), ..., A(n,n)]$
Now since we have proved the thm over $\mathbb{C}$ it works over $\mathbb{R}$ in particular. Hence, $q_{i,j}(A(1,1), ..., A(n,n)) = 0, \ \forall (A(1,1), ..., A(n,n)) \in \mathbb{R}^{n^2}$
We deduce $q_{i,j}$ is the zero polynomial, and this is independent of the field we want to prove Cayley-Hamilton on. $\blacksquare$
I'm not quite sure why it follows that the theorem holds on $\mathbb{R}$? To prove it over $\mathbb{C}$ we explicitly use the property that every matrix is trigonalizable. Now we view the characteristic polynomial as a formal expression with $n^2$ variables which as a $\mathbb{C} \rightarrow \mathbb{C}$ fucntion is the zero function then also as a $\mathbb{R} \rightarrow \mathbb{R}$ function. And this concludes the proof? I find the final steps a bit shady, I really don't understand how we can simply generalize to every field...
A few things here; I don't know if you are familiar with field extension, if you are not just consider (this is not the real definition) that a field extension $\mathbb{L}$ of $\mathbb{K}$ is a field which verifies $\mathbb{L} \supset \mathbb{K}$, just as $\mathbb{C} \supset \mathbb{R}$.
Given a field extension we can always transform a $\mathbb{K}$-vector space $V$ into a $\mathbb{L}$-vector space $V_\mathbb{L}$ by setting: $$V_\mathbb{L} = \mathbb{L} \otimes_\mathbb{K} V$$ This tensor product is "functorial", which means each linear morphism can be extended to $V_\mathbb{L}$.
We also get a similar operation with the dual space: $$V_\mathbb{L}^* =\hom_\mathbb{K}(V,\mathbb{L})$$
But this is not very important if you don't get it: you can always get back to a matrix representation with the injections: $$V \simeq \mathbb{K}^n \longrightarrow \mathbb{L}^n \simeq V_\mathbb{L} $$ $$\hom_\mathbb{K} (V,V) \simeq \mathcal{M}_{n}(\mathbb{K}) \longrightarrow \mathcal{M}_{n}(\mathbb{L}) \simeq \hom_\mathbb{L} (V_\mathbb{L},V_\mathbb{L}) $$ Things are easier.
It seems you already know this result: if an endomorphism $u$ of $V$ is extended to $u_{\mathbb{L}}$, endomorphism of $V_\mathbb{L}$, $$\chi^\mathbb{K}_u = \chi^\mathbb{L}_{u_\mathbb{L}} \in \mathbb{K}[x]$$ Which can also be stated as: the characteristic polynomial does not depend on the field extension. This is the same things that you've done with your matrices.
So let's get back to our theorem of Cayley-Hamilton: I will generalize it for any field. I think it is already clear for you that if $u$, or if you prefer $A$ your matrix, is trigonalizable the theorem is basically free and obvious. So what we do here is for any $u$, $\chi_u$ is not necessarily split. So we need to choose a field extension $\mathbb{L}$ in which it is (since the polynomial does not depend on the choice of extension).
What we do with $\mathbb{R}$ is to choose $\mathbb{C}$ but in general we can define the splitting field of a polynomial $P$, we just take its decomposition : $$P = P_1^{\alpha_1} \cdots P_r^{\alpha_r} $$ and take $$Q = P_1 \cdots P_r$$ If your field is perfect, we can easily define it with $P$ and $P\wedge P'$ but I'll let you do it.
Then we take $\mathbb{L} = \mathbb{K}[x] / (Q)$, you can easily see that $\mathbb{L}$ is a field on which $P$ is split. If you don't get this part, stick to $\mathbb{R}$ and $\mathbb{C}$.
Once you have your field $\mathbb{L}$, you know that $u_\mathbb{L}$ or $A_{\mathbb{L}}$ (the same endomorphism/matrix but in your extended space) is trigonalizable since $\chi^\mathbb{L}_u = \chi^\mathbb{K}_u$ is split in $\mathbb{L}$ (your ambiant field). Hence $u_\mathbb{L}$ (or $A_\mathbb{L}$) is trigonalizable, thus: $$\chi^\mathbb{K}_u (u_\mathbb{L}) = \chi^\mathbb{L}_u (u_\mathbb{L}) =0$$ But $v \mapsto v_\mathbb{L}$ is an injective algebra morphism, hence $$0 = \chi_u (u_\mathbb{L}) = (\chi_u (u))_\mathbb{L}$$ thus $\chi_u (u)=0$. Here you go!