If a square matrix $A$ has eigenvalue $1$, then $I-A$ is singular?

Question

I am a bit confused about the title. Consider the following

$$I-A = I - U^{-1}JU = U^{-1}U - U^{-1}JU = U^{-1}(I-J)U$$

However, for example if $$J = \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$$ then $$I-J = \begin{bmatrix}0 & -1 \\ 0 & 0 \end{bmatrix} $$

this is not a Jordan block and how to deal with $-1$? It seems we cannot include $-1$ to $U$; otherwise, we cannot get $U$ and $U^{-1}$.

How to show this? thanks!

Answer 1

Your thinking is too advanced. If $v$ is an eigenvector of $A$ with eigenvalue $1$, then $(I-A)v=0$, meaning $I-A$ has non-trivial kernel and is singular.