Cayley-Hamilton...

Say $A$ is a square matrix over an algebraically closed field. Say $m$ is the minimal polynomial and $p$ is the characteristic polynomial.

Of course C-H implies that $m|p$. Conversely, if we can show $m|p$ then C-H follows; the question is whether one can give a "simple", "elementary" or "straightforward" proof that $m|p$.

Note. What I really want is a proof such that I feel I actually understand the whole thing. Hence in particular no Jordan form allowed.

Edit. When I posted this is was an honest question that I didn't know the answer to. I think I got it; if anyone wants to say they believe the argument below (or not) that would be great.

First, it's clear that linear factors of $m$ must divide $p$:

If $m(\lambda)=0$ then $p(\lambda)=0$.

Because $m(t)=(t-\lambda)r(t)$, so $(A-\lambda)r(A)=0$. Minimality of $m$ shows that $r(A)\ne0$, hence $A-\lambda$ is not invertible, hence $p(\lambda)=0$.

If we could show that $(t-\lambda)^k|m$ implies $(t-\lambda)^k|p$ we'd be set. Some possible progress on that, first restricted to a simple special case:

If $t^2|m(t)$ then $\dim(\ker(A^2))\ge 2$.

Proof: Say $X=K^n$ is the underlying vector space. Say $m(t)=t^2q(t)$. Let $$Y=q(A)X,$$ $$B=A|_Y.$$ Then $Y\subset\ker(A^2)$. Say $d=\dim(Y)$.

Now $B^2=0$, and it follows easily that $B^d=0$. But $B\ne0$, hence $d\ge2$.

Similarly

If $(t-\lambda)^k|m$ then $\dim(\ker(A-\lambda)^k)\ge k$.

So we only need

If $\dim(\ker(A-\lambda)^k)\ge k$ then $(t-\lambda)^k|p$.

Which I gather is true, but only by hearsay; I'm sort of missing what it "really means" to say $t^2|p$.

Wait, I think I got it. Say $$m(t)=(t-\lambda)^kq(t),$$ $$q(\lambda)\ne0.$$ The "kernel lemma" shows that $$X=\ker((A-\lambda)^k)\oplus\ker(q(A))=X_1\oplus X_2.$$

Each $X_j$ is $A$-invariant, so we can define $$B_j=A|_{X_j}.$$Since similar matrices have the same determinant we can use any basis we like in calculating the determinant $p(t)$; if we use a basis compatible with the decomposition $X=X_1\oplus X_2$ it's clear that $$p_A=p_{B_1}p_{B_2},$$so we need only show that $$p_(t)=(t-\lambda)^k.$$ In fact it's actually enough to show $(t-\lambda)^k|p_{B_1}$, and that's clear:

Lemma. If $B$ is a $d\times d$ nilpotent matrix then $p_B(t)=t^d$.

Proof: We're still assuming $K$ is algebraically closed; $B$ cannot have a non-zero eigenvalue.

So if $d=\dim(\ker((A-\lambda)^k)$ then $$p_{B_1}(t)=(t-\lambda)^d;$$we've already shown that $d\ge k$, so $(t-\lambda)^k|p$.

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