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How to prove that any annihilating polynomial $p(x)\in \Bbb C[x]$ of an $n \times n$ matrix $A$ with entries from $\Bbb C$ must contain all irreducible factors of the characteristic polynomial $\chi_A(x)$. In other words, why can't we annihilate a polynomial $q(x) \in \Bbb C[x]$ by plugging $A$ if $q$ lacks an irreducible factor from $\chi_A(x)$.

I feel that, we have to prove there exists a commen factor (a least commen factor later to be called as minimal polynomial $m_A(x)$) for each annihilating polynomial $p (x)\in \Bbb C[x]$.

Is it enough proving that $\chi_A(x)$ and $m_A(x)$ possess same set of irreducible factors to say any annihilating polynomial must contain all these irreducible factors? How it is?

For instance, $p(x)=x^2-1$ is an annihilating polynomial for the $2\times 2$ identity matrix $I_2$, but it's not the chara. polynomial while $p$ possess the commen factor $x-1$.

How to prove in general?

Thanks in advance.

Riaz
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2 Answers2

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This simply stems from the definition of the minimal polynomial: it is the monic generator of the principal ideal of all annihilating polynomials.

Anne Bauval
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$\chi_A(x) = g^{n_1}_{\lambda_1}(x) g^{n_2}_{\lambda_2}(x)....g^{n_L}_{\lambda_L}(x)$ where $g_{\lambda_i}(x)$ is the irreducible polynomial which has $\lambda_i$ as root. Since we are talking about complex numbers: $g_{\lambda_i}(x) = (x-\lambda_i)$.

Hence : $\chi_A(x) = (x-\lambda_1)^{n_1} (x-\lambda_2)^{n_2} .... (x-\lambda_L)^{n_L}$.

Now let $m(x)$ be minimal polynomial of $\chi_A(x)$. Hence $m(x) g(x) = \chi_A(x)$.

If wlog $(x-\lambda_1) \nmid m(x)$ and $m(x) = (x-\lambda_2)^{m_2} .... (x-\lambda_L)^{m_L}$ then let $v_{\lambda_1}$ be the eigen vector corresponding to eigenvalue $\lambda_1$. Then $m(A)v_{\lambda_1} = (\lambda_1-\lambda_2)^{m_2} .... (\lambda_1-\lambda_L)^{m_L} v_{\lambda_1} \neq 0$ a contradiction to $m(A) = 0$.

Hence we have $(x-\lambda_i) | m(x)$ for every $\lambda_i$ eigenvalue and $(x-\lambda_i) | m(x)$ iff $(x-\lambda_i) | \chi_A(x)$.

Since $I = \{f(x) : f \in \mathbb{C}[x], f(A) = 0\}$ has a polynomial of least degree, call it $p(x)$ then let $f(x) \in I$ then by division algorithm $f(x) = w(x)p(x)+r(x)$ with $deg(r) < deg(p)$ hence we have $r(A) = 0$ contradicting the minimality of degree of $p(x)$ and hence $r(x) = 0$ and hence $I$ generated by a single element $p(x)$, $I = <p(x)>$ i.e. $f(x) \in I$ iff $p(x) | f(x)$. Its clear that $m(x) = p(x)$ assuming both $m(x),p(x)$ are monic polynomials.

Let $q(A) = 0$. Hence $m(x) \ | \ q(x)$. Hence $(x-\lambda_i) \ | \ m(x) \ | \ q(x)$ and hence $(x-\lambda_i) \ | \ q(x)$.

Balaji sb
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  • Can you make the statement '' hence we have $r(A) = 0$ contradicting the minimality of degree of $p(x)$ and hence $r(x) = 0$'' clear to me, Thanks in advance. – Riaz Dec 23 '22 at 08:28
  • Since $deg(r)<deg(p)$ and $r(A) = 0$ means $r(x) \in I$ but with degree of $r(x)$ being smaller than $p(x)$ but $p(x)$ is chosen as the element from $I$ with least degree. This is a constradiction. Cheers ! – Balaji sb Dec 23 '22 at 08:50
  • So I can conclude $r(A) \neq 0$, how to say ''hence $r(x)=0$''? – Riaz Dec 23 '22 at 08:53
  • $f(x) = w(x)p(x)+r(x)$. So $f(A) = w(A)p(A)+r(A)$. Since $f(A) = p(A) = 0$ the expression $f(A) = w(A)p(A)+r(A)$ implies $0 = 0 + r(A)$. Hence $r(A) = 0$. – Balaji sb Dec 23 '22 at 09:01
  • Ya...its clear now, thnks a lot... – Riaz Dec 23 '22 at 09:11
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    For future readers: According to Riaz's refusal of the proposed duplicate, only the second part of this answer (beginning with "Since $I = {f(x) : f \in \mathbb{C}[x], f(A) = 0}$ has a polynomial of least degree") is relevant. The first part was proved in the proposed duplicate. The second part details the definition of the minimal polynomial as the monic generator of that principal ideal $I.$ – Anne Bauval Dec 23 '22 at 13:35