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I was solving the problem from linear algebra and was able to complete it almost but I have the last step which I cannot solve by myself but I have spent like 3-4 hours trying to crack it.

Suppose $V$ is a finite dimensional vector space and $T:V\to V$ is a linear operator such that it has the following matrix in some basis: $$\begin{bmatrix} 0 & 1 & 0 &\cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ a_0 & a_1 & a_2 & \cdots & a_{n-1} \end{bmatrix}$$

Show that characteristic polynomial of this operator is the same as the minimal polynomial. Find the characteristic polynomial.

My approach: I was able to find the characteristic polynomial which is $$P(t)=(-1)^n(t^n-a_{n-1}t^{n-1}-\dots-a_2t^2-a_1t-a_0).$$

But I don't know how to show that minimal polynomial coincides with characteristic polynomial. Moreover, I think that the converse of this problem is also true.

Would be very thankful if you can show the solution. Please avoid using Jordan canonical form and Rational canonical form and other difficult topics since this problem is before those topics and the solution should be not so difficult.

RFZ
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    Let $A$ be your matrix and $v=(1,0,0,...,0)$. Then $vA^k=(0,0,...,1,...,0)$ where the $1$ is in the position $k+1$, for $k=1,2,...,n-1$. Those vectors are linearly independent. So, no polynomial in $A$ of degree not larger than $n-1$ will vanish at $v$. Therefore, the minimal polynomial has degree $n$. – MoonLightSyzygy Dec 27 '19 at 16:46
  • @MoonLightSyzygy, Hmm. Maybe you will a separate answer? I will appreciate it! BTW, what about the converse? – RFZ Dec 27 '19 at 16:51
  • Which converse? – MoonLightSyzygy Dec 27 '19 at 16:52
  • @MoonLightSyzygy, I mean if the characteristic and minimal polynomials are the same then there is some basis of $V$ such that matrix of operator $T$ has the above form. – RFZ Dec 27 '19 at 17:00
  • Without using Jordan or Rational canonical forms you have to partially disguise them in the argument. You pretty much prove them in the process. – MoonLightSyzygy Dec 27 '19 at 17:44
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    Does this answer your question? The characteristic and minimal polynomial of a companion matrix – Conifold Dec 27 '19 at 17:46
  • The "converse" that you are looking for is equivalent to the second corollary to the "Cyclic Decomposition Theorem" of section 7.2 of Hoffman and Kunze. I see no easy way to prove this result without the cyclic decomposition theorem or something equally powerful (such as the existence of a Jordan normal form) – Ben Grossmann Dec 27 '19 at 18:00
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    The "corollary" as stated in H+K states that a transformation will have a cyclic vector if and only if the characteristic and minimal polynomials of the matrix coincide. – Ben Grossmann Dec 27 '19 at 18:03
  • @MoonLightSyzygy, Regarding your first comment. How did you conclude that minimal polynomial has degree $n$? One thing which confuses me: you are multiplying $v$ to $A^k$ from the left. Could you clarify those moments? – RFZ Dec 27 '19 at 18:10
  • @Omnomnomnom, however this is very weird since this problem comes in before those topics. – RFZ Dec 27 '19 at 18:13
  • If $p$ is a polynomial of degree not larger than $n-1$, then $vp(A)$ is a non-trivial linear combination of the vectors $(1,0,...,0),(0,1,...,0),...,(0,0,...,1)$. Therefore $vp(A)$ is non-zero. Hence $p(A)$ is not the zero matrix. Note that just like I am multiplying from the left, the vectors are also rows. – MoonLightSyzygy Dec 27 '19 at 18:13
  • @MoonLightSyzygy, in the above by $p$ you mean minimal polynomial, right? – RFZ Dec 27 '19 at 18:15
  • @ZFR No, I meant what is written. $p$ is an arbitrary polynomial of degree not larger than $n-1$. The argument is showing that one will never have $p(A)=0$ for such polynomials. It follows from that that the degree of the minimal polynomial is $\geq n$. – MoonLightSyzygy Dec 27 '19 at 18:16
  • Have a look at https://en.wikipedia.org/wiki/Companion_matrix – amd Dec 27 '19 at 19:06
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    @ZFR but you are not supposed to prove the converse for this problem, so you're all set to answer the question. If you mean that it is weird that the converse is much harder to prove, I would say that this is a typical sort of asymmetry. – Ben Grossmann Dec 27 '19 at 19:48
  • @Omnomnomnom, yeah you are right! I am not supposed to prove the converse of the problem. I will try to prove to prove it when my knowledge on linear algebra will be better. Hopefully very soon i'll learn Jordan canonical forms. You said that the conversation can be proved using Jordan CF, right? – RFZ Dec 27 '19 at 19:59
  • @ZFR That's right – Ben Grossmann Dec 27 '19 at 20:00
  • @Omnomnomnom, could you give a link for the proof of this? It would be nice to learn it in the near future! – RFZ Dec 27 '19 at 20:02
  • Of the converse? There's no quick proof, as I've said, unless you assume the cyclic decomposition theorem or Jordan canonical form. Once you understand the cyclic decomposition theorem, your statement is an obvious converse. There are plenty of references for Jordan canonical form. For the cycle decomposition theorem, I recommend either Hoffman and Kunze or Dummit and Foote. – Ben Grossmann Dec 27 '19 at 20:58

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