Showing that matrix is invertible using eigenvalues 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
 A: You have $A^2-4A-4I=0$ for any $\vec{x}$. So, if $\lambda$ is an eigenvalue of $A$, then for some corresponding eigenvector, you can rewrite the equation as ${\lambda}^{2}\vec{x}-4\lambda\vec{x}-4\vec{x}=0$. Factoring $\vec{x}$ out, you get  $({\lambda}^{2}-4\lambda-4)\vec{x}=0$. Since you know that an eigenvector cannot be the zero vector, you know that ${\lambda}^{2}-4\lambda-4$ has to equal zero. From here on, find roots to the equation -- you can use Viete's formulas; it factors as $(\lambda-2(1-\sqrt(2))(\lambda-2(1+\sqrt(2))=0$, giving you two eigenvalues of $A$. So, since $0$ is not an eigenvalue of $A$, you know that $A$ is invertible (and so is $A^{2}$). From the original equation, you also know that $A^{2}=4A+4I$, so $A+I=\frac{A^{2}}{4}$. Since $A^{2}$ is invertible, and dividing a matrix by its scalar does not affect its invertibility (determinant can't become 0, preservation of dimensions etc. remains the same), you have that $A+I$ is equal to an invertible matrix. Hence, $A+I$ is invertible.
A: Hint: What does the initial information tell you about the eigenvalues of $A$?
A: 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 $A$. 
