The Question:

Let $f(x) = x^n+\alpha_{n-1}x^{n-1}+\alpha_{n-2}x^{n-2} + \cdots + \alpha_0$, and let $A$ be the matrix

\begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 \\ -\alpha_0 & -\alpha_1 & -\alpha_2 & -\alpha_3 & \cdots & -\alpha_{n-2} & -\alpha_{n-1} \end{pmatrix}

(i) Find the characteristic polynomial

(ii) Find the minimal polynomial

My Attempt:

(i) No problem here, I found that $\chi _A (x) = (-1)^n f(x)$.

(ii) No idea here, I only know that $m_A(x) \, | \, \chi _A (x)$.

Assuming that $f \in \bar {\Bbb F} [x]$, we can factorize $f$ into linear factors $$f(x) = (\lambda_1 - x)(\lambda_2 - x) \cdots (\lambda_n - x)$$

where the $\lambda_i$ are the eigenvalues of $A$, and then I am still stuck.

Any hints?


OK, after some experiments with smaller matrices, it seems that the minimal polynomial is always the full polynomial, i.e. $m_A(x) = f(x)$.

I have tried induction on the size of the matrix, but it does not seem to work either.

  • $\begingroup$ Hint: If $f$ has a repeated root, what is the dimension of the corresponding eigenspace? $\endgroup$ May 17 '18 at 15:46
  • $\begingroup$ Wow good question. I have no clue either! Have you tried writting f = (x-1)(x-1)(x-2), recovering the matrix, and see what is going on in that case? I would like to know the answer too :) $\endgroup$
    – Student
    May 17 '18 at 16:07
  • $\begingroup$ Look at the number of pivots of $A-\lambda I$ and determine the corresponding Jordan blocks. $\endgroup$ May 17 '18 at 16:17
  • $\begingroup$ I get that if $\lambda$ is a root of $f$ with multiplicity $k$, then the eigenspace for $\lambda$ is $≤k$, but I am not sure how to relate this to Jordan Blocks? $\endgroup$ May 17 '18 at 17:31

I prefer to see this by considering the space $V$ of row vectors of length $n$, with the standard basis $$ v_{0} = (1, 0, \dots, 0), \quad v_{1} = (0, 1, \dots, 0),\quad \quad\dots,\quad v_{n-1} = (0, 0, \dots, 1). $$ The matrix $A$ acts on $V$ by multiplication on the right. Clearly $v_{i} A = v_{i+1}$ for $0 \le i < n-1$, so that $v_{0} A^{i} = v_{i}$ for $0 < i \le n-1$.

Take any polynomial $$ c = c_{0} + c_{1} x + \dots + c_{n-1} x^{n-1} $$ of degree less than $n$, and suppose $$ 0 = c(A) =c_{0} I + c_{1} A + \dots + c_{n-1} A^{n-1}, $$ where $I$ is the $n \times n$ identity matrix.

Then $$ 0 = v_{0} c(A) = v_{0} (c_{0} I + c_{1} A + \dots + c_{n-1} A^{n-1}) = c_{0} v_{0} + c_{1} v_{1} + \dots + c_{n-1} v_{n-1}. $$ Since the $v_{i}$ are linearly independent, all $c_{i}$ are zero, so that the minimal polynomial has degree at least $n$. Since the minimal polynomial divides the characteristic polynomial, and the latter has degree $n$, the two must coincide.


Look at what happens to the matrix as you take powers. Those ones on the superdiagonal move up a diagonal, there are still zeros above them, and more or less anything goes below them. Thus, in order to get a non-trivial linear combination of powers of this matrix to be zero, you need to push the diagonal of ones out of the matrix entirely...otherwise the highest power appearing will have a super-super-....-super diagonal (possibly just the upper right corner) of ones that no other power can subtract out. It's fairly easy to see that you need to take the $n^{th}$ power to clear all the ones. Thus, the minimal polynomial must have degree at least $n$. Since it divides the characteristic polynomial, which also has degree $n$, they must be equal.

  • $\begingroup$ Thank you for your answer. I actually kinda of had this idea at some point,but it seems to me that there are too many subtleties to prove. $\endgroup$ May 19 '18 at 11:07

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