I need help solving the following exercise:

Let $K$ be a field, $a_0, a_1, ..., a_{n-1} \in K$ and $$A= \begin{pmatrix}0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{n-1} \end{pmatrix}$$ Show that the minimal polynomial is given by $F(x) = x^n + \sum_{i=0}^{n-1}a_i x^i$.

I proved by induction that the characteristic polynomial of $A$ is $(-1)^nF(x)$. However, I don't know how to proceed showing that that $F$ is indeed the minimal polyonomial, i.e. the smallest polynomial satisfying $F(A) = 0$. Any help appreciated.

  • $\begingroup$ Are you familiar with Jacobson canonical form? $\endgroup$ – Itay4 May 22 '17 at 20:14
  • $\begingroup$ Sadly not, not yet. The answer by C. Falcon is a little too advanced as well, I am looking for a simpler approach... $\endgroup$ – Staki42 May 22 '17 at 20:17
  • $\begingroup$ Are you sure this is the desired result? I believe the characteristic polynomial and the minimal polynomial of this kind of matrix should be equal. $\endgroup$ – Itay4 May 22 '17 at 20:23
  • $\begingroup$ The exercise should be correct as stated. What you say is probably what needs to be proved :D $\endgroup$ – Staki42 May 22 '17 at 20:25
  • $\begingroup$ @lappen68 I try to exposed an easier approach in a new answer, give me some feedback. :) $\endgroup$ – C. Falcon May 22 '17 at 20:43

Another approach as you find my previous one too advanced.

For all $i\in\{1,\ldots,n\}$, let define the following column vector: $$e_i:={}^\intercal(\delta_{i,j})_{j\in\{1,\ldots,n\}}.$$ Notice that for all $i\in\{1,\ldots,n-1\}$, one has: $$Ae_i=e_{i+1}.$$ From there, one gets that for all $i\in\{1,\ldots,n-1\}$: $$A^ie_1=e_i.$$ Let $\displaystyle G:=t^d+\sum_{k=0}^{d-1}b_kX^k$ be a monic polynomial of degree $d<n$ such that $G(A)=0$, then: $$0=G(A)e_1=e_d+\sum_{k=0}^{d-1}b_ke_{k+1},$$ which is a contradiction since $(e_k)_{k\in\{1,\ldots,n\}}$ is a basis and hence is free. Therefore, $F$ is the monic polynomial of the least degree which vanishes on $A$.

  • $\begingroup$ This one is way easier to understand and looks correct :) Thanks! $\endgroup$ – Staki42 May 22 '17 at 20:45
  • $\begingroup$ I am not too satisfied with this proof, as I used Cayley-Hamilton to show that $F(A)=0$. One can do it without using this theorem. $\endgroup$ – C. Falcon May 22 '17 at 20:47
  • $\begingroup$ That is true, but the Cayley-Hamilton theorem is something I may use and whose proof I know, while your other approach uses things I am not aware of :) $\endgroup$ – Staki42 May 22 '17 at 20:51
  • 1
    $\begingroup$ I mean, knowing that $A^ie_1=e_i$ probably implies that $F(A)=0$. There are some computations to do but is perfectly fine to use Cayley-Hamilton. In a few years, you will love the other approach, it gives some insight on why we are studying such matrices, they are called companion matrices. They are linked to $K[X]$-modules and reduction of endomorphisms for which the multiplication by $X$ holds a central place. $\endgroup$ – C. Falcon May 22 '17 at 20:58
  • $\begingroup$ Thank you for these insights. For now, the other approach is a little too advanced, but I believe you that it is more beatiful. $\endgroup$ – Staki42 May 22 '17 at 21:28

Let define $E=K[X]/(F)$, then $E$ is a vector space of degree $\deg(F)$ over $K$, a basis is given by: $$\underline{e}:=\left(\overline{1},\overline{X},\ldots,\overline{X}^{\deg(F)-1}\right).$$ This follows from euclidean divisions by $F$ in $K[X]$. Notice that $A$ is the matrix in $\overline{e}$ of the linear map $L\colon E\rightarrow E$ defined by: $$L(v)=\overline{X}v.$$ Now, since $\overline{F}=0$, then for all $i\in\mathbb{N}$, one has: $$\overline{F}\overline{X}^i=0.$$ Hence, $F(A)=0$. Furthermore, let $G$ be a polynomial of degree less that $\deg(F)$, then: $$G(A)(1)\neq 0,$$ since $\overline{G}\neq 0$. Finally, $F$ is the minimal polynomial of $A$.


One part is immediate: no non nonzero polynomial $P$ with $\deg P<n$ annihilates $A$; for this is suffices to observe that with $e_1,\ldots,e_n$ the standard basis one has $A^ke_1=e_{1+k}$ for $k<n$, so if $P=\sum_{i=0}^{n-1}p_iX^i$ then $P[A]e_1=\sum_{i=0}^{n-1}p_ie_{1+i}\neq0$ and in particular $P[A]\neq0$. Also $A^ne_1=Ae_n=-a_0e_1-\cdots-a_{n-1}e_n=-(a_0+a_1A+\cdots+a_{n-1}A^{n-1})e_1$, so your polynomial $F=a_0+a_1X+\cdots+a_{n-1}X^{n-1}+X^n$ is clearly the unique monic polynomial of degree$~n$ such that $F[A]e_1=0$.

Now to conclude that indeed $F[A]=0$, there are two paths. Either you know about the Cayley-Hamilton theorem which tell you there always exists a (monic) annihilating polynomial of degree$~n$ (namely the characteristic polynomial), so necessarily $F$ must be it. Note that your computation of the characteristic polynomial is not used here; its result follows from the argument. The other path avoids using the Cayley-Hamilton theorem, and shows directly that $F[A]e_{1+k}=0$ for $k=1,\ldots,n-1$ (then $F[A]$ kills all the $e_i$ and must be zero), namely $$F[A]e_{1+k}=F[A](A^ke_1)=(X^kF)[A]e_1=A^k(F[A]e_1)=A^k0=0.$$ Now your computation may serve to prove the Cayley-Hamilton theorem for this case (and this can be in turn be used to prove its general case of the Cayley-Hamilton theorem, but I won't go into that here).


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