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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?

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  • $\begingroup$ 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
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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.

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