Prove $T \in L(V)$ is invertible $\iff$ Constant term of $m_T(x)$ is not zero. 
Prove that $T \in L(V)$ is invertible if and only if the constant term of the minimal polynomial of $T$ is not zero.

My attempt: I will use contrapositive argument, contrapositive of "$a\implies b$" is "Not $b \implies$ Not $a$". So it is enough to show that constant term is zero in $m_T(x)$ $\iff$ $T$ is not invertible. 
Constant term of $m_T(x)$is Zero $\iff$ $m_T(x) = xp(x) \iff$ $0$ is a root of characteristic polynomial ( since zeroes of minimial and charcteristic polynomial for T are the same) $\iff$ An eigen value of $T$ is $0 \iff$ det$([T]) = 0$ (Since it is product of all eigen values)$\iff$ T is not invertible. 
Is this proof correct? Are there any other ways to solve this problem? 
 A: This is correct.
Alternatively, suppose $m_T(0) = 0$, i.e., the constant term of $m_T$ is $0$, and $T$ is invertible. Let $p(x) = m_T(x)/x$. Then
$$T(p(T)) = m_T(T) = 0.$$
Since $T$ is invertible, this implies that $p_T(T) = 0$, contradicting that $m_T$ is the minimal polynomial of $T$.
On the other hand, suppose $m_T(0) \neq 0$. If there exists some nonzero vector $v \in \ker(T)$, then
$$m_T(T)(v) = m_T(0)v \neq 0.$$
This is a contradiction since $m_T(T) = 0$. Therefore no such vector exists, hence $T$ is invertible.
A: I assume $V$ is a vector space over the field $F$.
Let
$m_T(x) = \displaystyle \sum_0^{\deg m_T} m_i x^i \in F[x] \tag 1$
be the minimal polynomial of
$T \in L(V); \tag 2$
we have
$m_T(T) =  \displaystyle \sum_0^{\deg m_T} m_i T^i = 0. \tag 3$
Now suppose $T$ is invertible, and that
$m_0 = m_T(0) = 0; \tag 4$
then $m_T(x)$ may be written
$m_T(x) = \displaystyle \sum_1^{\deg m_T} m_i x^i = x\sum_1^{\deg m_T} m_i x^{i - 1}, \tag 5$
and in light of (3) we have
$T\displaystyle \sum_1^{\deg m_T} m_i T^{i - 1} = m_T(T) = 0; \tag 6$
with $T$ invertible this yields
$\displaystyle \sum_1^{\deg m_T} m_i T^{i - 1} = (T^{-1}T) \sum_1^{\deg m_T} m_i T^{i - 1}$
$= T^{-1} \left (T \displaystyle \sum_1^{\deg m_T} m_i T^{i - 1} \right ) = T^{-1} m_T(T) = 0, \tag 7$
which shows that $T$ satisfies the polynomial
$p_T(x) = \displaystyle \sum_1^{\deg m_T} m_i x^{i - 1} \tag 8$
of degree
$\deg p_T(x) = \deg m_T - 1; \tag 9$
but now the fact that
$p_T(T) = 0 \tag{10}$
contradicts the hypothesis that $m_T(x)$ is the minimal polynomial of $T$; thus
$m_0 \ne 0. \tag{11}$
Now suppose that
$m_0 \ne 0; \tag{12}$
then (3) may be written in the form
$T\displaystyle \sum_1^{\deg m_T} m_i T^{i - 1} = -m_0I, \tag{13}$
whence
$T \left( -\dfrac{1}{m_0} \displaystyle \sum_1^{\deg m_T} m_i T^{i - 1} \right ) = I, \tag{14}$
which shows that $T$ is invertible with
$T^{-1} = -\dfrac{1}{m_0} \displaystyle \sum_1^{\deg m_T} m_i T^{i - 1}. \tag{15}$
A: Let $m_T(x) = \alpha_0+\alpha_1x+\cdots+\alpha_kx^k$, where $\alpha_k \neq 0$. Assume, for a contradiction, that $T$ is invertible and that $\alpha_0 = 0$. Then we would have $$0 = \alpha_1T+\cdots+\alpha_kT^k = (\alpha_1I+\cdots+\alpha_kT^{k-1})T.$$ Since we assumed that $T$ is invertible, multiply both sides of this equation by $T^{-1}$, the result would be
$$0=\alpha_1I+\cdots+\alpha_kT^{k-1}.$$ This implies that the polynomial $n_T(x) = \alpha_1+\cdots+\alpha_kx^{k-1}$ is a polynomial with $n_T(T) = 0$. But $\text{deg}(n_T) = k-1<k=\text{deg}(m_T)$, contradicting the fact that $m_T$ is the minimal polynomial.
Conversely, assume that $\alpha \neq 0$. Then $$0 = \alpha_0I+\cdots+\alpha_kT^k$$ $$\alpha_0I=-\alpha_1T-\cdots-\alpha_kT^k$$ $$I = -\frac{\alpha_1}{\alpha_0}T-\cdots-\frac{\alpha_k}{\alpha_0}T^k = \left(-\frac{\alpha_1}{\alpha_0}I-\cdots-\frac{\alpha_k}{\alpha_0}T^{k-1}\right)T.$$ This shows that $T^{-1} = -\frac{\alpha_1}{\alpha_0}I-\cdots-\frac{\alpha_k}{\alpha_0}T^{k-1}$.
