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Let $A,B \in M_n(R)$ being fixed matrices. Define $T\in \mathcal{L}(M_n)$ as

$$T(C)=AC-CB.$$

Prove that $T$ is invertible iff $gcd(m_A,m_B) \text{~}1$ where $m_X$ represents the minimal polynomial of $X$.

First, the condition is equivalent to $AC=CB \Leftrightarrow C=0$. I tried to use that $AC=CB \Rightarrow p(A)C=Cp(B)$ for every polynomial $p$ and use $m_A$ and $m_B$ but I couldn't make any further progress.

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    $\begingroup$ $\gcd(m_A,m_B)=1$ implies there exist $h(x),g(x)$ such that $m_A(x)h(x)+m_B(x)g(X)=1$; and by Cayley-Hamilton, $m_A(A)=0$, $m_B(B)=0$. So if $AC=CB$ then $C = IC = (m_A(A)h(A) + m_B(A)g(A))C = m_B(A)g(A)C = Cm_B(B)g(B) = 0$. $\endgroup$ – Arturo Magidin Oct 3 '18 at 4:00
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Taking $p=m_A$ gives $Cm_A(B)=0$. But also $Cm_B(B)=0$ as $m_B(B)=0$.

As $\gcd(m_A,m_B)=0$ there are polynomials $u$ and $v$ with $m_Au+m_Bv=1$. Then $$C=Cm_A(B)u(B)+Cm_B(B)v(B)=0.$$

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    $\begingroup$ $\gcd$ equal to $1$, not $0$... $\endgroup$ – Arturo Magidin Oct 3 '18 at 4:09

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