We have classified symplectic matrices in the previous section. Now we are going to classify regular symplectic pairs. In [8] Mehrmann and Poloni rearranged the blocks of given symplectic pair ingeniously, which makes a connection between regular symplectic pairs and symplectic matrices. By this rearrangement some properties of Lagrangian subspaces and symplectic matrices can be applied to classify symplectic pairs. For succinct statements, we first give the following definitions.
Definition 4.1. Let (A, B) =
([ A11 A12 A21 A22
] ,
[ B11 B12 B21 B22
])
∈ SPn. The matrix S(A,B) ∈ C4n×4n is defined by
S(A,B) ..=
B11∗ B21∗ −B12∗ −B22∗
A∗12 A∗22 −A∗11 −A∗21
B12∗ B22∗ B11∗ B21∗ A∗11 A∗21 A∗12 A∗22
.
Definition 4.1 is due to the work in [8] mentioned above. In fact, we shall prove that the matrix S(A,B) is symplectic later in Proposition 4.3. From the above definition, we see two 2n× 2n symplectic matrices merging into one 4n × 4n symplectic matrix. To connect the two different dimensional spaces, we need the following definitions.
Definition 4.2. Let v1, v2 ∈ {0, 1}n. v1⊕ v2 ∈ {0, 1}2n is defined by (v1 ⊕ v2)j =
{ (v1)j, for 1≤ j ≤ n, (v2)j−n, for n + 1≤ j ≤ 2n.
Definition 4.3. Let Π1, Π2 ∈ Pnand v1, v2 ∈ {0, 1}nsuch that Π1 = Πv1 and Π2 = Πv2. Π1 ⊕ Π2 ∈ P2n is defined by
Π1⊕ Π2 = Πv1⊕v2.
Now we prove some lemmas which will be used in our main theorem. Lemma 2.13 shows some connection between an orthogonal basis of a Lagrangian subspace of C2n and a symplectic matrix in C2n×2n. With the help of Lemma 2.13, Lemma 3.2 can be applied in Lemma 4.1 to show that for any Lagrangian subspace basis matrix U , the rows of U can be rearranged to obtain a nonsingular block we need. The proof of Lemma 4.1 is mainly referred to Theorem 3.1 in [8].
Lemma 4.1. If the columns of U ∈ C2n×n span a Lagrangian subspace of C2n, then there exists some Π ∈ Pn such that ΠU =
[ YΠ ZΠ
]
, where YΠ, ZΠ ∈ Cn×n and YΠ is nonsingular.
Proof. Let U = QR be a QR factorization, where Q∈ C2n×n is an orthogonal matrix and R ∈ Cn×n is an upper triangle matrix. Since the columns of U span a Lagrangian subspace, U is of full column rank and thus R is nonsingular. (If R is singular, then there exists some column vector v ̸= 0 such that Rv = 0. It follows that Uv = QRv = 0.
This contradicts that U is of full column rank.) Hence Q = U R−1 is also of full column rank and its columns span a Lagrangian subspace. Let Q be partitioned as Q =
[ Q1 Q2
]
where Q1, Q2 ∈ Cn×n. Since the columns of Q are orthogonal and span a Lagrangian subspace, by Lemma 2.13, [
Q1 −Q2
Q2 Q1 ]
is orthogonal and symplectic. Let Rα(Q1)∈ MI(Q1), then by Corollary 2.11, the rows of Rα′(Q2) together with the rows of Rα(Q1) constitute an n× n nonsingular matrix.
If we pick v ∈ {0, 1}n such that
vi =
{ 0, i∈ α, 1, i /∈ α,
then with the associated swap matrix Πv ∈ Pn, we have that ΠvQ = [ Q′1
Q′2 ]
with Q′1 nonsingular. Hence in ΠvU = ΠvQR =
[ Q′1 Q′2
] R =
[ Q′1R Q′2R
]
we also have that Q′1R is nonsingular.
Lemma 4.2. Let U = [ I
X ]
∈ C2n×n where I, X ∈ Cn×n and I is an identity matrix.
If the columns of U span a Lagrangian subspace, then X is Hermitian.
Proof. Since the columns of [ I
X ]
span a Lagrangian subspace, by Proposition 2.12 we know that I∗X = X∗I. That is, X = X∗. Therefore X is Hermitian.
Lemma 4.1 shows that we can rearrange a Lagrangian subspace basis matrix to obtain a nonsingular block. Combining this to Lemma 4.2, it shows that by multiplying the inverse of the nonsingular block, we can obtain a Hermitian matrix in the other block. Now we shall prove that S(A,B) is symplectic we claim after Definition 4.1. This proof mainly refers to Theorem 6.1 in [8].
Proposition 4.3. If (A, B) ∈ SPn, then S(A,B) ∈ C4n×4n is symplectic. That is, S(A,B) ∈ S2n.
Proof. Since (A, B) is a symplectic pair, (A, B) satisfies the condition AJ A∗ = BJ B∗, Equation (4.1) can be rewritten as
[ B11 A12 B12 A11
Moreover, we see that the 2n× 4n matrix
[ B11 A12 B12 A11 B21 A22 B22 A21
]
is of full row rank, for otherwise we can find a nonzero vector w such that
[ B11 A12 B12 A11
= 0. By shuffling the columns we have
w∗
is of full column rank. Combining this result with equation (4.2), we conclude that the columns of
[ B11 A12 B12 A11 B21 A22 B22 A21
]∗
span a Lagrangian subspace of C4n. Then by Lemma 2.13,
S(A,B) =
is a symplectic matrix.
Before proving Proposition 4.5, which shows close relationship between (A, B) ∈ SPn and S(A,B)∈ S2n, we need an easily seen lemma.
Lemma 4.4. If ΠS =
The close relationship between a regular symplectic pair (A, B) and its correspond-ing symplectic matrix S(A,B) is more than Proposition 4.3 tells. The following proposi-tion gives us an insight to see why we can classify regular symplectic pairs with some results obtained from symplectic matrices.
Proposition 4.5. Let (A, B) ∈ SPn and Π1, Π2 ∈ Pn . Then (A, B) ∈ SPΠ2,Π1 if and and S22′ is nonsingular. By Lemma 4.4 S11′ is also nonsingular. Moreover, by Propo-sition 2.14, we see that the columns of
[ B11 A12 B12 A11 B21 A22 B22 A21
]∗
span a Lagrangian subspace. From equation (4.3) we can derive that
(Π1⊕ Π2) of Lagrangian subspace, the columns of
[ I X
]
also spans a Lagrangian subspace. By Lemma 4.2 we see that X is Hermitian. Let X =
Rewrite equation (4.4) we have [ diag(bv1) diag(v1)
−diag(v1) diag(bv1)
] [ B11∗ B21∗ B12∗ B22∗
]
=
[ I O
X11 X12 ]
S11′ (4.6)
and [
diag(bv2) diag(v2)
−diag(v2) diag(bv2)
] [ A∗12 A∗22 A∗11 A∗21
]
=
[ O I
X21 X22 ]
S11′ . (4.7)
Equation (4.6) can be written as Π1B∗ =
[ I O
X11 X12 ]
S11′ . Take the conjugate trans-pose of both sides and by (4.5) we have
BΠ∗1 = S11′∗
[ I X11∗ O X12∗
]
= S11′∗
[ I X11 O X12∗
]
. (4.8)
By Proposition 2.7 we know that Π−11 = Π∗1. Hence we have B = S11′∗
[ I X11 O X21
]
Π1. (4.9)
Now we turn to equation (4.7). Take the conjugate transpose of both sides we have [ A12 A11
A22 A21 ]
Π∗2 = S11′∗
[ O X21∗ I X22∗
]
= S11′∗
[ O X12 I X22
]
. (4.10)
Rewrite the left side of (4.10) as following [ A12 A11
A22 A21 ]
Π∗2 =
[ A12 A11
A22 A21
] [ O I I O
] [ O I I O
] Π∗2
=
[ A11 A12 A21 A22
] [ O I I O
]
Π∗2. (4.11)
Combining (4.10) and (4.11) and moving some terms from left to right we then have
By (4.9), (4.12), and Definition 1.2, we conclude that (A, B)∼
for some Hermitian matrix X =
[ X11 X12
for some Hermitian matrix X =
[ X11 X12
for some nonsingular matrix M . From equations (4.6), (4.8), and (4.9) we see that equation (4.13) can be rewritten as
[ diag(bv1) diag(v1)
From equations (4.7), (4.10), (4.11), and (4.12) we see that equation (4.14) can be Combining equations (4.15) and (4.16) and rearranging the blocks we obtain
Now we proceed to prove the main theorem of this section.
Theorem 4.6 (Classification of Regular Symplectic Pairs). (i) For each regular sym-plectic pair (A, B) ∈ SPn, there exist swap matrices Π1, Π2 ∈ Pn and a Hermitian
Proof. We first prove assertion (i). This proof mainly refers to [8]. Let (A, B) ∈ SPn. By Proposition 4.3, S(A,B) ∈ S2n. Then by Lemma 3.2, there exists Π ∈ P2n such that S(A,B) ∈ SΠ. Let Π1, Π2 ∈ Pn such that Π1 ⊕ Π2 = Π. By Proposition 4.5, (A, B)∈ SPΠ2,Π1. That is,
(A, B)∼
([ X12 O X22 I
] Π∗2,
[ I X11 O X12∗
] Π1
) .
for some Hermitian matrix X =
[ X11 X12 X12∗ X22
]
∈ C2n×2n. Assertion (ii) is simply a corollary of assertion (i). Since for each regular symplectic pair (A, B), there eixst Π1, Π2 ∈ Pn such that (A, B) ∈ SPΠ1,Π2, we have SPn ⊂ ∪
Π1,Π2∈PnSPΠ1,Π2. On the other hand, by Definition 1.4 we know that each element of SPΠ1,Π2 is an element of SPn. Hence SPn = ∪
Π1,Π2∈PnSPΠ1,Π2. Since there are 22n elements in Pn × Pn, we break SPn into 22n classes. However, this classification is not minimal. We will give an example later. Finally we are going to prove that each SPΠ1,Π2 is an open set in SPn. The metric we adopt here is defined by
∥(A1, B1)−(A2, B2)∥..=
( ∑
1≤i,j≤2n
| (A1)ij − (A2)ij |2 +| (B1)ij − (B2)ij |2 )1/2
(4.20)
for (A1, B1), (A2, B2)∈ SPn. Let Π1, Π2 ∈ Pnbe fixed and assume that (A, B)∈ SΠ1,Π2. By Proposition 4.5, S(A,B) ∈ SΠ2⊕Π1 ⊂ S2n. Then by Theorem 3.3 (iv), SΠ2⊕Π1 is relatively open to S2n. That is, there exists some δ > 0 such that S ∈ S2n together with ∥S − S(A,B)∥ < δ implying that S ∈ SΠ2⊕Π1. By the definitions of metric we adopted (see equations (3.6) and (4.20)) and Definition 4.1 of S(A,B), we see that
∥S(C,D) − S(A,B)∥2 = 2∥(C, D) − (A, B)∥2 (4.21) for any (C, D) ∈ SPn with S(C,D) ∈ S2n. If we choose d = δ/√
2. Then ∥(C, D) − (A, B)∥ < d implies that ∥S(C,D)− S(A,B)∥ < δ, and thus S(C,D) ∈ SΠ2⊕Π1. Applying Proposition 4.5 again, we conclude that (C, D)∈ SΠ1,Π2.
Comparing Theorems 3.3 and 4.6, we can see in Theorem 4.6 there is no correspon-dent statement to Theorem 3.3 (iii), “for each SPΠ1,Π2, there exists (A, B) ∈ SPΠ1,Π2
such that (A, B) /∈ SΠ′1,Π′2 for all Π′1 ̸= Π1 or Π′2 ̸= Π2.” In fact, this statement could not be true in P1. We give an example as following.
Example 4.1. Let I =
[ 1 0 0 1
]
and J =
[ 0 1
−1 0 ]
. Then SPI,J ⊂ SP1 is such a class that each element in SPI,J must also be in another class.
Proof. There are only four classes in SP1, i.e., SPI,I,SPI,J,SPJ,I, and SPJ,J. Suppose that (A, B) ∈ SP1 such that (A, B) ∈ SPI,J but (A, B) /∈ SPI,I, (A, B) /∈ SPJ,I, and (A, B) /∈ SPJ,J. By Proposition 4.5, we have that
S(A,B) ∈ SI⊕J,S(A,B) ∈ S/ I⊕I,S(A,B) ∈ S/ J⊕I, andS(A,B) ∈ S/ J⊕J. (4.22) For convenience, we just consider real entries here. Let A =
[ a b
. Doing some computatione we have that
By equations (4.22) to (4.26) and Definition 1.4 of SΠ, we have that f h
, we discuss in four cases.
(Case 1) If
This would lead a contradiction to the regularity assumption.
Since (A, B) is a symplectic pair, we see that ah = cf . But this contradicts equation (4.27) that
f h
So we conclude that such (A, B) does not exist and that each element in SPI⊕J must be also in some other class.