Why is $1-\lambda$ an eigenvalue of $I -A$?

Question

someone posed this problem to me and it may be lack of sleep but i can't really figure it out. he said it was an easy problem too.

ok so i have tried just assuming A is a 2x2 matrix so the determinant of the matrix that equals zero in the end looks like $$\begin{pmatrix}1-a_{11}-\lambda & -a_{12}\\-a_{21} &1-a_{22}-\lambda\end{pmatrix}$$

so then i tried to get the determinant of it but it is this long polynomial that needs to be condensed, i can't help but think i am going the wrong way about it.

Answer 1

Suppose $Ax=\lambda x$, then $(I-A)x = Ix - Ax = x - \lambda x = (1-\lambda)x$.