Let $A\in\Bbb{K}^{n\times n}$ and $\mu_A$ the minimal polynomial of $A$. How can I show that $A$ is invertible and only if $\mu_A(0)\neq0$? 
Let $\mathbb K$ be a field, $n\in \mathbb N$, $A\in \mathbb{K}^{n\times n}$ and $\mu_A \in \mathbb{K}[X]$ the minimal polynomial of $A$. How can I show that $A$ is invertible and only if $\mu_A(0)\neq0$?

 A: Assume $\mu_A(0)\ne 0$ first. We have
$\mu_A(X) = \displaystyle \sum_0^{\deg \mu_A} m_i X^i, \; m_i \in \Bbb K; \tag 1$
thus
$m_0 = \mu_A(0) \ne 0; \tag 2$
since
$\displaystyle \sum_0^{\deg \mu_A} m_i A^i = \mu_A(A) = 0, \tag 3$
we may write
$\displaystyle \sum_1^{\deg \mu_A} m_i A^i = -m_0 I \ne 0, \tag 4$
which we may re-arrange to obtain
$A \left (-\dfrac{1}{m_0} \displaystyle \sum_1^{\deg \mu_A} m_i A^{i - 1} \right ) = I, \tag 5$
which shows that
$A^{-1} = -\dfrac{1}{m_0} \displaystyle \sum_1^{\deg \mu_A} m_i A^{i - 1}, \tag 6$
so (2) implies $A$ is invertible.  Going the other way, assuming $A^{-1}$ exists, from (3) we may write
$\displaystyle \sum_1^{\deg \mu_A} m_i A^i = -m_0 I, \tag 7$
whence
$\displaystyle \sum_1^{\deg \mu_A} m_i A^{i - 1} = -m_0 A^{-1}; \tag 8$
now if
$m_0 = \mu_A(0) = 0, \tag 9$
then 
$\displaystyle \sum_1^{\deg \mu_A} m_i A^{i - 1} = 0; \tag{10}$
but the degree of 
$\displaystyle \sum_1^{\deg \mu_A} m_i X^{i - 1} \tag{11}$
is $\deg \mu_A - 1$, contradicting the minimality of $\mu_A(X)$; we conclude then that (8) cannot bind and that (2) holds.
A: The condition $\mu_A[0]=0$ means precisely that $X$ divides $\mu_A$. Assume first that this is the case, so that $\mu_A=XQ$ for some polynomial$~Q$, which is nonzero (since $\mu_A$ is). Because $\deg(Q)<\deg(\mu_A)$, it cannot be that $Q$ is an annihilating polynomial of$~A$, so the image $V$ of $Q[A]$ is a subspace of nonzero dimension. But $ 0=\mu_A[A]=(XQ)[A]=A\circ Q[A]$ shows that $V\subseteq\ker(A)$, whence $A$ cannot be invertible. Conversely, suppose that $A$ is not invertible, so it has some eigenvector$~v$ with eigenvalue$~0$ (that is, a nonzero element $v\in\ker(A)$). Then for any polynomial$~P$ one has $P[A](v)=P[0]v$ (since $A$ acts on $v$ as the scalar$~0$), in particular $0=0(v)=\mu_A[A](v)=\mu_A[0]v$, whence $\mu_A[0]=0$ since $v\neq0$.
