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?


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.

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