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Suppose $A \in \mathcal{M}_{n}(\mathbb{C})$ is an $n$ by $n$ matrix over the field $\mathbb{C}$. Prove that the following two statements are equivalent.

  1. The (monic) characteristic polynomial of $A$ is equal to the minimal polynomial of $A$.
  2. For any $X \in \mathcal{M}_{n}(\mathbb{C})$ satisfying $AX=XA$, there exists a polynomial $p$ such that $X=p(A)$.

Comments: I found that (2)=>(1) can be done by assuming that the Jordan form of $A$ can be split into two Jordan blocks corresponding to the same eigenvalue $\lambda$. Letting $D=\text{diag}\{1,1,\cdots,1,0,0,\cdots,0\}$ and setting $B=PDP^{-1}$, where $P$ is the Jordan basis of $A$, we can easily show that $AB=BA$ but $B$ is not a polynomial of $A$.

But for the (1)=>(2) direction, I still have no idea yet. Could someone help?

Yen-Chi Lee
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  • It is about the Jordan blocks; the conditions are unchanged if we demand $A$ in Jordan form. For a 2 by 2 block, call it $A= \left( \begin{array}{cc} \lambda&1 \ 0& \lambda \end{array} \right) $ and $ X=\left( \begin{array}{cc} p&q \ r&s \end{array} \right), $ what are the exact conditions on $p,q,r,s$ that make these blocks commute? – Will Jagy Nov 09 '21 at 02:40
  • For a 3 by 3 block, call it $A= \left( \begin{array}{ccc} \lambda&1&0 \ 0& \lambda&1 \ 0&0&\lambda \end{array} \right) $ and $ X=\left( \begin{array}{ccc} p&q&t \ r&s&u \ v&w&z \end{array} \right), $ what are the exact conditions..... for this part of the experiments, there is no loss in setting the $\lambda = 0$ – Will Jagy Nov 09 '21 at 02:48
  • Thanks for your hint. I followed your approach, and found that $AX=XA$ exactly implies $X=aI + bN+cN^2$ for 3 dimension, and can be verified to be a polynomial of $A$. But for the case that $A$ has two or more different eigenvalues, I cannot reproduce the result. Say $A=\begin{bmatrix} \alpha&1&0&0\ 0&\alpha&0&0\ 0&0&\beta&1\ 0&0&0&\beta \end{bmatrix}$. – Yen-Chi Lee Nov 09 '21 at 06:03
  • Right; more is needed then. You now have an answer by one of the matrix experts on MSE. One aspect of characteristic and minimal polynomials coinciding is that each eigenvalue gets just one Jordan block. – Will Jagy Nov 09 '21 at 17:21

1 Answers1

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As $A$ is non-derogatory, $\mathbb C^n$ has an $A$-cyclic basis $\mathcal B=\{v,Av,A^2v,\ldots,A^{n-1}v\}$. Let $Xv=c_0v+c_1Av+\cdots+c_{n-1}A^{n-1}v$ and let $f$ be the polynomial $c_0+c_1x+\cdots+c_{n-1}x^{n-1}$. Since $A$ and $X$ commute, $XA^kv=A^kXv=A^kf(A)v=f(A)A^kv$ for each basis vector $A^kv$. Therefore $X=f(A)$ on $\mathcal B$ and in turn, $X=f(A)$ on $\mathbb C^n$.

PS, In your comments I see that you are interested in a proof using Jordan forms. Denote by $J_m(\lambda)$ a Jordan block of size $m$ for an eigenvalue $\lambda$. Suppose $A=J_{m_1}(\lambda_1)\oplus\cdots\oplus J_{m_r}(\lambda_r)$ where $\lambda_1,\ldots,\lambda_r$ are distinct. As you may already know, it can be shown that when $X$ commutes with such an $A$, it must be in the form of $X_1\oplus\cdots\oplus X_r$ where $X_i=g_i\left(J_{m_i}(\lambda_i)\right)$ for some polynomial $g_i$.

It remains to construct a polynomial $f$ from $g_1,\ldots,g_r$ that makes $X=f(A)$. For each fixed $i$, let $p_i(x)=\prod_{k\ne i}(x-\lambda_k)^{m_k}$. Since all $\lambda_i$s are distinct, $p_i$ and $(x-\lambda_i)^{m_i}$ are coprime. Therefore, by Bézout's lemma, there exist two polynomials $a_i$ and $b_i$ such that $$ a_i(x)p_i(x)+b_i(x)(x-\lambda_i)^{m_i}=1. $$ It follows that $(a_ip_i)\left(J_{m_j}(\lambda_j)\right)$ is equal to $I_{m_i}$ when $j=i$, or $0$ when $j\ne i$. Therefore \begin{aligned} (a_ip_ig_i)(A) &=(a_ip_ig_i)\left(J_{m_1}(\lambda_1)\right) \oplus\cdots\oplus (a_ip_ig_i)\left(J_{m_r}(\lambda_r)\right)\\ &=0\oplus\cdots\oplus0\oplus\underbrace{X_i}_{\text{$i$-th}}\oplus0\oplus\cdots\oplus0. \end{aligned} Consequently, $X=\sum_{i=1}^r(a_ip_ig_i)(A)$.

user1551
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  • Thanks for your reply! I can understand your argument for a single invariant-subspace of $A$. But in my understanding, a non-derogatory $A$ can have two or more invariant-subspaces corresponding to different eigenvalues; in this case, we could have two or more cyclic vectors, say $v_1$ and $v_2$. Following your construction, for each $X$ we obtain two polynomials $f_1$ and $f_2$ (the coefficients may be different?) with $X=f_1(A)$ on $B_1$ and $X=f_2(A)$ on $B_2$. Is there any method to further combine these two polynomials into a single polynomial defined on the whole vector space $C^n$? – Yen-Chi Lee Nov 10 '21 at 00:59
  • @Yen-ChiLee I collected some equivalent statements and links to some proofs at https://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 One of the inks is to my MO question, originally titled When Animals Attack https://mathoverflow.net/questions/65796/when-matrices-commute – Will Jagy Nov 10 '21 at 03:11
  • @Yen-ChiLee Whether $A$ has more than one generalised eigenspaces is unimportant. As long as $A$ is non-derogatory, it has a cyclic vector $v$ that generates a basis of $\mathbb C^n$. This is usually proved in textbooks in the sections about companion matrices. – user1551 Nov 10 '21 at 06:48
  • @user1551 Thank you for the detailed explanation! – Yen-Chi Lee Nov 10 '21 at 07:19
  • @WillJagy Thank you for the helpful links! – Yen-Chi Lee Nov 10 '21 at 07:20