# 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. An Answer has appeared that shows $$m|p$$ in a very simple way - simply demolishes what I wrote below.

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 $$p_{{B_1}}(t)=(t-\lambda)^k.$$ 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 $$(t-\lambda)^k|p$$.

Hmm. Maybe that doesn't look all that simple. It's nonetheless the sort of thing I wanted, because I can give a one-line summary making it at least comprehensible:

One-line summary: Since $$m$$ splits, the kernel lemma (a simple consequence of the fact that $$K[t]$$ is a PID) shows that $$A$$ is the direct sum of operators $$B_j$$ such that $$B_j-\lambda_j$$ is nilpotent. So it's enough to prove C-H for nilpotent operators, which is not hard.

• but ..but..what's the matter with Jordan forms? :( – Alvin Lepik Sep 28 at 16:33
• @AlvinLepik Jordan forms are great, just less elementary than what I was hoping for here. – David C. Ullrich Sep 28 at 19:03

Let $$\lambda$$ be any eigenvalue of a minimal counterexample $$A$$ and choose a basis so $$A=\begin{pmatrix}\lambda&*\\0&B\end{pmatrix}.$$
Let $$m(x)$$ and $$n(x)$$ be the minimal polynomials of $$A$$ and $$B$$, respectively. Let $$q(x)$$ be the characteristic polynomial of $$B$$, where we can suppose that $$n(x)|q(x)$$.
Then $$(A-\lambda)n(A)=\begin{pmatrix}0&*\\0&*\end{pmatrix}\begin{pmatrix}n(\lambda)&*\\0&0\end{pmatrix}=0$$ and therefore $$m(x)|n(x)(x-\lambda)|q(x)(x-\lambda)=p(x)$$.
• Not that it matters, but if $Ae_1=\lambda e_1$ then $A=\begin{pmatrix}\lambda&*\\0&B\end{pmatrix}.$ – David C. Ullrich Sep 28 at 18:58
• @darijgrinberg That's no problem! Whether or not $p(A)=0$ is not changed by replacing $K$ by its algebraic closure. So wlog $K$ is algebraically closed. So $m$ has a zero, and it's easy to show that $m(\lambda)=0$ implies $\lambda$ is an eigenvalue. – David C. Ullrich Sep 29 at 11:29