If $\|G\| < 1$, then show that $I-G$ is invertible.
This can be proved by contradiction: If $I-G$ is singular, then 1 is an eigenvalue of $I-I+G$. So if the matrix norm is induced the 2-norm, $\| G\|$ is at least 1 since the largest singular value of a matrix is not less than its eigenvalue in absolute value.
I have two questions: 1) Why does 1 have to be an eigenvalue of $I-A$ if A is singular? and why is the largest singular value of a matrix not less than its eigenvalue in absolute value?
2) I am learning iterative methods in class and i don't believe we have come acrossed eigenvalues yet; is there any other ways to prove this?