Sufficient condition for eigenvalues Suppose that I have a matrix $A$ and  I want to find its eigenvalues. I then set up the equation
$$Av=\lambda v$$
which becomes 
$$(A-\lambda I)v =0$$
Now I now that this is true if the determinant of $A-I$ is 0, and this is a necessary condition, but how do I know that this is sufficient for finding all eigenvalues?

You messed up on the details.   So,  starting at $$Av=\lambda v$$, we subtract and insert the identity operator: $$Av-\lambda Iv=0$$.  Now, using linearity of matrix addition, we can "factor out" the vector, to be left with $$(A-\lambda I)v=0$$.
This says that $v$ is in the null space of $A-\lambda I$.  If we want a nontrivial solution (which we need a non 0 vector for eigenvector), we need the null space to be nontrivial, this is the case if and only if the operator is not invertible.  A matrix/operator is not invertible if and only if the determinant is zero,  so it is necessary and sufficient that $\lambda$ is an eigenvalue if $$\det (A-\lambda I)=0$$
