Um. No, not determinant. Individual eigenvalues. The ordinary dot product of two column vectors $v,w$ is given by the matrix product $v^T w = w^T v$ because the transpose of a 1 by 1 matrix is itself. So suppose your $A$ has a real eigenvalue $\lambda,$ with an eigenvector $v.$ We have $Av = \lambda v,$ and
$$ \lambda v^T v = v^T (\lambda v) = v^T (Av) = (Av)^T v = v^T A^T v = -v^T A v = - \lambda v^T v. $$ Now $v \neq 0,$ so $v^T v \neq 0.$ Thus
$$ \lambda v^T v = - \lambda v^T v $$ means $\lambda = 0.$
So, the only possible real eigenvalue is $0.$ In particular, $1$ is never an eigenvalue, we always have $Av \neq v,$ and $(I-A)v \neq 0.$ Put more simply, $0$ is not an eigenvalue of $(I-A),$ which is thus nonsingular.