Eigen values of $I+A$ if $A^2=0$ Let $A\in M_{n\times n}(\mathbb{C})$. Suppose that $A^2=0$. Show that $\lambda$ is an eigenvalue of $A+I$ if and only if $\lambda=1$
As $A^2=0$, $A^2-t^2I=-t^2I$ i.e., $(A+tI)(A-tI)=-t^2I$ this imply $A-tI$ is invertible iff $t\neq 0$.
So, $(A+I)-\lambda I=A-(\lambda-1)I$ is invertible iff $\lambda-1=0$ i.e., $\lambda=1$. So, $1$ is the only eigen value of 
$I+A$.
After posting this question i got this idea.. Let me know if this is fine..
 A: Suppose $\lambda$ is an eigenvalue of $A+I$ and $v \neq 0$ is an eigenvector of $A + I$ with respect to $\lambda$. Then
$$Av = (\lambda - 1)v \implies 0 = A^2v = (\lambda - 1)^2 v \implies (\lambda - 1)^2 = 0 \implies \lambda = 1.$$
This shows that if $\lambda$ is an eigenvalue of $A+I$, $\lambda$ has to be $1$.
On the other hand, we have
$$A^2 = 0 \implies \det(A)^2 = \det(A^2) = 0 \implies \det(A) = 0,$$
which shows $\lambda = 1$ is indeed an eigenvalue of $A+I$.
A: Since $A^2=0$, we have
$$
(I+A)^n=I+nA
$$
. Let $v$ be an eigenvector of $I+A$, then
$$\begin{align}
(I+A)^nv &= (I+nA)v \\
\lambda^nv &= v+nA v \\
\frac{(\lambda^n-1)}{n}v &= Av
\end{align}$$
for all $n\in \Bbb N$. Since $v\ne 0$, the term on the left implies that $\lambda=1$ or else we'd have a contradiction ($Av$ not well defined).
A: 
$\lambda$ is an eigenvalue of $A$ if and only if $\lambda+a$ is an eigenvalue of $A+aI$.

Sketch of proof:
$$
Av=\lambda v\iff Av+av=\lambda v+av
$$

If $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$.

Sketch of proof: if $Av=\lambda v$, then
$$
A^{n+1}v=A^n(Av)=A^n(\lambda v)=\lambda(A^nv)=\lambda(\lambda^nv)=
\lambda^{n+1}v
$$
So the only eigenvalue of $A$ is $0$, and the result follows.
A: If $\vec v$ is a nonzero vector, then $$A^2 \vec v = 0_{n\times n}\cdot \vec v = 0\cdot \vec v,$$ where $0_{n\times n}$ is the $n\times n$ zero matrix. Hence every (nonzero) vector is an eigenvector belonging to the eigenvalue $0$.
The eigenvalues of $A^2$ are the eigenvalues of $A$, squared. Hence. The eigenvalues of $A$ are all $0$.
The eigenvalues of $A+I$ are the eigenvalues of $A$ plus one; hence there is only one eigenvalue of $A+I$, namely $0+1=1$.
