Question about eigenvalues of partitioned matrices So, I received this question to solve a while ago, and frankly, I'm a bit confused, since it's been a while since I've done linear algebra. Since the two questions depend on each other, I would like a detailed solution for both, please. The question is listed below:Let $A$ be an n x n nonsingular matrix and $I_n$ denote the n x n identity matrix. Also let M be the 2nx2n matrix given by:
$$M=\pmatrix{I_n&A\\ A^{-1}&I_n}$$
a) Show that if $λ$ is an eigenvalue of $M$ then either $λ = 0$ or $λ = 2$.
b) Is $M$ diagonalizable? If so, find a diagonalizer for $M$.
 A: Suppose $v$ is an eigenvector of $M$ with eigenvalue $\lambda$. Write $v=\begin{pmatrix}
v_1\\v_2
\end{pmatrix}$ where $v_i$ are columns of length $n$. Then $$Mv=\begin{pmatrix}
v_1+Av_2\\
A^{-1}v_1+v_2
\end{pmatrix}=\lambda \begin{pmatrix}
v_1\\v_2
\end{pmatrix}.$$
Hence \begin{eqnarray} v_1+Av_2&=& \lambda v_1\\ A^{-1}v_1+v_2&=&\lambda v_2\\\end{eqnarray}
Thus $Av_2=(\lambda-1)v_1$ and, assuming $\lambda\neq 1$, $A^{-1}\frac{Av_2}{(\lambda-1)}+v_2=v_2(\frac{1}{\lambda-1}+1)=\frac{\lambda}{\lambda-1}v_2=\lambda v_2$.
Hence $\frac{\lambda}{\lambda-1}=\lambda$. Thus $\lambda(\lambda-2)=0$. 
We still have to deal with the case $\lambda=1$. So suppose $\lambda=1$, then \begin{eqnarray} Av_2&=&  0\\ A^{-1}v_1&=& 0\\\end{eqnarray}But since $A$ is invertbile this implies that $v_1=0=v_2$. Thus $v=0$, but $v\neq 0$ as it is an eigenvector. Thus $\lambda\neq 1$. 
Can you take it from here?
A: Let $v = \begin{pmatrix} x \\ y \end{pmatrix}$ where $x,y$ are $n \times 1$ column vectors. For $v$ to be an eigenvector of $M$ with eigenvalue $\lambda$, we must have $Mv = \lambda v$. Written explicitly, this becomes
$$ \begin{pmatrix} I_n & A \\ A^{-1} & I_n \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + Ay \\ A^{-1} x + y \end{pmatrix} = \begin{pmatrix} \lambda x \\ \lambda y \end{pmatrix}. $$
Start with the equation $A^{-1} x + y = \lambda y$ and multiply this by $A$ to get $x + Ay = \lambda Ay = \lambda x$. If $\lambda \neq 0$, this implies that $x = Ay$ and so $x + Ay = x + x = 2x$ which implies that $\lambda = 2$. Hence, there are $n$ linearly independent eigenvectors of $M$ with eigenvalue $2$ given by 
$$\begin{pmatrix} Ay \\ y \end{pmatrix}, \,\,\, y \in \mathbb{F}^n. $$
What happens if $\lambda = 0$? Then we get the equation $x + Ay = 0$ so $Ay = -x$ and if this occurs, the second equation is satisfied automatically (multiply by $A^{-1}$ to see why). Thus, we got also $n$ linearly independent eigenvectors of $M$ with eigenvalue $0$ given by
$$ \begin{pmatrix} -Ay \\ y \end{pmatrix}, \,\,\, y \in \mathbb{F}^n. $$
From this we see that $A$ is diagonalizable with eigenvalues $0,2$ and since we found a basis of eigenvectors, we get
$$ P^{-1} A P = \begin{pmatrix} 2I & 0 \\ 0 & 0 \end{pmatrix} $$
where 
$$ P = \begin{pmatrix} A & -A \\ I & I \end{pmatrix}. $$
You can check that
$$ P^{-1} = \frac{1}{2} \begin{pmatrix} A^{-1} & I \\ -A^{-1} & I \end{pmatrix} $$
and that the equation above indeed holds.
