# Condition for linear transformation being invertible

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.

• $\gcd(m_A,m_B)=1$ 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 $C = IC = (m_A(A)h(A) + m_B(A)g(A))C = m_B(A)g(A)C = Cm_B(B)g(B) = 0$. – Arturo Magidin Oct 3 '18 at 4:00

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 $$C=Cm_A(B)u(B)+Cm_B(B)v(B)=0.$$
• $\gcd$ equal to $1$, not $0$... – Arturo Magidin Oct 3 '18 at 4:09