# Characteristic polynomial of a A agrees with its minimal polynomial if and only if all matrices that commutes with A is a polynomial of A

Let $$A \in \mathcal{M}_n (\mathbb{C})$$. Over $$\mathbb{C}$$, show that the following two statements are equivalent:

1. Characteristic polynomial of a $$A$$ agrees with its minimal polynomial;
2. All matrices that commutes with $$A$$ is a polynomial of $$A$$.

For 1 > 2, from 1 I can only get that $$A$$ has a Jordan canonical form that each eigenvalue has only one Jordan block (i.e. maximum size), and I have no idea afterwards.

2 > 1 is a complete no-go for me.

Any hints or solutions are appreciated.

## 1 Answer

$$(2)\to (1)$$ Suppose for the sake of contradiction that (1) does not hold. Let $$J=T^{-1}AT$$ be a Jordan canonical form of $$A$$ and $$T$$ be a non-singular matrix. By Claim from Ben Grossmann’s answer, $$J$$ has distinct Jordan cells $$J_1$$ and $$J_2$$ corresponding to the same eigenvalue $$\lambda$$. Let $$R_k$$ be the set of rows occupied by $$J_k$$ for $$k=1,2$$. Let $$D=\|d_{ij}\|$$ be a diagonal matrix such that $$d_{ii}$$ equals $$1$$, if $$i\in R_1$$, and equals $$0$$, otherwise. Let $$B=\|b_{ij}\|$$ be any polynomial of $$J$$. Then $$b_{ii}=b_{jj}$$ for any $$i\in R_1$$ and $$j\in R_2$$, so $$D$$ is not a polynomial of $$J$$. On the other hand, it is easy to check that $$DJ=JD$$. Then a matrix $$TDT^{-1}$$ commutes with $$A$$, but $$TDT^{-1}$$ is not a polynomial of $$A$$, a contradiction.

$$(1)\to (2)$$ Gerry Myerson proved here that it suffices to show that there exists a vector $$v$$ such that vectors $$A^0v,A^1v,\dots,A^{n-1}v$$ are linearly independent. Let $$\chi(x) = (x - \lambda_1)^{d_1} \cdots (x - \lambda_k)^{d_k}$$ be the characteristic polynomial of $$A$$ and $$\lambda_i$$’s are distinct. For each $$i=1,\dots, k$$ put $$\chi_i(x)=\chi(x)/(x-\lambda_i)$$ and $$\bar\chi_i(x)=\chi(x)/(x-\lambda_i)^{d_i}$$. Since $$\chi_i(x)$$ is not a minimal polynomial of $$A$$, there exists a vector $$w_i$$ such that $$\chi_i(A)w_i\ne 0$$. Let $$\varphi_i(x)$$ be a polynomial of minimal degree such that $$\varphi_i(A)w_i=0$$. The minimality of degree of $$\varphi_i(x)$$ easily implies that $$\varphi_i(x)|\chi(x)$$. Put $$v_i=\bar\chi_i(A)w_i$$ and $$v=v_1+\dots+v_k$$. Suppose for the sake of contradiction that there exists a polynomial $$\psi(x)$$ with minimal degree $$\deg \psi(x) <\deg \chi(x)$$ such that $$\psi(A)v=0$$. The minimality of degree of $$\psi(x)$$ easily implies that $$\psi(x)|\chi(x)$$. Since $$\deg\psi(x)<\deg\chi(x)$$, there exists $$i=1,\dots, k$$ such that $$\psi(x)|\chi_i(x)$$. Then $$0=\chi_i(A)v=\chi_i(A)v_i=\chi_i(A)\bar\chi_i(A)w_i.$$ The minimality of degree of $$\varphi_i(x)$$ easily implies that $$\varphi_i(x)| \chi_i(x)\bar\chi_i(x)$$. Since $$\varphi_i(x)|\chi(x)$$, we have that $$\varphi_i(x)|(\chi_i(x)\bar\chi_i(x),\chi(x))=\chi_i(x)$$. But then $$\chi_i(A)w_i=\varphi_i(A)w_i=0$$, a contradiction.

• Why is $$\chi_i(A)v = \chi_i(A) v_i \quad ?$$ And what is the possible intuition behind these construction? – macton Jan 22 at 7:29
• Sorry for the ping, but I want to really make sure that I understand before sending out the 50 points. – macton Jan 22 at 18:17
• @macton The identity $\chi_i(A)v=\chi_i(A)v_i$ follows from the identity $\chi_i(A)v_j=0$ when $i\neq j$. To see why this latter identity is true, note that $\chi_i(A)v_j=\chi_i(A)\bar{\chi}_j(A)w_j$ ; now the polynomial $P=\chi_i\bar{\chi}_j$ is a multiple of $\chi$, so $P(A)=0$. – Ewan Delanoy Jan 23 at 12:43
• @EwanDelanoy got it, thanks! – macton Jan 24 at 7:06