Is this a valid proof of "A matrix A is not invertible iff 0 is an eigenvalue of A"? Problem: A matrix $A$ is not invertible iff $0$ is an eigenvalue of $A$
My attempt: 
Suppose $\det(A)=0$ ($A$ is not invertible).
Then $\det(A-\lambda I)=0$ iff $\lambda =0$.
$\therefore A$ is not invertible iff $\lambda = 0$.
I was wondering if I'm using valid logical induction steps to prove this.
 A: It is not valid as you wrote it, but rearranging the same idea could make it correct.  
You wrote "$\det(A-\lambda I)=0$ iff $\lambda =0$".  That is not true in general, and is equivalent to the statement that $0$ is the only eigenvalue of $A$.  You conclude "$A$ is not invertible iff $\lambda = 0$," which is not a clear statement as you haven't quantified $\lambda$. 
However, it is true that $A$ is not invertible if and only if $\det(A) = 0$ if and only if $\det(A-0 I)=0$ if and only if $0$ is an eigenvalue for $A$.  Therefore, $A$ is not invertible if and only if $0$ is an eigenvalue for $A$.  That is valid as long as you are allowed to cite theorems that give you each of the equivalencies above.
Alternatively, perhaps you have a theorem that tells you that a matrix is invertible if and only if its nullspace is trivial.  Having nontrivial nullspace is equivalent to having $0$ as an eigenvalue, directly for the definitions of what those words mean: both mean there exists $v \neq 0$ such that $Av=0$.
