Prove that $A+I$ is invertible if $A$ is nilpotent 
Possible Duplicate:
Units and Nilpotents 

Given $A^{2012}=0$ prove that $A+I$ is invertible and find an expression for $(A+I)^{-1}$ in terms of $A$. ($I$ is the identity matrix).
 A: You can easily prove that if $A^n=0$:
$$\left(A+I\right)\left(I-A+A^2-...+(-1)^n A^{n-1}\right) = I +(-1)^{n-1} A^n = I$$
Thus proving that $A+I$ is invertible for any nilpotent $A$.
A: Hint: the series $\sum_{j=0}^{+\infty}(-1)^jA^j$ has only finitely many terms. In fact, it works it there is a positive integer $p$ such that $A^p=0$. 
A: A matrix $A$ is nilpotent if and only if all its eigenvalues are zero. It is not hard also to see that the eigenvalues of $A+I$ will all be equal to $1$ (when we add $I$ to any matrix, we just shift its spectrum by 1). Thus $A+I$ is invertible,
since all its eigenvalues are non-zero. 
A: More generally: A (square) matrix $A$ is invertible if and only if $\lambda=0$ is not an eigenvalue.
Independently of this, we have that if $\lambda$ is an eigenvalue of $A$, then $\lambda+\mu$ is an eigenvalue of $A+\mu I$: if $\mathbf{x}$ is an eigenvector of $A$ corresponding $\lambda$, then $(A+\mu I)\mathbf{x} = A\mathbf{x}+\mu I\mathbf{x} = \lambda\mathbf{x}+\mu\mathbf{x} = (\lambda+\mu)\mathbf{x}$.
Also independently of all of this, if $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$. (Careful, though, this one is not reversible: a rotation of $90^{\circ}$ of the plane has no eigenvalues over $\mathbb{R}$, but its square has eigenvalue $-1$. The square of the identity has $(-1)^2$ as an eigenvalue, but the identity does not have $-1$ as an eigenvalue).
So: 
$$\begin{align*}
A+I\text{ is invertible}&\iff 0\text{ is not an eigenvalue of }A+I\\
&\iff -1\text{ is not an eigenvalue of }A.
\end{align*}$$
And if $A^n=0$ for some $n\gt 0$, then $-1$ is not an eigenvalue of $A$.
