The symplectic characteristic polynomial

Kohei Ichizuka Department of Mathematics, Saitama University, 255 Shimo-Okubo, Sakura-Ku, Saitama 338-8570, Japan [email protected]

Abstract.

We introduce the notion of the symplectic characteristic polynomial of an endomorphism of a symplectic vector space. This is a polynomial in two variables and can be considered as a generalization of the characteristic polynomial of the endomorphism in the context of symplectic linear algebra. One of the goal of this paper is to prove that the symplectic characteristic polynomial is a complete symplectic invariant of symplectically diagonalizable endomorphisms.

Key words and phrases:

Characteristic polynomial, Symplectic similarity, Symplectically diagonalizability

2020 Mathematics Subject Classification:

15A20, 15A21

1. Introduction

Let $(V,\omega)$ be a finite dimensional symplectic vector space over a field $\mathbb{K}$ . It is a fundamental question to classify endomorphisms of $V$ with respect to symplectic similarity. Endomorphisms which are self-adjoint, anti-self-adjoint or preserve the symplectic form $\omega$ have been classified when the characteristic of $\mathbb{K}$ is not $2$ (see [5, Theorem 4], [1, Theorem 2.2], [4, Section 9]). Such an endomorphism is symplectically normal (Definition 2.1 (2)) and the classification of symplectically normal endomorphisms is not known even though they are diagonalizable. An endomorphism is symplectically diagonalizable if and only if the endomorprhism is symplectically normal and diagonalizable ([2, Theorem 13]). A motivation of the paper is to classify symplectically diagonalizable endomorphisms up to symplectic similarity.

Let us recall several facts on the characteristic polynomial of an endomorphism $M$ of a vector space $V$ . Cayley-Hamilton theorem implies that $V$ admits the direct sum decomposition into the generalized eigenspaces of $M$ when all eigenvalues of $M$ are in $\mathbb{K}$ . The characteristic polynomial is a complete invariant of diagonalizable endomorphisms. Moreover, if the number of eigenvalues of $M$ is equal to the dimension of $V$ , then $M$ is diagonalizable.

Now we consider a symplectic vector space $(V,\omega)$ . Let $M$ be an endomorphism of $V$ . Suppose that $M$ is symplectically normal and that all eigenvalues of $M$ are in $\mathbb{K}$ . It can be proved by using the same idea in ([3, Lemma 1]) that $V$ is the symplectically orthogonal sum of the symplectic subspaces associated with two eigenvalues of $M$ . This implies that we have projections onto these subspaces. On the other hand, if one considers the endomorphism $(M-sE)(M^{*\omega}-sE)$ of the $\mathbb{K}(s)$ -vector space $V\otimes_{\mathbb{K}}\mathbb{K}(s)$ , we have other projections, whose images are the generalized eigenspaces of $({M}-sE)({M}^{*\omega}-sE)$ . Here $E$ is the identity map and $\mathbb{K}(s)$ is the field of rational functions over $\mathbb{K}$ .

In this paper, we show that these generalized eigenspaces and projections correspond to the previous symplectic subspaces and projections respectively. We introduce the symplectic characteristic polynomial as the Pfaffian of $(M-sE)^{*}\omega-t\omega$ . We prove that the symplectic characteristic polynomial is a complete invariant of symplectically diagonalizable endomorphisms up to symplectic similarity. Moreover, we give a sufficient condition for symplectically diagonalizability.

The paper is organized as follows. Section 2 contains several fundamental facts on symplectic linear algebra. In Section 3, we define the symplectic characteristic polynomial of an endomorphism and prove several properties. In Section 4, we study the characteristic polynomial of the endomorphism associated with two endomorphisms $M$ , $N$ of a (not necessary symplectic) vector space. In Section 5, we prove that the symplectic characteristic polynomial is a complete invariant of symplectically diagonalizable endomorphisms and give a sufficient condition on symplectically diagonalizability.

2. Preliminaries

We first introduce the following notation.

•

We denote the field of rational functions over $\mathbb{K}$ by $\mathbb{K}(s)$ .
•

The set of endomorphisms of a vector space $V$ is denoted by $\operatorname{End}(V)$ . By abuse of notation, $E$ denotes the identity map of $V$ and $O$ denotes the zero map of $V$ .
•

The characteristic polynomial of an endomorphism $M$ is denoted by $\varphi_{M}(t)$ .

•

Let $M,N\in\operatorname{End}(V)$ and $\lambda,\mu\in\mathbb{K}$ . We define the subspaces $V_{M}(\lambda)$ , $\tilde{V}_{M}(\lambda)$ by

	$\displaystyle V_{M}(\lambda)$	$\displaystyle=\operatorname{Ker}(M-\lambda E),$
	$\displaystyle\tilde{V}_{M}(\lambda)$	$\displaystyle=\operatorname{Ker}(M-\lambda E)^{n}\ \text{where $\dim_{\mathbb{K}}V=n$}.$

The space $V_{M}(\lambda)$ (resp. $\tilde{V}_{M}(\lambda)$ ) is an eigenspace (resp. a generalized eigenspace ) if it is not zero.

We also define the subspaces $V_{M,N}(\lambda,\mu)$ , $\tilde{V}_{M,N}(\lambda,\mu)$ of $V$ by

	$\displaystyle V_{M,N}(\lambda,\mu)$	$\displaystyle=\left(V_{M}(\lambda)\cap V_{N}(\mu)\right)+\left(V_{M}(\mu)\cap V_{N}(\lambda)\right),$
	$\displaystyle\tilde{V}_{M,N}(\lambda,\mu)$	$\displaystyle=\left(\tilde{V}_{M}(\lambda)\cap\tilde{V}_{N}(\mu)\right)+\left(\tilde{V}_{M}(\mu)\cap\tilde{V}_{N}(\lambda)\right).$

•

Let $M\in\operatorname{End}(V)$ and $\omega$ a bilinear form. Let $\mathbb{F}$ be a extension field of $\mathbb{K}$ . We use the same letter $M$ (resp. $\omega$ ) for the natural extension of $M$ (resp. $\omega$ ) to the endomorphism (resp. the bilinear form) of $V\otimes_{\mathbb{K}}\mathbb{F}$ .

Now we recall several fundamental facts on symplectic linear algebra.

Let $V$ be a vector space over a field $\mathbb{K}$ and let $\dim_{\mathbb{K}}V=2n$ . A bilinear form $\omega$ of $V$ is said to be symplectic if

•

$\omega$ is alternating, that is, $\omega(v,v)=0$ for all $v\in V$ and
•

$\omega$ is non-degenerate, that is, a linear map

$\omega^{\flat}:V\rightarrow V^{*}\quad v\mapsto\omega(\cdot,v)$

is an isomorphism.

We call $(V,\omega)$ a symplectic vector space. A representation matrix of an alternating matrix with respect to some bases is anti-symmetric and the diagonal entries are zero. This matrix has determinant zero when the size of the matrix is odd and a symplectic vector space must be even-dimensional.

The set of automorphisms which preserve $\omega$ is denoted by $\operatorname{Sp}(V,\omega)$ . Since the Pfaffian of a matrix which represents $\omega$ is not zero, any automorphism in $\operatorname{Sp}(V,\omega)$ has determinant $1$ .

Let $(V,\omega)$ be a symplectic vector space and let $M\in\operatorname{End}(V)$ . The symplectic adjoint endomorphism $M^{*\omega}$ of $M$ is an endomorphism of $V$ defined by

M^{*\omega}=(\omega^{\flat})^{-1}M^{*}\omega^{\flat}

where $M^{*}$ is the dual map of $M$ . It is easily shown that the following properties hold for any $M,N\in\operatorname{End}(V)$ .

•

$\omega(Mv,w)=\omega(v,M^{*\omega}w)$ for all $v,w\in V$ .
•

$M\in\operatorname{Sp}(V,\omega)$ if and only if $M^{*\omega}=M^{-1}$ .
•

$(M^{*\omega})^{*\omega}=M.$
•

$(MN)^{*\omega}=N^{*\omega}M^{*\omega}$ .
•

The characteristic polynomial of $M$ is equal to that of $M^{*\omega}$ .

Definition 2.1.

(1)

An endomorphism $M$ is said to be symplectically similar to an endomorphism $N$ if there exists $P\in\operatorname{Sp}(V,\omega)$ such that $P^{-1}MP=N$ .
(2)

An endomorphism $M$ is said to be symplectically normal if $MM^{*\omega}=M^{*\omega}M$ .

Let $W$ be a subspace of $V$ . We define the symplectically orthogonal subspace $W^{\perp\omega}$ by

W^{\perp\omega}=\{v\in V|\ \omega(v,w)=0\quad\forall w\in W\}.

As in the case of an inner product, we have the following.

Proposition 2.2.

Let $M\in\operatorname{End}(V)$ and let $W$ be a subspace of $V$ . Then

$(1)$

$\dim W+\dim W^{\perp\omega}=\dim V$ .
$(2)$

$(W^{\perp\omega})^{\perp\omega}=W$ .
$(3)$

$\operatorname{Ker}M=(\operatorname{Im}M^{*\omega})^{\perp\omega}$ .
$(4)$

$M(W)\subset W$ implies that $M^{*\omega}(W^{\perp\omega})\subset W^{\perp\omega}.$
$(5)$

If $\lambda\neq\mu$ , then $\tilde{V}_{M}(\lambda)\subset\tilde{V}_{M^{*\omega}}(\mu)^{\perp\omega}.$

Proof.

$(1)$ It follows that $W^{\perp\omega}=\operatorname{Ker}\iota_{W}^{*}\omega^{\flat}$ where $\iota_{W}$ is the inclusion. Since a linear map $\iota_{W}^{*}\omega^{\flat}:V\rightarrow W^{*}$ is surjective, we have $(1)$ . $(2)$ It is clear that $W\subset(W^{\perp\omega})^{\perp\omega}$ . Applying $(1)$ for $W^{\perp\omega}$ , we have $\dim(W^{\perp\omega})^{\perp\omega}=\dim W$ . Hence, we get $(2)$ . $(3)$ It is easy to see that $\operatorname{Ker}M\subset(\operatorname{Im}M^{*\omega})^{\perp\omega}$ . Let $v_{0}\in(\operatorname{Im}M^{*\omega})^{\perp\omega}$ . Then $\omega(Mv_{0},v)=\omega(v_{0},M^{*\omega}v)=0$ for all $v\in V$ . This implies that $v_{0}\in\operatorname{Ker}M$ . Hence, we have $(3)$ . The items $(4)$ are easily shown. $(5)$ Suppose that $\lambda\neq\mu$ . Since $(M-\mu E)^{2n}|_{\tilde{V}_{M}(\lambda)}$ is an automorphism, $\tilde{V}_{M}(\lambda)=\operatorname{Im}(M-\mu E)^{2n}\cap\tilde{V}_{M}(\lambda)\subset\operatorname{Im}(M-\mu E)^{2n}$ . Therefore, by $(2)$ and $(3)$ , we get $(5)$ . ∎

A subspace $W$ of $V$ is said to be

•

symplectic if $W\cap W^{\perp\omega}=\{0\}$ ,
•

isotropic if $W\subset W^{\perp\omega}$ ,
•

coisotropic if $W^{\perp\omega}\subset W$ ,
•

Lagrangian if $W=W^{\perp\omega}$ .

Since $(W^{\perp\omega})^{\perp\omega}=W$ , we have that $W$ is symplectic if and only if $W^{\perp\omega}$ is symplectic.

The item $(1)$ in Proposition 2.2 shows that if $W$ is isotropic, then $\dim_{\mathbb{K}}W\leq n$ and that $W$ is Lagrangian if and only if $W$ is isotropic and $\dim_{\mathbb{K}}W=n$ .

Proposition 2.3.

Let $S\in\operatorname{End}(V)$ . Suppose that $S^{*\omega}=S$ . Then $\tilde{V}_{S}(\lambda)$ is a symplectic subspace of $(V,\omega)$ .

Proof.

It follows from the item $(2)$ and $(3)$ in Proposition 2.2 that $\tilde{V}_{S}(\lambda)^{\perp\omega}=\operatorname{Im}(S-\lambda E)^{2n}$ . This implies that $\tilde{V}_{S}(\lambda)\cap\tilde{V}_{S}(\lambda)^{\perp\omega}=\{0\}$ . ∎

Lemma 2.4.

Let $M\in\operatorname{End}(V)$ and let $W=\tilde{V}_{{M},{M}^{*\omega}}(\lambda,\mu)$ . Suppose that $\lambda\neq\mu$ and that $W$ is a symplectic subspace of $(V,\omega)$ . Then $\tilde{V}_{M}(\lambda)\cap\tilde{V}_{M^{*\omega}}(\mu)$ , $\tilde{V}_{M}(\mu)\cap\tilde{V}_{M^{*\omega}}(\lambda)$ are Lagrangian subspaces of $(V,\omega)$ .

Proof.

The item $(5)$ in Proposition 2.2 gives that $\tilde{V}_{M}(\lambda)\cap\tilde{V}_{M^{*\omega}}(\mu)$ , $\tilde{V}_{M}(\lambda)\cap\tilde{V}_{M^{*\omega}}(\mu)$ are isotropic subspaces of $(W,\omega|_{W})$ . Since $\tilde{V}_{M}(\lambda)\cap\tilde{V}_{M}(\mu)=\{0\}$ , we have

\dim_{\mathbb{K}}(\tilde{V}_{M}(\lambda)\cap\tilde{V}_{M^{*\omega}}(\mu))=\dim_{\mathbb{K}}(\tilde{V}_{M}(\mu)\cap\tilde{V}_{M^{*\omega}}(\lambda))=\frac{1}{2}\dim_{\mathbb{K}}W.

Therefore, $\tilde{V}_{M}(\lambda)\cap\tilde{V}_{M^{*\omega}}(\mu)$ , $\tilde{V}_{M}(\mu)\cap\tilde{V}_{M^{*\omega}}(\lambda)$ are Lagrangian subspaces of $(W,\omega|_{W})$ . ∎

A basis $\mathcal{B}=(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ of $V$ is said to be symplectic if

\omega(e_{i},e_{j})=0,\ \omega(f_{i},f_{j})=0,\ \omega(e_{i},f_{j})=\delta_{ij},\quad i,j\in\{1,\dots,n\}.

Proposition 2.5.

There exists a symplectic basis of $(V,\omega)$ .

Proof.

We proof by induction on $n$ . Since $\omega$ is non-degenerate, there exist $e_{1},f_{1}\in V$ such that $\omega(e_{1},f_{1})=1$ . Hence, the assertion holds for $n=1$ . We assume the assertion holds for $n-1$ . Let $W$ be a subspace spanned by $e_{1},f_{1}$ . Then $W$ is symplectic and so is $W^{\perp\omega}$ . The induction hypothesis shows that there exists a symplectic basis $(e_{2},\dots,e_{n},f_{2},\dots,f_{n})$ of $(W^{\perp\omega},\omega|_{W^{\perp\omega}})$ . It is clear that $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ is a symplectic basis of $(V,\omega)$ . ∎

Lemma 2.6.

Let $L_{1},L_{2}$ be Lagrangian subspaces of $(V,\omega)$ . Suppose that $L_{1}\cap L_{2}=\{0\}$ and $(e_{1},\dots,e_{n})$ is a basis of $L_{1}$ . Then there exists a basis $(f_{1},\dots,f_{n})$ of $L_{2}$ such that $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ is a symplectic basis of $(V,\omega)$ .

Proof.

The condition $L_{1}\cap L_{2}=\{0\}$ implies that there exist $\lambda_{1},\dots,\lambda_{n}\in V^{*}$ such that $\lambda_{i}(e_{j})=\delta_{ij}$ and that $\ \lambda_{i}|_{L_{2}}=0.$ Since $\omega$ is non-degenerate, there exists $f_{i}\in V$ such that $\lambda_{i}=\omega^{\flat}(f_{i})$ . It follows from $\lambda_{i}|_{L_{2}}=0$ that $f_{i}\in(L_{2})^{\perp\omega}=L_{2}$ . It is clear that $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ is a symplectic basis. ∎

Let $\mathcal{B}$ be a symplectic basis of $(V,\omega)$ . Then the matrix $[\omega]_{\mathcal{B}}$ which represents $\omega$ with respect to $\mathcal{B}$ is

[\omega]_{\mathcal{B}}=\left(\begin{array}[]{cc}O&E_{n}\\ -E_{n}&O\end{array}\right)

where $E_{n}$ is the identity matrix of order $n$ and $O$ is the zero matrix. If $M\in\operatorname{End}(V)$ is represented by

[M]_{\mathcal{B}}=\left(\begin{array}[]{cc}A_{n}&B_{n}\\ C_{n}&D_{n}\end{array}\right)

where $A_{n}$ , $B_{n}$ , $C_{n}$ , $D_{n}$ are matrices of order $n$ , then $M^{*\omega}$ is represented by

[M^{*\omega}]_{\mathcal{B}}=[\omega]_{\mathcal{B}}^{-1}[M]_{\mathcal{B}}^{T}[\omega]_{\mathcal{B}}=\left(\begin{array}[]{cc}D_{n}^{T}&-B_{n}^{T}\\ -C_{n}^{T}&A_{n}^{T}\end{array}\right).

Here $T$ denotes the transpose of a matrix. Hence, if $\dim_{\mathbb{K}}V=2$ , then $MM^{*\omega}=(\det M)E$ , $M+M^{*\omega}=(\operatorname{tr}M)E$ .

Let $P\in\operatorname{End}(V)$ and let $\mathcal{B}=(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ be a symplectic basis of $(V,\omega)$ . Then $(Pe_{1},\dots,Pe_{n},Pf_{1},\dots,Pf_{n})$ is a symplectic basis if and only if $P\in\operatorname{Sp}(V,\omega)$ . This means that

Proposition 2.7.

An endomorphism $M$ is symplectically similar to an endomorphism $N$ if and only if there exist symplectic bases $\mathcal{B}$ , $\mathcal{B^{\prime}}$ such that $[M]_{\mathcal{B}}=[N]_{\mathcal{B^{\prime}}}$ .

An endomorphism $M$ is said to be symplectically diagonalizable if there exists a symplectic basis consisting of eigenvectors of $M$ . It is easy to see that if an endomorphism $M$ is symplectically diagonalizable, then $MM^{*\omega}=M^{*\omega}M$ and $M$ is diagonalizable. Theorem 5.4 shows that the converse is true.

Let $M\in\operatorname{End}(V).$ We define the bilinear form $\omega_{M}$ of $V$ by

\omega_{M}(v,w)=\omega(v,Mw)\quad v,w\in V.

It is easy to see that $\omega_{MM^{*\omega}}=(M^{*\omega})^{*}\omega$ . We remark that $\omega_{MM^{*\omega}}$ , $\omega_{M+M^{*\omega}}$ are alternating for any $M\in\operatorname{End}(V)$ .

Let $A\in\operatorname{End}(V)$ and assume that $\omega_{A}$ is alternating. Let $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ be a symplectic basis. Set $(v_{1},\dots,v_{2n})=(e_{1},f_{1},\dots,e_{n},f_{n})$ . Define the matrix $\Omega_{A}$ by $\Omega_{A}=(\omega_{A}(v_{i},v_{j}))_{i,j}$ . Since $\det P=1$ for $P\in\operatorname{Sp}(V,\omega)$ , the Pfaffian $\operatorname{pf}(\Omega_{A})$ does not depend on the choice of symplectic bases. We denote $\operatorname{pf}(\Omega_{A})$ by $\operatorname{pf}_{\omega}(\omega_{A})$ . It is easily shown that $\operatorname{pf}_{\omega}(\omega)=1$ .

Definition 2.8.

Let $A\in\operatorname{End}(V)$ and assume that $\omega_{A}$ is alternating. The polynomial $\psi_{A}(t)$ is defined by $\psi_{A}(t)=\operatorname{pf}_{\omega}(\omega_{A-tE})$ ,

Proposition 2.9.

Let $A\in\operatorname{End}(V)$ and assume that $\omega_{A}$ is alternating. The polynomial $\psi_{A}(t)$ has the following properties.

$(1)$

The square $\psi_{A}(t)^{2}$ is the characteristic polynomial $\varphi_{A}(t)$ .
$(2)$

The polynomial $\psi_{A}(t)$ is an invariant of $A$ under symplectic similarity transformation, that is, $\psi_{P^{-1}AP}(t)=\psi_{A}(t)$ for $P\in\operatorname{Sp}(V,\omega)$ .
$(3)$

Let $\psi_{A}(t)=a_{0}+\dots+a_{n-1}t^{n-1}+a_{n}t^{n}$ . Then

$a_{0}=\operatorname{pf}_{\omega}(\omega_{A}),\ 2a_{n-1}=(-1)^{n-1}\operatorname{tr}{A},\ a_{n}=(-1)^{n}.$
$(4)$

$\psi_{A}(A)=O.$

Proof.

It is clear from the definition of $\psi_{A}(t)$ that $(1)$ holds.

Let $P\in\operatorname{Sp}(V,\omega)$ . Since $\omega_{P^{-1}AP}=P^{*}\omega$ , we get that $\omega_{P^{-1}AP}$ is alternating. The item $(1)$ implies that $\psi_{P^{-1}AP}(t)^{2}=\psi_{A}(t)^{2}$ . Hence, we have $(2)$ .

It follows from [5, Theorem 3] that there exists a symplectic basis $\mathcal{B}$ such that

[A]_{\mathcal{B}}=\left(\begin{array}[]{cccc}D_{n}^{T}&O\\ O&D_{n}\end{array}\right)

where $D_{n}$ is a matrix of order $n$ . Then

[\omega_{A-tE}]_{\mathcal{B}}=[\omega]_{\mathcal{B}}[A-tE]_{\mathcal{B}}=\left(\begin{array}[]{cccc}O&D_{n}-tE_{n}\\ -(D_{n}-tE_{n})^{T}&O\end{array}\right).

This implies that $\operatorname{pf}([\omega_{A-tE}]_{\mathcal{B}})=(-1)^{\frac{n(n-1)}{2}}\det(D_{n}-tE_{n})$ . Hence, we have $\psi_{A}(t)=\det(D_{n}-tE_{n}).$ Since $\det D_{n}=\psi_{A}(0)=\operatorname{pf}_{\omega}(\omega_{A})$ and $2\operatorname{tr}D_{n}=\operatorname{tr}A$ , we get $(3)$ . Cayley-Hamilton theorem gives that $\psi_{A}(D_{n})=O$ . Therefore,

\displaystyle[\psi_{A}(A)]_{\mathcal{B}}

\displaystyle=\left(\begin{array}[]{cc}\psi_{A}(D_{n})^{T}&O\\ O&\psi_{A}(D_{n})\end{array}\right)=O,

which completes the proof. ∎

Proposition 2.10.

Let $M\in\operatorname{End}(V)$ . Then we have $\psi_{MM^{*\omega}}(t)=\psi_{M^{*\omega}M}(t)=\operatorname{pf}_{\omega}(M^{*}\omega-t\omega)$ and $\psi_{MM^{*\omega}}(0)=\det M$ .

Proof.

By the item $(2)$ in Proposition 2.9, we get $\psi_{MM^{*\omega}}(t)^{2}=\psi_{M^{*\omega}M}(t)^{2}$ . The item $(4)$ in Proposition 2.9 shows that the coefficient of $t^{n}$ in $\psi_{MM^{*\omega}}(t)$ and the coefficient of $t^{n}$ in $\psi_{M^{*\omega}M}(t)$ are $(-1)^{n}$ . Hence, $\psi_{MM^{*\omega}}(t)=\psi_{M^{*\omega}M}(t)$ . Since $\omega_{M^{*\omega}M}=M^{*}\omega$ , we have $\psi_{M^{*\omega}M}(t)=\operatorname{pf}_{\omega}(M^{*}\omega-t\omega)$ . This implies that $\psi_{MM^{*\omega}}(0)=\det M\cdot\operatorname{pf}_{\omega}(\omega)=\det M$ . ∎

3. The symplectic characteristic polynomial

Let $(V,\omega)$ be a $2n$ -dimensional symplectic vector space over a field $\mathbb{K}$ .

Definition 3.1.

The symplectic characteristic polynomial $\varphi_{M}^{\omega}(s,t)$ of an endomorphism $M$ of $(V,\omega)$ is defined by

\varphi^{\omega}_{M}(s,t)=\psi_{({M}-sE)({M}^{*\omega}-sE)}(t).

Here $\psi_{({M}-sE)({M}^{*\omega}-sE)}(t)$ is a polynomial defined in Definition 2.8.

Remark 3.2.

It follows from Proposition 2.10 that

(3.1)

\psi_{({M}-sE)({M}^{*\omega}-sE)}(t)=\psi_{(M^{*\omega}-sE)(M-sE)}(t)=\operatorname{pf}_{\omega}(M-sE)^{*}\omega-t\omega).

Remark 3.3.

The definition above is motivated by the following example. Let $\dim_{\mathbb{K}}V=4$ and $M\in\operatorname{End}(V)$ . Assume that $M$ has four distinct eigenvalues and that $M$ is symplectically diagonalizable, that is, there exists a symplectic basis $\mathcal{B}$ such that $[M]_{\mathcal{B}}$ is a diagonal matrix $\operatorname{diag}(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})$ . Define the subgroup $G$ of the symmetric group $\mathfrak{S}_{4}$ as follows: An element $\sigma\in\mathfrak{S}_{4}$ is in $G$ if there exists a symplectic basis $\mathcal{B}^{\prime}$ such that $[M]_{\mathcal{B}^{\prime}}=\operatorname{diag}(\lambda_{\sigma(1)},\lambda_{\sigma(2)},\lambda_{\sigma(3)},\lambda_{\sigma(4)})$ . Since $\operatorname{tr}MM^{*\omega}=\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{4}$ is independent on the choice of symplectic bases, $G$ consists of eight elements

	$\displaystyle\left(\begin{array}[]{cccc}1&2&3&4\\ 1&2&3&4\end{array}\right),\ \left(\begin{array}[]{cccc}1&2&3&4\\ 2&3&4&1\end{array}\right),\ \left(\begin{array}[]{cccc}1&2&3&4\\ 3&4&1&2\end{array}\right),\ \left(\begin{array}[]{cccc}1&2&3&4\\ 4&1&2&3\end{array}\right),$
	$\displaystyle\left(\begin{array}[]{cccc}1&2&3&4\\ 1&4&3&2\end{array}\right),\ \left(\begin{array}[]{cccc}1&2&3&4\\ 2&1&4&3\end{array}\right),\ \left(\begin{array}[]{cccc}1&2&3&4\\ 3&2&1&4\end{array}\right),\ \left(\begin{array}[]{cccc}1&2&3&4\\ 4&3&2&1\end{array}\right),$

which implies that $G$ is isomorphic to the dihedral group of square. Therefore, we find that $\{\{\lambda_{1},\lambda_{3}\},\{\lambda_{2},\lambda_{4}\}\}$ is independent on the choice of symplectic bases. This set determines uniquely the polynomial

p(s,t)=\{(\lambda_{1}-s)(\lambda_{3}-s)-t\}\{(\lambda_{2}-s)(\lambda_{4}-s)-t\}.

On the other hand,

		$\displaystyle[({M}-sE)({M}^{*\omega}-sE)]_{\mathcal{B}}$
	$\displaystyle=$	$\displaystyle\operatorname{diag}((\lambda_{1}-s)(\lambda_{3}-s),(\lambda_{2}-s)(\lambda_{4}-s),(\lambda_{1}-s)(\lambda_{3}-s),(\lambda_{2}-s)(\lambda_{4}-s)).$

This implies that $\varphi_{({M}-sE)({M}^{*\omega}-sE)}(t)=p(s,t)^{2}.$ The item $(4)$ in Proposition 2.9 shows that the coefficient of $t^{2}$ in $\psi_{({M}-sE)({M}^{*\omega}-sE)}(t)$ is $1$ . Hence, we have $\psi_{({M}-sE)({M}^{*\omega}-sE)}(t)=p(s,t).$

We show properties of the symplectic characteristic polynomial.

Proposition 3.4.

Let $M\in\operatorname{End}(V)$ . The symplectic characteristic polynomial $\varphi_{M}^{\omega}(s,t)$ satisfies the following.

$(1)$

The polynomial $\varphi^{\omega}_{M}(0,t)$ is the polynomial $\psi_{MM^{*\omega}}(t)$ .
$(2)$

The polynomial $\varphi_{M}^{\omega}(s,0)$ is the characteristic polynomial $\varphi_{M}(s)$ .
$(3)$

$\varphi_{M}^{\omega}(s,(\sigma-s)(\tau-s))=\psi_{MM^{*\omega}}(\sigma\tau)+\dots+\psi_{M+M^{*\omega}}(\sigma+\tau)(-s)^{n}.$
$(4)$

If $\varphi^{\omega}_{M}(s,(\lambda-s)(\mu-s))=0$ for $\lambda,\mu\in\mathbb{K}$ , then $\lambda$ , $\mu$ are eigenvalues of $M$ , $\lambda\mu$ is an eigenvalue of $MM^{*\omega}$ and $\lambda+\mu$ is an eigenvalue of $M+M^{*\omega}$ .
$(5)$

The symplectic characteristic polynomial is an invariant of $M$ under symplectic similarity transformation and taking the symplectic adjoint, that is, $\varphi^{\omega}_{P^{-1}MP}(s,t)=\varphi^{\omega}_{M}(s,t)$ for $P\in\operatorname{Sp}(V,\omega)$ and $\varphi^{\omega}_{M^{*\omega}}(s,t)=\varphi^{\omega}_{M}(s,t)$ .
$(6)$

The square $\varphi^{\omega}_{M}(s,t)^{2}$ is the characteristic polynomial $\varphi_{({M}-sE)({M}^{*\omega}-sE)}(t)$ .

(7)

Let $\varphi^{\omega}_{M}(s,t)=a_{0}(s)+\dots+a_{n-1}(s)t^{n-1}+a_{n}(s)t^{n}.$ Then

	$\displaystyle a_{0}(s)$	$\displaystyle=\varphi_{M}(s),\ 2a_{n-1}(s)=(-1)^{n-1}\{\operatorname{tr}MM^{*\omega}-2(\operatorname{tr}M)s+ns^{2}\},$
	$\displaystyle\ a_{n}(s)$	$\displaystyle=(-1)^{n}.$

$(8)$

$\varphi^{\omega}_{M}(s,({M}-sE)({M}^{*\omega}-sE))=O.$

Proof.

It is clear from the definition that $(1)$ holds. The item $(2)$ follows from Proposition 2.10.

$(3)$ The easy calculation shows that

(M^{*\omega}-sE)^{*}\omega-(\sigma-s)(\tau-s)\omega=\omega_{MM^{*\omega}-\sigma\tau E}-s\omega_{M+M^{*\omega}-(\sigma+\tau)E}.

This implies that

		$\displaystyle\varphi_{M}^{\omega}(s,(\sigma-s)(\tau-s))=\operatorname{pf}_{\omega}\left(\omega_{MM^{\omega}-\sigma\tau E}-s\omega_{\{M+M^{\omega}-(\sigma+\tau)E\}}\right)$
	$\displaystyle=$	$\displaystyle\psi_{MM^{\omega}}(\sigma\tau)+\dots+\psi_{M+M^{\omega}}(\sigma+\tau)(-s)^{n}.$

$(4)$ Assume that $\varphi^{\omega}_{M}(s,(\lambda-s)(\mu-s))=0$ for $\lambda,\mu\in\mathbb{K}$ . Then by $(2)$ , we have $\varphi_{M}(\lambda)=\varphi^{\omega}_{M}(\lambda,0)=0$ . From $(3)$ , we get $\psi_{MM^{*\omega}}(\lambda\mu)=0$ , $\psi_{M+M^{*\omega}}(\lambda+\mu)=0$ . This implies that $\varphi_{MM^{*\omega}}(\lambda\mu)=0$ , $\varphi_{M+M^{*\omega}}(\lambda+\mu)=0$ .

$(5)$ Let $P\in\operatorname{Sp}(V,\omega)$ . It is easy to see that

(P^{-1}MP-sE)(P^{-1}M^{*\omega}P-sE)=P^{-1}(M-sE)(M^{*\omega}-sE)P.

Hence, from the item $(1)$ in Proposition 2.9, we have $\varphi^{\omega}_{P^{-1}MP}(s,t)=\varphi^{\omega}_{M}(s,t)$ . The equality (3.1) implies that $\varphi^{\omega}_{M^{*\omega}}(s,t)=\varphi^{\omega}_{M}(s,t)$ .

The items $(6)$ – $(8)$ follows from the items $(2)$ – $(4)$ in Proposition 2.9 respectively. ∎

The symplectic characteristic polynomial $\varphi^{\omega}_{M}(s,t)$ of an endomorphism $M$ is determined by the characteristic polynomial $\varphi_{M}(t)$ if $M^{*\omega}=M$ , $M^{*\omega}=-M$ or $M^{*\omega}=M^{-1}$ . Precisely, we have the following proposition.

Proposition 3.5.

Let $M\in\operatorname{End}(V)$ . Then

\varphi^{\omega}_{M}(s,t)^{2}=\begin{dcases}\varphi_{M}(s-t^{\frac{1}{2}})\varphi_{M}(s+t^{\frac{1}{2}})&\text{if $M^{*\omega}=M$},\\ \varphi_{M^{2}}(s^{2}-t)&\text{if $M^{*\omega}=-M$},\\ s^{2n}\varphi_{M+M^{-1}}(s^{-1}(s^{2}-t+1))&\text{if $M^{*\omega}=M^{-1}$}.\end{dcases}

Proof.

This is verified from the fact that $\varphi^{\omega}_{M}(s,t)^{2}=\varphi_{({M}-sE)({M}^{*\omega}-sE)}(t)$ . ∎

4. The characteristic polynomial associated with two endomorphisms

We have seen in Proposition 3.4 that the square of the symplectic characteristic polynomial $\varphi^{\omega}_{M}(s,t)$ of an endomorphism $M$ is the characteristic polynomial $\varphi_{({M}-sE)({M}^{*\omega}-sE)}(t)$ . This section is devoted to the study of the characteristic polynomial $\varphi_{({M}-sE)({N}-sE)}(t)$ for any endomorphisms $M,N$ of a (not necessarily symplectic) vector space.

Let $V$ be a finite dimensional vector space over a field $\mathbb{K}$ with $\dim_{\mathbb{K}}V=n$ .

Proposition 4.1.

Let $M,N\in\operatorname{End}(V)$ . Then

•

$\varphi_{({M}-sE)({N}-sE)}({0})=\varphi_{M}(s)\varphi_{N}(s)$ ,
•

$\varphi_{({M}-sE)({N}-sE)}({(\sigma-s)(\tau-s)})=\varphi_{MN}(\sigma\tau)+\dots+\varphi_{M+N}(\sigma+\tau)(-s)^{n}$ .

Proof.

It is clear from the definition. ∎

Proposition 4.2.

Let $M,N\in\operatorname{End}(V)$ . If $M$ , $N$ are simultaneously triangularizable, then

\varphi_{({M}-sE)({N}-sE)}({t})=\prod_{i=1}^{n}\{(\lambda_{i}-s)(\mu_{i}-s)-t\},\quad\lambda_{i},\mu_{i}\in\mathbb{K}.

In particular, a polynomial $(\lambda_{i}-s)(\mu_{i}-s)$ is an eigenvalue of $({M}-sE)({N}-sE)$ .

Proof.

Since $M$ and $N$ are simultaneously triangularizable, there exists a basis $\mathcal{B}$ such that

[M]_{\mathcal{B}}=\left(\begin{array}[]{ccccc}\lambda_{1}&&*\\ &\ddots\\ 0&&\lambda_{n}\end{array}\right),\quad[N]_{\mathcal{B}}=\left(\begin{array}[]{ccccc}\mu_{1}&&*\\ &\ddots\\ 0&&\mu_{n}\end{array}\right).

Therefore, $\varphi_{({M}-sE)({N}-sE)}({t})=\prod_{i=1}^{n}\{(\lambda_{i}-s)(\mu_{i}-s)-t\}$ . ∎

Proposition 4.3.

Let $M,N\in\operatorname{End}(V)$ . Suppose that $MN=NM$ . The following are equivalent.

$(1)$

All eigenvalues of $M$ and all eigenvalues of $N$ are in $\mathbb{K}$ .

(2)

The characteristic polynomial $\varphi_{({M}-sE)({N}-sE)}({t})$ is of the form

\varphi_{({M}-sE)({N}-sE)}({t})=\prod_{i=1}^{n}\{(\lambda_{i}-s)(\mu_{i}-s)-t\},\quad\lambda_{i},\mu_{i}\in\mathbb{K}.

Proof.

It is clear that $(2)$ implies $(1)$ . If all eigenvalues of $M$ , $N$ are in $\mathbb{K}$ , then $M$ , $N$ are simultaneously triangularizable. Hence, Proposition 4.2 gives $(2)$ . ∎

Proposition 4.4.

Let $M,N\in\operatorname{End}(V)$ and $\lambda,\mu\in\mathbb{K}$ . If $(\lambda-s)(\mu-s)$ is an eigenvalue of $({M}-sE)({N}-sE)$ , then $\lambda\mu$ is an eigenvalue of $MN$ and $NM$ and $\lambda+\mu$ is an eigenvalue of $M+N$ . Moreover, if the characteristic polynomial of $M$ is equal to that of $N$ , then $\lambda,\mu$ are eigenvalues of $M$ and $N$ .

Proof.

It is easy to see from Proposition 4.1. ∎

By using this proposition, we give an example of endomorphisms $M$ , $N$ which satisfy that $\varphi_{({M}-sE)({N}-sE)}({(\lambda-s)(\mu-s)})\neq 0$ for all $\lambda,\mu\in\mathbb{K}$ .

Example 4.5.

Let $V=\mathbb{K}^{2}$ . Define two matrices $M$ , $N$ by

M=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right),\ N=\left(\begin{array}[]{cccc}0&0\\ 1&0\end{array}\right).

It is easily shown that $\varphi_{M}(t)=\varphi_{N}(t)=t^{2}$ and that $\varphi_{M+N}(t)=(1-t)(1+t).$ Hence, Proposition 4.4 shows that $\varphi_{({M}-sE)({N}-sE)}({(\lambda-s)(\mu-s)})\neq 0$ for all $\lambda,\mu\in\mathbb{K}$ . We note that $\varphi_{({M}-sE)({N}-sE)}(t)=s^{4}-(1+2s^{2})t+t^{2}$ .

It is well-known that $(\lambda-s)(\mu-s)$ is an eigenvalue of $({M}-sE)({N}-sE)$ if and only if $\tilde{V}_{({M}-sE)({N}-sE)}((\lambda-s)(\mu-s))\neq\{0\}$ . We show that if $MN=NM$ , then

•

the space $\tilde{V}_{({M}-sE)({N}-sE)}((\lambda-s)(\mu-s))$ is generated by vectors in $\tilde{V}_{M,N}(\lambda,\mu)$ ,
•

the space $\tilde{V}_{M,N}(\lambda,\mu)$ is equal to $\tilde{V}_{MN}(\lambda\mu)\cap\tilde{V}_{M+N}(\lambda+\mu)$

We first give a lemma.

Lemma 4.6.

Let $M,N\in\operatorname{End}(V)$ and $\lambda,\mu,\in\mathbb{K}$ . Suppose that $MN=NM$ . Then

	$\displaystyle\tilde{V}_{M,N}(\lambda,\mu)$	$\displaystyle\subset\tilde{V}_{MN}(\lambda\mu)\cap\tilde{V}_{M+N}(\lambda+\mu),$
	$\displaystyle(\tilde{V}_{MN}(\lambda\mu)\cap\tilde{V}_{M+N}(\lambda+\mu))\otimes_{\mathbb{K}}\mathbb{K}(s)$	$\displaystyle\subset\tilde{V}_{(M-sE)(N-sE)}((\lambda-s)(\mu-s)).$

Proof.

To show the first implication, it suffices to see that $\tilde{V}_{M}(\lambda)\cap\tilde{V}_{N}(\mu)\subset\tilde{V}_{MN}(\lambda\mu)\cap\tilde{V}_{M+N}(\lambda+\mu)$ . Let $W=\tilde{V}_{M}(\lambda)\cap\tilde{V}_{N}(\mu)$ and $M_{1}=M|_{W}$ , $N_{1}=N|_{W}$ . Since $M_{1}$ , $N_{1}$ are simultaneously triangularizable, there exists a basis $\mathcal{B}_{1}$ of $W$ such that

[M_{1}]_{\mathcal{B}_{1}}=\left(\begin{array}[]{cccc}\lambda&&*\\ &\ddots\\ 0&&\lambda\end{array}\right),\ [N_{1}]_{\mathcal{B}_{1}}=\left(\begin{array}[]{cccc}\mu&&*\\ &\ddots\\ 0&&\mu\end{array}\right).

Hence, $W=\tilde{V}_{M_{1}N_{1}}(\lambda\mu)\cap\tilde{V}_{M_{1}+N_{1}}(\lambda+\mu)\subset\tilde{V}_{MN}(\lambda\mu)\cap\tilde{V}_{M+N}(\lambda+\mu)$ .

Let $Z=(\tilde{V}_{MN}(\lambda\mu)\cap\tilde{V}_{M+N}(\lambda+\mu))\otimes_{\mathbb{K}}\mathbb{K}(s)$ . It is easy to see that

\tilde{V}_{({M}-sE)({N}-sE)}((\lambda-s)(\mu-s))=\tilde{V}_{MN-s(M+N)}(\lambda\mu-s(\lambda+\mu)).

Since $MN|_{Z}$ , $(M+N)|_{Z}$ are simultaneously triangularizable, in the same way as above, we get $Z\subset\tilde{V}_{MN-s(M+N)}(\lambda\mu-s(\lambda+\mu))$ . ∎

Theorem 4.7.

Under the same assumption as the lemma above, we have

	$\displaystyle\tilde{V}_{M,N}(\lambda,\mu)$	$\displaystyle=\tilde{V}_{MN}(\lambda\mu)\cap\tilde{V}_{M+N}(\lambda+\mu),$
	$\displaystyle\tilde{V}_{M,N}(\lambda,\mu)\otimes_{\mathbb{K}}{\mathbb{K}(s)}$	$\displaystyle=\tilde{V}_{({M}-sE)({N}-sE)}((\lambda-s)(\mu-s)).$

Proof.

By Lemma 4.6, it suffices to prove that

\tilde{V}_{({M}-sE)({N}-sE)}((\lambda-s)(\mu-s))\subset\tilde{V}_{M,N}(\lambda,\mu)\otimes_{\mathbb{K}}\mathbb{K}(s).

Let $W=\tilde{V}_{({M}-sE)({N}-sE)}((\lambda-s)(\mu-s))$ and let $M_{1}=M|_{W}$ , $N_{1}=N|_{W}$ . Then

(4.1)

\varphi_{M_{1}}(s)\varphi_{N_{1}}(s)=(\lambda-s)^{k}(\mu-s)^{k}.

In the case where $\lambda=\mu$ , this implies that $W=\tilde{V}_{M_{1}}(\lambda)\cap\tilde{V}_{N_{1}}(\lambda)=\tilde{V}_{M_{1},N_{1}}(\lambda,\lambda)\subset\tilde{V}_{M,N}(\lambda,\lambda)\otimes_{\mathbb{K}}\mathbb{K}(s).$ Suppose that $\lambda\neq\mu$ . Then the equality (4.1) implies that $W$ is the sum of $\tilde{V}_{M_{1},N_{1}}(\lambda,\mu)$ , $\tilde{V}_{M_{1}}(\lambda)\cap\tilde{V}_{N_{1}}(\lambda)$ , $\tilde{V}_{M_{1}}(\mu)\cap\tilde{V}_{N_{1}}(\mu)$ . Applying Lemma 4.6 for $M_{1}$ , $N_{1}$ , we get

\tilde{V}_{M_{1}}(\lambda)\cap\tilde{V}_{N_{1}}(\lambda)\subset\tilde{V}_{(M_{1}-sE)(N_{1}-sE)}(\lambda-s)(\lambda-s))=\{0\}.

In the same way, $\tilde{V}_{M_{1}}(\mu)\cap\tilde{V}_{N_{1}}(\mu)=\{0\}$ . Hence, we get $W=\tilde{V}_{M_{1},N_{1}}(\lambda,\mu)\subset\tilde{V}_{M,N}(\lambda,\mu)\otimes_{\mathbb{K}}\mathbb{K}(s)$ . ∎

As a consequence of Theorem 4.7, we have the following.

Corollary 4.8.

Let $M,N\in\operatorname{End}(V)$ . Suppose that $MN=NM$ and that

\varphi_{({M}-sE)({N}-sE)}({t})=\prod_{i=1}^{k}\{(\lambda_{i}-s)(\mu_{i}-s)-t\}^{m_{i}}

so that $(\lambda_{i}-s)(\mu_{i}-s)\neq(\lambda_{j}-s)(\mu_{j}-s)$ for $i\neq j$ . Then we have the following.

•

The space $V$ is the direct sum of the $m_{i}$ -dimensional $(M,N)$ -invariant subspaces $\tilde{V}_{M,N}(\lambda_{i},\mu_{i})$ .
•

Let $P_{1},\dots,P_{k}$ be projections with respect to $V=\bigoplus_{i=1}^{k}\tilde{V}_{M,N}(\lambda_{i},\mu_{i})$ and let $Q_{1},\dots,Q_{k}$ be projections with respect to

$V\otimes_{\mathbb{K}}\mathbb{K}(s)=\bigoplus_{i=1}^{k}\tilde{V}_{({M}-sE)({N}-sE)}((\lambda_{i}-s)(\mu_{i}-s)).$

Then $P_{i}=Q_{i}.$

Proposition 4.9.

Let $M,N\in\operatorname{End}(V)$ and $\lambda,\mu\in\mathbb{K}$ . Suppose that $MN=NM$ . Then $(\lambda-s)(\mu-s)$ is an eigenvalue of $({M}-sE)({N}-sE)$ if and only if $V_{M,N}(\lambda,\mu)\neq\{0\}$ .

Proof.

Suppose that $V_{M,N}(\lambda,\mu)\neq\{0\}$ . Let $v\in(V_{M,N}(\lambda,\mu))\otimes_{\mathbb{K}}\mathbb{K}(s)$ . Then we get $({M}-sE)({N}-sE)v=(\lambda-s)(\mu-s)v$ . Hence, $(\lambda-s)(\mu-s)$ is an eigenvalue of $({M}-sE)({N}-sE)$ .

The only if part follows from Theorem 4.7 and the fact that $V_{M,N}(\lambda,\mu)=\{0\}$ is equivalent to $\tilde{V}_{M,N}(\lambda,\mu)=\{0\}$ . ∎

5. Applications of the symplectic characteristic polynomial

In this section, we prove several results for symplectically normal endomorphisms which are concerned with the symplectic characteristic polynomial.

Let $(V,\omega)$ be a $2n$ -dimensional symplectic vector space over a field $\mathbb{K}$ .

Proposition 5.1.

Let $M\in\operatorname{End}(V)$ . Suppose that $MM^{*\omega}=M^{*\omega}M$ . The following are equivalent.

$(1)$

All eigenvalues of $M$ are in $\mathbb{K}$ .
$(2)$

The characteristic polynomial $\varphi^{\omega}_{M}(s,t)$ is of the form

$\varphi^{\omega}_{M}(s,t)=\prod_{i=1}^{n}\{(\lambda_{i}-s)(\mu_{i}-s)-t\},\quad\lambda_{i},\mu_{i}\in\mathbb{K}.$

Proof.

This follows from $\varphi^{\omega}_{M}(s,t)^{2}=\varphi_{({M}-sE)({M}^{*\omega}-sE)}(t)$ and Proposition 4.3. ∎

Let $M\in\operatorname{End}(V)$ . Suppose that all eigenvalues of $M$ are in $\mathbb{K}$ . By using the same idea of the proof of Lemma 1 in [3], we can show that $V$ is the symplectically orthogonal direct sum of the symplectic subspaces $\tilde{V}_{{M},{M}^{*\omega}}(\lambda,\mu)$ associated with two eigenvalues $\lambda,\mu$ of $M$ . We reformulate this result with the symplectic characteristic polynomial and give a short proof.

Lemma 5.2.

Let $M\in\operatorname{End}(V)$ . Suppose that $MM^{*\omega}=M^{*\omega}M$ . If the symplectic characteristic polynomial $\varphi^{\omega}_{M}(s,t)$ is of the form

\varphi^{\omega}_{M}(s,t)=\prod_{i=1}^{k}\{(\lambda_{i}-s)(\mu_{i}-s)-t\}^{m_{i}},\quad\lambda_{i},\mu_{i}\in\mathbb{K}

so that $(\lambda_{i}-s)(\mu_{i}-s)\neq(\lambda_{j}-s)(\mu_{j}-s)$ for $i\neq j$ , then $V$ is the symplectically orthogonal direct sum of the $2m_{i}$ -dimensional symplectic subspaces $\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})$ . Precisely, we have the following.

$(1)$

$\dim_{\mathbb{K}}\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})=2m_{i}$ .
$(2)$

$V=\bigoplus_{i=1}^{k}\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})$ .
$(3)$

$\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})\subset\left(\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{j},\mu_{j})\right)^{\perp\omega}$ for $i\neq j$ .
$(4)$

$\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})$ is an $(M,M^{*\omega})$ -invariant symplectic subspace of $(V,\omega)$ .

Proof.

Let $M_{s}^{\omega}=(M-sE)(M^{*\omega}-sE)$ . By Theorem 4.7, we have

(5.1)

\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})\otimes_{\mathbb{K}}\mathbb{K}(s)=\tilde{V}_{M_{s}^{\omega}}((\lambda_{i}-s)(\mu_{i}-s)).

Since $\varphi^{\omega}_{M}(s,t)^{2}=\varphi_{({M}-sE)({M}^{*\omega}-sE)}(t)$ , this implies $(1)$ and $(2)$ . The item $(5)$ in Proposition 2.2 shows that $\tilde{V}_{M_{s}^{\omega}}((\lambda_{i}-s)(\mu_{i}-s))\subset\left(\tilde{V}_{M_{s}^{\omega}}((\lambda_{j}-s)(\mu_{j}-s))\right)^{\perp\omega}.$ Moreover, Proposition 2.3 gives that $\tilde{V}_{M_{s}^{\omega}}((\lambda_{i}-s)(\mu_{i}-s))$ is a symplectic subspace of $(V\otimes_{\mathbb{K}}\mathbb{K}(s),\omega)$ . Hence, from (5.1), we get $(3)$ and $(4)$ . ∎

In the symplectic case, we have a stronger result than Proposition 4.9.

Proposition 5.3.

Let $M\in\operatorname{End}(V)$ and $\lambda,\mu\in\mathbb{K}$ . Suppose that $MM^{*\omega}=M^{*\omega}M$ . Then $\varphi^{\omega}_{M}(s,(\lambda-s)(\mu-s))=0$ if and only if $V_{M}(\lambda)\cap V_{M^{*\omega}}(\mu)\neq\{0\}$ .

Proof.

By Proposition 4.9, it is enough to show that $V_{{M},{M}^{*\omega}}(\lambda,\mu)\neq\{0\}$ implies $V_{M}(\lambda)\cap V_{M^{*\omega}}(\mu)\neq\{0\}$ . The case where $\lambda=\mu$ is clear. Suppose that $\lambda\neq\mu$ . Lemma 5.2 shows that $\tilde{V}_{{M},{M}^{*\omega}}(\lambda,\mu)$ is a symplectic subspace of $(V,\omega)$ . Hence, by Lemma 2.4, we have $\tilde{V}_{M}(\lambda)\cap\tilde{V}_{M^{*\omega}}(\mu)\neq\{0\}$ . Since $MM^{*\omega}=M^{*\omega}M$ , this implies $V_{M}(\lambda)\cap V_{M^{*\omega}}(\mu)\neq\{0\}$ . ∎

It is proved in [2, Theorem 13] that if an endomorphism is symplectically normal and diagonalizable, then the endomorphism is symplectically diagonalizable. The following theorem is an improvement of this fact.

Theorem 5.4.

Let $M\in\operatorname{End}(V)$ . Suppose that $MM^{*\omega}=M^{*\omega}M$ and that $M$ is diagonalizable. Then there exists a symplectic basis $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ such that

Me_{i}=\lambda_{i}e_{i},\ Mf_{i}=\mu_{i}f_{i},\quad i\in\{1\dots,n\}

where the symplectic characteristic polynomial $\varphi^{\omega}_{M}(s,t)$ is of the form

\varphi^{\omega}_{M}(s,t)=\prod_{i=1}^{n}\{(\lambda_{i}-s)(\mu_{i}-s)-t\},\quad\lambda_{i},\mu_{i}\in\mathbb{K}.

Proof.

We proof by induction on $n$ . The item $(4)$ in Lemma 5.2 shows that $V_{{M},{M}^{*\omega}}(\lambda_{1},\mu_{1})$ is a symplectic subspaces of $(V,\omega)$ . There exist $e_{1},f_{1}\in V$ such that

e_{1}\in V_{M}(\lambda_{1})\cap V_{M^{*\omega}}(\mu_{1}),\ f_{1}\in V_{M}(\mu_{1})\cap V_{M^{*\omega}}(\lambda_{1}),\ \omega(e_{1},f_{1})=1.

Indeed, the case where $\lambda=\mu$ follows from Proposition 2.5 and the case where $\lambda\neq\mu$ follows from Lemma 2.4 and 2.6. Hence, the assertion holds for $n=1$ . We assume that the assertion holds for $n-1$ . Let $W$ be a subspace generated by $e_{1},f_{1}$ . Then $W$ is an $(M,M^{*\omega})$ -invariant symplectic subspace of $(V,\omega)$ and by the item $(4)$ in Proposition 2.2, so is $W^{\perp\omega}$ . Let $M_{1}=M|_{W^{\perp\omega}}$ and $\omega_{1}=\omega|_{W^{\perp\omega}}$ . Since $M_{1}^{*\omega_{1}}=M^{*\omega}|_{W^{\perp\omega}}$ , we have $M_{1}M_{1}^{*\omega_{1}}=M_{1}^{*\omega_{1}}M_{1}$ . It is clear that $M_{1}$ is diagonalizable. The induction hypothesis shows that there exists a symplectic basis $(e_{2},\dots,e_{n},f_{1},\dots,f_{n})$ of $(W^{\perp\omega},\omega|_{W^{\perp\omega}})$ such that $Me_{i}=\lambda_{i}e_{i},\ Mf_{i}=\mu_{i}f_{i}$ for $i\in\{2,\dots,n\}$ . It is clear that $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ is a symplectic basis which is desired. ∎

Corollary 5.5.

The symplectic characteristic polynomial is a complete invariant with respect to symplectic similarity for symplectically diagonalizable endomorphisms. In particular, a symplectically diagonalizable endomorphism $M$ is symplectically similar to the symplectic adjoint endomorphism $M^{*\omega}$ .

Proof.

Let $M$ , $N$ be symplectically diagonalizable endomorphisms. Suppose that $\varphi^{\omega}_{M}(s,t)=\varphi^{\omega}_{N}(s,t)$ . Theorem 5.4 shows that there exist symplectic bases $\mathcal{B}$ , $\mathcal{B}^{\prime}$ such that $[M]_{\mathcal{B}}=[N]_{\mathcal{B}^{\prime}}$ . Hence, Proposition 2.7 gives that $M$ is symplectically similar to $N$ . ∎

It is well-known that if the number of distinct eigenvalues of an endomorphism is equal to $\dim V$ , then the endomorphism is diagonalizable. In the symplectic case, we have the following two theorems.

Theorem 5.6.

Let $M\in\operatorname{End}(V)$ . Suppose that $MM^{*\omega}=M^{*\omega}M$ and that the symplectic characteristic polynomial $\varphi^{\omega}_{M}(s,t)$ is of the form

\varphi^{\omega}_{M}(s,t)=\prod_{i=1}^{n}\{(\lambda_{i}-s)(\mu_{i}-s)-t\},\quad\lambda_{i},\mu_{i}\in\mathbb{K}

and $(\lambda_{i}-s)(\mu_{i}-s)\neq(\lambda_{j}-s)(\mu_{j}-s)$ for $i\neq j.$ Then there exists a symplectic basis $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ such that

	$\displaystyle MM^{*\omega}e_{i}$	$\displaystyle=\lambda_{i}\mu_{i}e_{i},$	$\displaystyle(M+M^{*\omega})e_{i}$	$\displaystyle=(\lambda_{i}+\mu_{i})e_{i},$
	$\displaystyle MM^{*\omega}f_{i}$	$\displaystyle=\lambda_{i}\mu_{i}f_{i},$	$\displaystyle(M+M^{*\omega})f_{i}$	$\displaystyle=(\lambda_{i}+\mu_{i})f_{i},$	$\displaystyle i\in\{1,\dots,n\}.$

Proof.

Let $W_{i}=\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})$ and let $M_{i}=M|_{W_{i}}$ , $\omega_{i}=\omega|_{W_{i}}$ . Lemma 5.2 shows that $\dim_{\mathbb{K}}W_{i}=2$ and that $W_{i}$ is symplectic. Hence, we get $\varphi^{\omega_{i}}_{M_{i}}(s,t)=\{(\lambda_{i}-s)(\mu_{i}-s)-t\}$ , which implies that $\varphi_{M_{i}}(t)=(\lambda_{i}-s)(\mu_{i}-s)$ . It follows from $M^{*\omega}|_{W_{i}}=(M_{i})^{*\omega_{i}}$ that

	$\displaystyle(MM^{*\omega})\|_{W_{i}}$	$\displaystyle=M_{i}M_{i}^{*\omega_{i}}=(\det M_{i})E=\lambda_{i}\mu_{i}E,$
	$\displaystyle(M+M^{*\omega})\|_{W_{i}}$	$\displaystyle=M_{i}+M_{i}^{*\omega_{i}}=(\operatorname{tr}M_{i})E=(\lambda_{i}+\mu_{i})E.$

Since $W_{i}$ is symplectic, there exist $e_{i},f_{i}\in W_{i}$ such that $\omega(e_{i},f_{i})=1$ . The item $(3)$ in Lemma 5.2 shows that $(e_{1},\dots,e_{n},f_{1}\dots,f_{n})$ is a symplectic basis, which completes the proof. ∎

Theorem 5.7.

Under the same assumption as above, we assume further that $\lambda_{i}\neq\mu_{i}$ for all $i$ . Then there exists a symplectic basis $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ such that

	$\displaystyle Me_{i}$	$\displaystyle=\lambda_{i}e_{i},$	$\displaystyle M^{*\omega}e_{i}$	$\displaystyle=\mu_{i}e_{i},$
	$\displaystyle Mf_{i}$	$\displaystyle=\mu_{i}f_{i},$	$\displaystyle M^{*\omega}f_{i}$	$\displaystyle=\lambda_{i}f_{i},$	$\displaystyle i\in\{1,\dots,n\}.$

Proof.

Let $W_{i}=\tilde{V}_{{M},{M}^{*\omega}}(\lambda_{i},\mu_{i})$ . Lemma 5.2 gives that $\dim_{\mathbb{K}}W=2$ and that $W_{i}$ is a symplectic subspace. Hence, Proposition 5.3 shows that

\dim_{\mathbb{K}}\left(V_{M}(\lambda_{i})\cap V_{M^{*\omega}}(\mu_{i})\right)=\dim_{\mathbb{K}}\left(V_{M}(\mu_{i})\cap V_{M^{*\omega}}(\lambda_{i})\right)=1.

Since $W_{i}$ is symplectic, there exist $e_{i},f_{i}\in V$ such that

e_{i}\in V_{M}(\lambda_{i})\cap V_{M^{*\omega}}(\mu_{i}),\ f_{i}\in V_{M}(\mu_{i})\cap V_{M^{*\omega}}(\lambda_{i}),\ \omega(e_{i},f_{i})=1.

The item $(3)$ in Lemma 5.2 shows that $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ is a symplectic basis which is desired. ∎

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