Minimal polynomial, determinants and invertibility I need to prove the following:

Let $A$ be a square matrix over a field. If the matrix $A$ is invertible, then the minimal polynomial $m_A$ satisfies $m_A(0) \neq 0$.

There is one definition I am unsure of or need help making more clear.
I will proceed with proof by contraposition:
We must show that if $m_A(0) = 0$ then $A$ is not invertible. By definition of minimum polynomial of $A$ we have:
$m_A(x) = x^r - \lambda_{r-1} x^{r-1} - \ldots - \lambda_1 x + \det(A)$. Not sure about the determinant term here
So, $m_A(0) = \det(A) = 0$. We know $\det(A) = 0 \implies A$ is not invertible.  
 A: Here's another method, if it helps:
We know that $m_A(A)=0,$ thus if $m_A(0)=0,$ i.e. if there is no constant term, then we can write $A^r-\lambda_{r-1}A^{r-1}+\cdots\pm\lambda_1 A=(A^{r-1}-\lambda_{r-1}A^{r-2}+\cdots\pm\lambda_1I)A=0.$ Since $A$ is invertible, multiplying on the right by $A^{-1}$ shows that $A$ satisfies a polynomial of lesser degree than its minimal polynomial, giving a contradiction.
A: You have a good shot, but there are two errors in your attempted proof.


*

*In general, the constant term of the characteristic polynomial of $A$ is a (matrix independent) multiple of $\det A$, but the constant term of the minimal polynomial is not.

*Even if we are talking about the characteristic polynomial, your sign of the constant term is not correct. The characteristic polynomial is $p_A(x)=\det(xI-A)$. Hence $p_A(0)=\det(-A)=(-1)^r\det A$. Although we don't really need the sign to answer this question here, it would be better if we get every detail right.


For a correct proof following your line of thought, see Jacob Schlather's answer. To learn a different perspective, you should read Andrew's answer too.
A: Here is another way. Note that $m_A(x) \mid \mathrm{char}_A(x)$ in particular if $m_A(0)=0$ then $0$ is an eigenvalue of $A$. So $A$ has non-trivial kernel and is thereby not invertible. 
