I have $A=\begin{pmatrix} 0 & \cdots & 0& -a_{0} \\ 1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}$, $A\in M_{n}$ and, as the title says, I want to know the minimal polynomial, but without using Hamilton Cayley (as I'm trying to do an alternative, inductive proof of the theorem).

I know that $m_{e_1} = x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$.

Maybe I can compute $m_{e_i}$ for the other canonical vectors, but it's ugly, and I'm searching for a faster, non-horrible way. Could someone help me?

  • $\begingroup$ OK, I got it now. As $e_1$ is cyclic, and $m_{e_1}(v) = 0 \forall v \in <e_1,Ae_1,A^2 e_1,\cdots>, m_A = m_{e_1}$ $\endgroup$ – Guillermo Mosse Oct 27 '16 at 21:16
  • $\begingroup$ But...I still don't know how to prove Hamilton Cayley...without using Frobenius form. We didn't see it in the course...I think there should be another way $\endgroup$ – Guillermo Mosse Oct 27 '16 at 21:17
  • $\begingroup$ Pedro, you are using Hamilton-Cayley, and I'm supposed to prove this without using it, as this is the first part of an alternative proof of the theorem. Remember that tomorrow during the exam! $\endgroup$ – Guillermo Mosse Oct 27 '16 at 21:53

You have checked that $P=x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ is the minimal degree monic polynomial such that $P[A](e_1)=0$, so you just need to show that $P[A](e_i)=0$ for every $i>1$ as well.

Now no doubt you noticed during your calculation that $e_i=A^{i-1}(e_1)$. Then $$P[A](e_i)=P[A](A^{i-1}(e_1)) = A^{i-1}(P[A](e_1))=A^{i-1}(0)=0, $$ because polynomials in $A$ commute with each other, and you are done.

In general, if some polynomial in $A$ kills a vector (here $v=e_1$), then it also kills its repeated images by$~A$, and their span. Stated differently, the kernel of the polynomial in $A$ is an $A$-stable subspace.


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.