TITLE: Showing that matrix is invertible using eigenvalues
QUESTION [5 upvotes]: Let $A$ be matrix from the  vector space of square $N \times N$ matrices.
With the inital information: $A^2-4A=4I$.
How does one show that $A+I$ is invertible?
(I need please a solution that involves eigenvalues)
Thank you

REPLY [2 votes]: Hint: What does the initial information tell you about the eigenvalues of $A$?

REPLY [2 votes]: Note that a square matrix $B$ is invertible if and only if $\lambda=0$ is not an eigenvalue of $B$.
Thus, 
$$\begin{align*}
A+I\text{ is invertible }&\Longleftrightarrow \lambda=0\text{ is not an eigenvalue of }A+I\\
&\Longleftrightarrow \text{there is no nonzero solution to }(A+I)x=0\\
&\Longleftrightarrow \text{there is no nonzero solution to }Ax +x = 0\\
&\Longleftrightarrow \text{there is no nonzero solution to }Ax = -x\\
&\Longleftrightarrow \lambda = -1\text{ is not an eigenvalue of }A.
\end{align*}$$
The "initial information" tells you that $A$ satisfies $t^2-4t-4$. So the minimal polynomial of $A$ divides $(t-2-2\sqrt{2})(t-2+2\sqrt{2})$.
What does that tell you about the eigenvalues of $A$?
Added. More generally, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda+\mu$ is an eigenvalue of $A+\mu I$. Because $Ax=\lambda x$ if and only if $(A-\mu I)x = (\lambda-\mu)x$. So $A+\mu I$ is invertible if and only if $-\mu$ is not an eigenvalue of $A$.