What is wrong with this "proof"? $1$ is always an eigenvalue for $I + A$ ($A$ is nilpotent)? 
Consider the nilpotent matrix $A$ ($A^k = 0$ for some positive $k$).
  It is well known that the only eigenvalue of $A$ is $0$.
Then suppose $\lambda$ is any eigenvalue of $I + A$ such that 
  $(I + A) \mathbf{v} = \lambda \mathbf{v}$ where ($\mathbf{v} \neq \mathbf{0}$).
Then $I \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies 
 \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies A \mathbf{v} =
(\lambda - 1) \mathbf{v}$
We know that $\lambda - 1 = 0$ because the only eigenvalue of a
  nilpotent matrix is $0$. Therefore $\lambda = 1$

This "proof" seems to indicate that $1$ is always an eigenvalue for the sum of the identity matrix with any nilpotent matrix, but I believe I have a counterexample that disproves this.
I believe my error was in assuming that $I + A$ has eigenvalues -- but I do not know how I could prove/disprove this.
If someone could help me see where I've gone wrong I would greatly appreciate it!

Here's an alternate argument (assuming the base field is algebraically closed):


*

*For a nilpotent matrix with $A^k=0$, all eigenvalues are $0$.
Proof:  Let $\lambda$ be an eigenvalue for $A$ and $v\not=0$ an eigenvector for $\lambda$.  Then
$$
0=0v=A^kv=\lambda^kv
$$
Since $v\not=0$, it must be that $\lambda^k=0$ so that $\lambda=0$

*$1$ is an eigenvalue for $I+A$.  Let $v$ be an eigenvector for $A$.  
Proof: We know that all eigenvalues for $A$ are $0$, so $Av=0$.  Therefore,
$$
(I+A)v=Iv+Av=v+0=v.
$$
Hence, $1$ is the eigenvalue for $v$ in $I+A$.

*You can prove more, every eigenvalue of $I+A$ is $1$.
Proof: Suppose that $\mu$ is an eigenvalue for $I+A$ and $w$ the corresponding eigenvector.  Then
$$
Aw=((I+A)-I)w=(I+A)w-Iw=\mu w-w=(\mu-1)w.
$$
Therefore, $w$ is an eigenvector for $A$ with eigenvalue $\mu-1$.  Since the only eigenvalues for $A$ are $0$, $\mu=1$.
