Finding the characteristic and minimal polynomials of this block matrix. Let $n>1$.
If I have $M = \begin{bmatrix} I_n & I_n \\
J_n(0) & -I_n \end{bmatrix} \in M_{2n}(\mathbb{C})$ where $J_n(0)$ is a size $n$ Jordan block matrix consisting of $0$'s in its diagonal, what are its characteristic polynomials and its minimal polynomials?
Here is what I got:
If we square $M$, we would get
$M^2 = \begin{bmatrix} J_n(1) & 0_n \\
0_n & J_n(1) \end{bmatrix}$.
Now, the minimal polynomial for a Jordan block matrix $J_n(\lambda)$ is $(x-\lambda)^n$.
Therefore $M$ must satisfy $(x-1)^{2n}$.
My guess is that this would be the minimal polynomial for $M$ itself. And since all the irreducible polynomials of the characteristic polynomial of $M$ must divide the minimal polynomial, then I say that $(x-1)^{2n}$ must be the characteristic polynomial of M.
Also, if this is true, here are my other claims:
1) M is not diagonalizable and 2) The only eigenvalue of M is 1.
If M is not diagonalizable, the algebraic and geometric multiplicities of $1$ are not equal. Though I am not sure we can easily find the geometric multiplicity of $1$ in this case.
Can someone affirm if all of my claims and premises above are true?
 A: Consider the eigenvalue equation
$$
\begin{bmatrix}I&I\\ J&-I\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}
=\begin{bmatrix}\lambda x\\ \lambda y\end{bmatrix}.
$$
This gives the two equations
\begin{align}\tag{*}
(\lambda-1)x-y&=0 \\ -Jx+(\lambda+1)y&=0.
\end{align}
If we multiply the first one by $\lambda+1$ and add, we get 
$$
(\lambda^2-1)x=Jx.
$$
Since the Jordan block $J$ only has $0$ as an eigenvalue, we get $\lambda^2-1=0$. So the only possible eigenvalues are $1$ and $-1$. 
When $\lambda=1$, the equations $(*)$ become $y=0$ and $Jx=0$. Then $x=\alpha e_1$ (where $e_1$ is the first vector of the canonical basis) and so the eigenspace of $\lambda=1$ consists of the vectors of the form 
$$
\begin{bmatrix}\alpha e_1\\0\end{bmatrix},\ \ \ \alpha\in\mathbb R.
$$
Thus the geometric multiplicity of the eigenvalue $1$ is $1$. 
When $\lambda=-1$, the system $(*)$ becomes $y=-2x$, $Jx=0$. So again $x=\alpha e_1$, and now $y=-2\alpha e_1$. The eigenvectors for $\lambda=-1$ are then
$$
\begin{bmatrix}\alpha e_1\\-2\alpha e_1\end{bmatrix},\ \ \ \alpha\in\mathbb R.
$$
and the geometric multiplicity of $\lambda=-1$ is also $1$. In particular, because of the deficit in the geometric multiplicities (they do not add to $2n$), we now know that $M$ is not diagonalizable. 
For the  characteristic polynomial (and also the minimal polynomial). What you found is that $$(M^2-I)^n=0.$$
So $p(t)=(t^2-1)^n$ is a monic polynomial of degree $n$ that annihilates $M$, and so it has to be a multiple of the minimal polynomial $m(t)$. 
Note that $\text{Tr}(M)=0$. This tells us that the algebraic multiplicities of $\lambda=1$ and $\lambda=-1$ are the same. Then the minimal polynomial is of the form 
$$
m(t)=(t^2-1)^k.
$$
Now, $$(M^2-I)^{n-1}=\begin{bmatrix}J_n(0)^{n-1}&0\\0&J_n(0)^{n-1}\end{bmatrix}\ne0, 
$$ so $k$ has to be equal to $n$. 
In summary,


*

*The characteristic (and minimal) polynomial of $M$ is $p(t)=(t^2-1)^n$. 

*The eigenvalues are $1$ and $-1$, both with algebraic multiplicity $n$ and geometric multiplicity $1$. 

*$M$ is not diagonalizable.
