I am reading a proof of the result: a square matrix $A$ is invertible if and only if $\lambda = 0$ is not an eigenvalue of $A$.
Proof goes as follows: Assume that $A$ is an $n\times n$ matrix and observe that $\lambda = 0$ is the solution of the characteristic equation of $A$ given by
$det (\lambda I - A) = \lambda^n + c_1 \lambda^{n-1} + \ldots c_n = 0 $.
On setting $\lambda = 0$ in above equation, we get $det(-A) = c_n$ or $(-1)^n det(A) = c_n$ Now it follows that $det(A) = 0$ if and ony if $a_n \neq 0$.
I am curious to know that if I want to write the characteristic equation of $A$ as $det( A - \lambda I)$ then what would be the general form of characteristic equation in terms of polynomial in $\lambda$.
Can I still write the characteristic equation of $A$ as
$det( A - \lambda I)$= $\lambda^n + c_1 \lambda^{n-1} + \ldots c_n = 0 $.
Please clarify my doubt. As writing in this form changes the fact that the constant term of characteristic polynomial is $(-1)^n det(A)$, where $n$ is the order of the matrix.
Thank you