14
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ but ..but..what's the matter with Jordan forms? :( $\endgroup$ – Alvin Lepik Sep 28 at 16:33
  • $\begingroup$ @AlvinLepik Jordan forms are great, just less elementary than what I was hoping for here. $\endgroup$ – David C. Ullrich Sep 28 at 19:03
5
$\begingroup$

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 $m(x)|n(x)(x-\lambda)|q(x)(x-\lambda)=p(x)$.

$\endgroup$
  • $\begingroup$ Not that it matters, but if $Ae_1=\lambda e_1$ then $A=\begin{pmatrix}\lambda&*\\0&B\end{pmatrix}.$ $\endgroup$ – David C. Ullrich Sep 28 at 18:58
  • $\begingroup$ The edits are fine - I was puzzled by a lot in the first version $\endgroup$ – David C. Ullrich Sep 28 at 18:59
  • 1
    $\begingroup$ This presumes the existence of an eigenvalue, which makes it hard to generalize and also not quite self-contained. $\endgroup$ – darij grinberg Sep 28 at 19:18
  • $\begingroup$ One can choose a root of the minimal polynomial since it's easy to show this is an eigenvalue. I originally did this in the proof and, in view of your comment, perhaps I should have retained this approach. $\endgroup$ – S. Dolan Sep 28 at 19:21
  • $\begingroup$ @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. $\endgroup$ – David C. Ullrich Sep 29 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.