Relation Between Determinant and Eigenvalue Well,
If $\det(A-\lambda I)=0$ and also $\lambda\neq 0$, how can you conclude that $\det(A)\neq 0$
Since $\det$ is not a linear operator, I do not see it as an immediate result, but may be I am missing something silly.
EDIT I mean all eigenvalues are nonzero, not just one.

you can't conclud it. 
$$\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} $$  is a counter exampel (with $\lambda=1$)
If $0$ is not an eigenvalue you know that $\det(A - 0 I)\neq 0$ (when it is $0$ it is an eigenvalue).
$$\det(A-0 I) = \det(A) \neq 0$$
