# Exercise 2, Section 6.3 of Hoffman’s Linear Algebra [duplicate]

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?

• Dec 23, 2022 at 19:55

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 !

• 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. Dec 24, 2022 at 8:14
• 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? Dec 24, 2022 at 8:28
• 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.
– Zag
Dec 24, 2022 at 9:14

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

$$x_{n+3}=ax_{n+2}+bx_{n+1}+cx_n$$

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.

• 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. Dec 24, 2022 at 9:13