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I am currently in the process of familiarizing myself with linear algebra, so there may be inaccuracies in my question.

The method of constructing nilpotent matrices is as follows:

  1. Randomly pick a monic polynomial $f(x)=x^n+a_{n-1}\cdot x^{n-1}+...+a_0$;
  2. Find the companion matrix $C$ associated with the polynomial $g(x)=f(x)^k$;
  3. Let the nilpotent matrix $B=f(C)$.

The above inspiration is derived from the answer of this question. I am uncertain whether taking the companion matrix $C$ as the input of $f(x)$ would result in $B^k=0$, due to some related thoerems that I have not yet come across.

Bill Dubuque
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X.H. Yue
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  • It looks like your polynomials are all linear. You should be able to rewrite this as a product of two matrices, one of coefficients and the other of the $z$s. Not sure if this helps solve the problem though since the matrices aren’t square. – Michael Burr Mar 20 '24 at 09:10
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    @Amateur_Algebraist I changed the description of my question. – X.H. Yue Mar 26 '24 at 08:44
  • @MichaelBurr I changed the description of my question. – X.H. Yue Mar 26 '24 at 09:23
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    @mingalex You should undo the last edit so the question is visible without viewing the history. The point of this website is to archive good questions and answers, not to answer questions. – Kepler's Triangle Mar 28 '24 at 12:35

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Yes, your matrix $\ B\ $ will always be nilpotent. This follows from the fact that the companion matrix $\ C\ $ of a polynomial $\ g(x)\ $ must satisfy the equation $\ g(C)=0\ .$ In fact $\ g(x)\ $ is its characteristic polynomial, as has been proved in this answer, and the fact that $\ g(C)=0\ $ then follows from the Cayley-Hamilton theorem. Thus, if $\ g(x)= f(x)^k\ $ where $\ k\ge2\ $, then \begin{align} 0&=g(C)\\ &=(f(C))^k\ . \end{align} Therefore $\ B=f(C)\ $ is nilpotent. It's also trivially nilpotent when $\ k=1\ ,$ because $\ B\ $ must be the zero matrix in that case, which is normally of no interest.

lonza leggiera
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  • Thanks a lot for your response! However, I cannot find the related thoerems or lemmas to demonstrate the fact you mentioned that "the companion matrix $C$ of a polynomial $g(x)$ must satisfy the equation $g(C)=0$". Could you further explain the fact? – X.H. Yue Mar 27 '24 at 07:37
  • @ming alex I've added a link to a proof that a polynomial $\ g\ $ is the characteristic polynomial of its companion matrix $\ C\ $, and noted that the identity $\ g(C)=0\ $ then follows from the Cayley-Hamilton theorem. – lonza leggiera Mar 27 '24 at 08:38
  • I have verified the correctness of this method though the Mathematica software. Thank you! – X.H. Yue Mar 28 '24 at 03:34
  • Hello again! I have another question based on your previous response. Is the companion matrix $C$ the unique non-zero solution of $g(x)$ such that $g(C)=0$? – X.H. Yue Apr 18 '24 at 06:43
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    No. If $\ g(x)\ $ has roots $\ r_1,r_2,\dots,r_s\ $ of multiplicities $\ m_1,m_2,\dots,m_s\ ,$ respectively, then a matrix $\ M\ $ will satisfy the equation $\ g(M)=0\ $ if and only if $\ r_1,r_2,\dots,r_s\ $ are its eigenvalues, and for each $\ i\ $ $\ m_i\ $ is the maximum size of the blocks of its Jordan Normal form corresponding to eigenvalue $\ r_i\ .$ There are an infinite number of such matrices. – lonza leggiera Apr 19 '24 at 05:02
  • I understand your idea! One additional question is: if there are two linearly independent coefficient vectors $\vec a={a_0,...,a_n},\vec a'={a'_0,...a'_n}$, and the nilpotent matrices $B$ and $B'$ are generated using $\vec a$ and $\vec a'$ respectively, in the case of nilpotent index $k=2$, is the probability of $B\cdot B'=0$ negligible over finite fields? Formally, $Pr[B\cdot B'=0|B \gets F(\vec a),B' \gets F(\vec a'),k=2]\leq \epsilon$, where $F:\mathbb{F}^n \to \mathbb{F}^{kn\times kn}$ is the method for generating matrices as described in my question, and $\epsilon$ is negligible val. – X.H. Yue Apr 19 '24 at 06:11