Let $a$, $b$, and $c$ be elements of a field $F$, and let $A$ be the following $3\times 3$ matrix over $F$: $$A=\begin{bmatrix}0 & 0& c\\ 1& 0& b\\ 0& 1& a\\ \end{bmatrix}$$ Prove that the characteristic polynomial for $A$ is $x^3-ax^2-bx-c$ and that this is also the minimal polynomial for $A$.

My attempt: Characteristic polynomial function of $A$ is $f:F\to F$ such that $f(x)=\det (xI_3-A)$, $\forall x\in F$. It’s easy to check, $\det (xI_3-A)=x^3-ax^2-bx-c$. We claim $f$ is minimal polynomial of $A$. Proof: we need to show (1) $f$ is monic, (2) $f(A)=0$, (3) If $g\in F[x]$ and $g(A)=0$, then $3=\deg (f)\leq \deg (g)$. Property (3)$\iff$$\nexists g\in F[x]$ such that $\deg (g)\lt 3$ and $g(A)=0$. Clearly $f$ is monic. By Cayley–Hamilton theorem, $f(A)=0$. Assume towards contradiction, $\exists g\in F[x]$ such that $\deg (g)\leq 2$ and $g(A)=0$. Let $g=px^2+qx+r$. Since we assign degree to $g$, it is implicitly non zero. Then $g(A)=pA^2+qA+r=0$, i.e. $$\begin{bmatrix}0 & pc& pac\\ 0& pb& pab\\ p& pa& pb+pa^2\\ \end{bmatrix}+ \begin{bmatrix}0 & 0& qc\\ q& 0& qb\\ 0& q& qa\\ \end{bmatrix}+\begin{bmatrix}r & 0& 0\\ 0& r& 0\\ 0& 0& r\\ \end{bmatrix}=\begin{bmatrix}0 & 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ \end{bmatrix}.$$ By definition of matrix addition, we have $p+0+0=0$, $0+q+0=0$, and $0+0+r=0$. So $p=q=r=0$ and $g=0$. Thus we reach contradiction. So $\nexists g\in F[x]$ such that $\deg (g)\lt 3$ and $g(A)=0$. Hence $f=x^3-ax^2-bx-c$ is minimal polynomial of $A$. Is my proof correct?


2 Answers 2


Given that your computations are correct, your approach indeed solves the problem. To simplify the presentation you could skip saying that $f$ is monic and reminding Cayley-Hamilton theorem, this is a very common argument.

However, I might suggest a more simpler approach. If $e_1, e_2, e_3$ are the vectors of the canonical basis of $F^3$, you get $Ae_1 = e_2$ and $A^2e_1 = Ae_2 = e_3$. Thus, if $f \in F[X]$ is such that $f(A) = 0$ and $f$ has degree $\leq 2$, you would get $[f(A)]\cdot e_1 = 0$ and thus a linear combination of $e_1, e_2, e_3$ that is equal to $0$, which implies by linear independence that $f = 0$.
In essence, it is quite similar to what you did but is lighter (you don't actually need to calculate $A^2e_2$ and $A^2e_3$ and that is kind of what you did)

If you want to go further, I would suggest that you go check out companion matrices and cyclic endomorphisms !

  • $\begingroup$ Thank you so for the answer. Your solution is really clever! We only care about first column of $f(A)$ (that’s what my solution suggest). Computing $A\cdot e_1$ and $A^2\cdot e_1=A\cdot A\cdot e_1$ is elementary. So we don’t have to unnecessary compute $A^2$. I will check companion matrices and cyclic endomorphisms. $\endgroup$
    – user264745
    Dec 24, 2022 at 8:14
  • $\begingroup$ I have one question, some book write characteristic polynomial as $\det (A-xI)$ and some as $\det=(xI-A)$. We know $\det (A-xI)=(-1)^n \det (xI-A)$. Which one is correct? $\endgroup$
    – user264745
    Dec 24, 2022 at 8:28
  • 1
    $\begingroup$ Those are juste two different conventions, some prefer to have a monic characteristic polynomial, some prefer to have it remind the form of the eigenspaces. It's just a matter of taste since the constant $(-1)^n$ doesn't change the properties of the characteristic polynomial (apart from it beeing monic), whatever the base field is. $\endgroup$
    – Zag
    Dec 24, 2022 at 9:14

If you know linear recurrences, this is related to a linear recurrent:


Specifically, if $\mathbf v=(x_0,x_1,x_2)$ then $(x_n,x_{n+1},x_{n+2})=\mathbf vA^n.$

Now, if you know the general formula for the $x_{n}$ in terms of the roots of $p(u)=u^3-au^2-bu-c$ and the number of repetitions, you can show that $p(A)=0,$ and you can show any polynomial of smaller degree, $q(u),$ can't have $q(A)=0$ by picking $\mathbf v$ so that $q(A)v\neq 0.$

Basically, the minimal polynomial of $A$ has to divide $p.$ But if if $p(u)$ has root $\lambda$ repeated $r$ times, and fewer times in $q(u),$ we can take $\mathbf v=(0^{r-1},\lambda,2^{r-1}\lambda^2),$ (where $0^{r-1}=1$ if $r=1,$ and $0$ otherwise.)

Ultimately, this will require us to work in the algebraic closure of $F.$

Of course, if you don't know the formula for such linear recurrences, this approach can be useful in reverse.

  • $\begingroup$ Thank you so much for the answer. To be honest, I don’t understand your proof. Hopefully in future I’ll do. Please don’t delete this post. $\endgroup$
    – user264745
    Dec 24, 2022 at 9:13

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