The antitriangular factorization of skew-symmetric matrices

1 Introduction

Skew-symmetric matrices are significantly less used than the symmetric ones. Many algorithms designed for the symmetric matrices have been transformed in the course of last two decades to work with the skew-symmetric and other structured matrices, to avoid using algorithms for general, nonstructured, matrices.

Mastronardi and Van Dooren in Mastronardi-VanDooren-13 showed that every symmetric and indefinite matrix $A \in R^{n \times n}$ can be transformed into the block antitriangular form by orthogonal similarities. More precisely, if $inertia (A) = (n_{-}, n_{0}, n_{+})$ , $n_{1} = min (n_{-}, n_{+})$ , $n_{2} = max (n_{-}, n_{+}) - n_{1}$

, there exists orthogonal matrix

Q \in R^{n \times n}

such that

M = Q^{T} A Q = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 & 0 0 & 0 & 0 & Y^{T} 0 & 0 & X & Z^{T} 0 & Y & Z & W \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦,

(1.1)

where $Y \in R^{n_{1} \times n_{1}}$ is nonsingular and lower antitriangular, $W \in R^{n_{1} \times n_{1}}$ is symmetric, $X \in R^{n_{2} \times n_{2}}$ is symmetric and definite, and $Z \in R^{n_{1} \times n_{2}}$ .

Bujanović and Kressner in Bujanovic-Kressner-16 derived computationally effective block algorithm that computes the block antitriangular factorization (1.1). Unfortunately that algorithm sometimes fails to detect the inertia. A new algorithm for the antitriangular factorization was presented in Laudadio-Mastronardi-VanDooren-16 .

Pestana and Wathen in Pestana-Wathen-14 simplified the algorithm for the special saddle point matrices

A = [\begin{matrix} H & B^{T} B & 0 \end{matrix}],

where $H \in R^{k \times k}$ is symmetric, but not necessary positive definite, and $B \in R^{m \times k}$ , $m \geq k$ .

In this paper we show that the skew-symmetric matrices have antitriangular form, while the skew-Hermitian have the block antitriangular form similar to the block antitriangular form of the real symmetric matrices.

In the next section of the paper we constructively prove that every skew-symmetric matrix can be transformed into the lower antritriangular form, and establish the connection between the number of nontrivial antidiagonals and the rank of the skew-symmetric matrix. In Section 3 we show that the antitriangular form can be reorganized to the multi-arrowhead form. Section 4 contains the results about block antitriangular form of the Hermitian, and, therefore, skew-Hermitian matrices.

2 Factorization of a skew-symmetric matrix into antitriangular form

In this section we constructively prove that every skew-symmetric matrix can be reduced to antitriangular form, by using orthogonal similarity transformations.

To this end we use Givens rotations, since Jacobi rotations $Q_{i j} : = Q (i, j, φ_{i j})$ cannot annihilate element at the position $(i, j)$ in a skew symmetric matrix $A$ . Suppose that $A_{i j}$ is a skew symmetric matrix of order $2$ , and $Q_{i j}$ is a rotation. Then we have

Q_{i j}^{T} A_{i j} Q_{i j} = [\begin{matrix} cos φ & sin φ - sin φ & cos φ \end{matrix}] [\begin{matrix} 0 & a_{i j} - a_{i j} & 0 \end{matrix}] [\begin{matrix} cos φ & - sin φ sin φ & cos φ \end{matrix}] = [\begin{matrix} 0 & a_{i j} - a_{i j} & 0 \end{matrix}] = A_{i j} .

Therefore, we will use Givens rotation $Q_{i j}$ to annihilate element at positions $(i, k)$ , and $(k, i)$ , $k \neq j$ , or at positions $(k, j)$ , and $(j, k)$ , $k \neq i$ .

Theorem 2.1.

Let $A \in R^{n \times n}$ be a skew-symmetric matrix. Then matrix $A$ can be factored as

A = Q M Q^{T},

where $Q$ is orthogonal, and $A$ is antitriangular matrix.

Proof.

The proof is by induction over the number of already annihilated antidiagonals of a skew-symmetric matrix $A$ .

Note that $A$ has zero on its position $(1, 1)$ , and this fact serves as basis of induction.

Suppose that after $k - 1$ annihilated antidiagonals $M_{k - 1}$ has the following form,

M_{k - 1} : = Q_{k - 1}^{T} A Q_{k - 1} = [\begin{matrix} M_{11} & M_{12} - M_{12}^{T} & M_{22} \end{matrix}],

(2.1)

where

M1,1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0⋯⋯0⋮\iddotsm2,k−1⋮\iddots\iddots⋮0−m2,k−1⋯mk−1,k−1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦,

(2.2)

while the matrices $M_{12}$ and $M_{22}$ are generally full. In matrix $Q_{k - 1}$ we keep the product of used rotations. If $n = k$ , we have completed the job. Otherwise, in the next step we annihilate $k$ th antidiagonal.

First we annihilate elements at positions $(1, k)$ and $(k, 1)$ by using rotation $Q_{k, k + 1}$ in the plane $(k, k + 1)$

that is equal to identity matrix except at crossings of the

k

th and

(k + 1)

th rows and columns, where

{ˆ Q}_{k, k + 1} = [\begin{matrix} cos φ_{k, k + 1} & - sin φ_{k, k + 1} sin φ_{k, k + 1} & cos φ_{k, k + 1} \end{matrix}] .

(2.3)

We may assume that the elements at positions $(1, k)$ and $(k, 1)$ are nonzero. Otherwise, we may skip this transformation.

Since the element at the position $(1, k)$ is transformed only from the right-hand side (and element at the position $(k, 1)$ only from the left-hand side), new elements at these positions are

	$m_{1 k}^{'}$	$= m_{1 k} cos φ_{k, k + 1} + m_{1, k + 1} sin φ_{k, k + 1},$
	$m_{k 1}^{'}$	$= - (m_{1 k} cos φ_{k, k + 1} + m_{1, k + 1} sin φ_{k, k + 1}) = - m_{1 k}^{'} .$

By choosing

cot φ_{k, k + 1} = - \frac{m_{1, k + 1}}{m_{1 k}},

(2.4)

from the basic identity for the trigonometric functions ${sin}^{2} φ_{k, k + 1} + {cos}^{2} φ_{k, k + 1} = 1$ , it is easy to derive that the sines and cosines in (2.1) (which annihilate $m_{1, k}^{'}$ ) are

sin φ_{k, k + 1} = \pm \frac{1}{\sqrt{1 + {cot}^{2} φ_{k, k + 1}}}, cos φ_{k, k + 1} = sin φ_{k, k + 1} cot φ_{k, k + 1},

where $cot φ_{k, k + 1}$ is defined by (2.4).

The next step is to annihilate elements at positions $(2, k - 1)$ and $(k - 1, 2)$ by using rotation in the plane $(k - 1, k)$ . This transformation will not destroy the zero pattern, since rows/columns $k - 1$ and $k$ already have zeros as the first elements in the corresponding row/column.

In the similar way all the elements of the $k$ th antidiagonal will be annihilated without destroying already introduced zeros.

After the annihilation in this step we obtain $M_{k}$ which has the same form as $M_{k - 1}$ from (2.1), but the matrix $M_{1, 1}$ , still antitriangular, has one row and column more than the matrix $M_{1, 1}$ from (2.2). This was the step of the induction.

We proceed with the annihilation of one antidiagonal after another until $k$ becomes $n$ . ∎

As one can expect, since the skew-symmetric matrices have eigenvalues in pairs

\pm λ i

, one ‘positive’ and one ‘negative’ on the imaginary axis, there is no submatrix

X

in the symmetric block antitriangular form (1.1), whose dimension corresponds to the difference between the number of positive and negative eigenvalues of the symmetric matrix.

If skew-symmetric matrix is given by its antitriangular factors, then the determinant of $A$ , where $n = 2 p$ is

	$det (A)$	$= det (Q M Q^{T}) = det (M)$
		$= (- 1)^{2 p + 1} \cdot (- 1)^{2 p} \dots (- 1)^{3} \cdot (- 1)^{2} \cdot (- 1)^{p} m_{1, 2 p}^{2} m_{2, 2 p - 1}^{2} \dots m_{p, p + 1}^{2}$
		$= (- 1)^{2 (p^{2} + p)} m_{1, 2 p}^{2} m_{2, 2 p - 1}^{2} \dots m_{p, p + 1}^{2} = m_{1, 2 p}^{2} m_{2, 2 p - 1}^{2} \dots m_{p, p + 1}^{2} .$

Therefore, $A$ (of even order) is singular if and only if at least one of the antidiagonal entries is equal to zero. If $A$

is of odd order, one of the zeroes of the main diagonal is on the antidiagonal, which proves well-known fact that the skew-symmetric matrix of odd order is always singular. Now suppose that

A

is of even order and singular, and the antidiagonal entry at the position

(ℓ, n - ℓ + 1)

ℓ \leq n - ℓ + 1

is zero. Obviously, due to skew-symmetry, element at the position

(n - ℓ + 1, ℓ)

is also zero. If there are more than one pair of zeroes on the antidiagonal, we start from the zero with the smallest difference of the column minus row index.

Now we are applying procedure similar to procedure of annihilation of the elements of the antidiagonal from the previous theorem, but starting with the annihilation of the element at position $(ℓ + 1, n - ℓ)$ by using rotation in the plane $(n - ℓ, n - ℓ + 1)$ . This rotation will also annihilate the element at the position $(n - ℓ, ℓ + 1)$ . We are proceeding with this annihilation process until all the elements on the antidiagonal between $(ℓ, n - ℓ + 1)$ and $(n - ℓ + 1, ℓ)$ are zeros.

If $A$ is of even order, after the previous sequence of transformations, our matrix has middle part of the antidiagonal equal to zero. From now on, procedure for the annihilation of nonzero elements on the antidiagonal is valid no matter if $A$ is of odd or even order. If $A$ is of odd order, first elements to be annihilated are at positions $(⌈ n / 2 ⌉ + 1, ⌈ n / 2 ⌉ - 1)$ and $(⌈ n / 2 ⌉ - 1, ⌈ n / 2 ⌉ + 1)$ by the rotation in the plane $(⌈ n / 2 ⌉ - 1, ⌈ n / 2 ⌉)$ . If $A$ is of even order, we proceed with the annihilation of the elements at positions $(n - ℓ + 2, ℓ - 1)$ and $(ℓ - 1, n - ℓ + 2)$ with rotation in the plane $(ℓ - 1, ℓ)$ . The process is finished when the elements at the position $(1, n)$ and $(n, 1)$ are annihilated by using the rotation in the plane $(1, 2)$ .

If all the elements on the first nontrivial antidiagonal of the obtained final matrix are nonzero, matrix has rank $n - 1$ . Otherwise, we continue the process until all diagonal elements of some antidiagonal are nonzero. Their count is the rank of the matrix.

The process of detecting rank is illustrated in the Figures 2.1 and 2.2. First of them is for the matrix of even, and the second for the matrix of odd order.

Figure 2.1: From left to right, top to bottom – annihilation of the antidiagonal of matrix of even order: the first figure is the state before annihilation, while the last is after the completion of the process for the first antidiagonal. Horizontal and vertical lines show the application of the Givens rotations from the left and right, respectively.

Figure 2.2: From left to right, top to bottom – annihilation of the antidiagonal of matrix of odd order: the first figure is the state before annihilation, while the last is after the completion of the process for the first antidiagonal. Horizontal and vertical lines show the application of the Givens rotations from the left and right, respectively.

This result is of purely theoretical nature, not the computational procedure to obtain the rank of the skew-symmetric matrix.

3 Multi-arrowhead form of a skew-symmetric matrix

In the previous section we transform full skew-symmetric matrix to an antritriangular form. From the antritriangular form of a skew-symmetric matrix it is easy to obtain a new form – multi-arrowhead form of a matrix.

Theorem 3.1.

Let $M \in R^{n \times n}$ be a skew-symmetric matrix in the antitriangular form. By two-sided permutations $P$ , matrix $M$ can be transformed into

M = P S P^{T},

where $S$ has the following multi-arrowhead form. If $n$ is odd, then

and if $n$ is even, then

S = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & s_{12} & 0 & s_{14} & 0 & \dots & 0 & s_{1 n} - s_{12} & 0 & 0 & s_{24} & 0 & \dots & 0 & s_{2 n} 0 & 0 & 0 & s_{34} & 0 & \dots & 0 & s_{3 n} - s_{14} & - s_{24} & - s_{34} & 0 & 0 & \dots & 0 & s_{4 n} 0 & 0 & 0 & 0 & 0 & \dots & 0 & s_{5 n} ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ 0 & 0 & 0 & 0 & 0 & \dots & 0 & s_{n - 1, n} - s_{1 n} & - s_{2 n} & - s_{3 n} & - s_{4 n} & - s_{5 n} & \dots & - s_{n - 1, n} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ .

In addition, if $n$ is odd with additional sequence of rotations at positions $(1, 2)$ , $(1, 4)$ , …, $(1, n - 1)$ the first row and the first column can become zero row and column.

Proof.

The required result is obtained by symmetric permutation $P^{T} M P$ , where

P = {\begin{matrix} [e_{k}, e_{k - 1}, e_{k + 1}, e_{k - 2}, e_{k + 2}, \dots, e_{1}, e_{n}], & if n = 2 k - 1,, & if n = 2 k . \end{matrix}

Remaining part of the proof for the skew-symmetric matrices of the odd order is straightforward. By rotation at position $(1, 2)$ we will annihilate the elements at positions $(1, 3)$ and $(3, 1)$ . Then we will use rotations at position $(1, 4)$ and annihilate the elements at positions $(1, 5)$ and $(5, 1)$ , and so on until the rotation at position $(1, n - 1)$ which will annihilate elements at positions $(1, n)$ and $(n, 1)$ . ∎

4 Factorization of a skew-Hermitian matrix into block antitriangular form

Skew-Hermitian matrices are complex generalizations of the skew-symmetric matrices, with the purely imaginary eigenvalues, but now they need not be in the complex-conjugate pairs. Therefore we can have surplus of ‘positive’ or ‘negative’ signs on the imaginary axis.

For example, if $Q$

is any unitary matrix, then matrix

A = a i I

, where

a \in R

, and

a \neq 0

cannot be transformed into the antitriangular form since

Q^{*} A Q = a i I

On the other hand, $H : = i A$ is Hermitian matrix if $A$ is skew-Hermitian. Therefore if $H$ can be transformed into block antitriangular form, relation between skew-Hermitian and Hermitian matrices will be used to obtain the block antitriangular form of $A$ .

If we look at the proof of Theorem 2.1 from Mastronardi-VanDooren-13 , that theorem is also valid for Hermitian matrices if in the statement of the Theorem orthogonal matrices are replaced by unitary and transpose is replaced by conjugate transpose. That proof relies on the properties of the nonnegative, nonpositive, neutral and null-spaces. In Gohberg-Lancaster-Rodman-05 , all the required properties are derived, not only for the complex Euclidean scalar products, but for the indefinite complex scalar products. Therefore, it is easy to prove the following theorem.

Theorem 4.1.

Let $A \in C^{n \times n}$ be a Hermitian indefinite matrix with $inertia (A) = (n_{-}, n_{0}, n_{+})$ , $n_{1} = min (n_{-}, n_{+})$ , $n_{2} = max (n_{-}, n_{+}) - n_{1}$ . Then, there exist unitary matrix $Q \in C^{n \times n}$ such that

M = Q^{*} A Q = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 & 0 0 & 0 & 0 & Y^{*} 0 & 0 & X & Z^{*} 0 & Y & Z & W \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦,

where $Y \in C^{n_{1} \times n_{1}}$ is nonsingular and lower antitriangular, $W \in C^{n_{1} \times n_{1}}$ is Hermitian, $X \in C^{n_{2} \times n_{2}}$ is Hermitian and definite, and $Z \in C^{n_{1} \times n_{2}}$ .

In the next Corollary we abuse notation for the inertia of the skew-Hermitian matrices. If the skew-Hermitian matrix $A$ has $n_{-}$ eigenvalues on the negative part of the imaginary axis, $n_{0}$ zeros as eigenvalues and $n_{+}$ eigenvalues on the positive part of the imaginary axis, we denote this by $inertia (A) = i (n_{-}, n_{0}, n_{+})$ .

Corollary 4.2.

Let $A \in C^{n \times n}$ be a skew-Hermitian matrix, and let $inertia (A) = i (n_{-}, n_{0}, n_{+})$ , such that neither $n_{-} = n$ , nor $n_{+} = n$ , and $n_{1} = min (n_{-}, n_{+})$ , $n_{2} = max (n_{-}, n_{+}) - n_{1}$ . Then, there exist unitary matrix $Q \in C^{n \times n}$ such that

M = Q^{*} A Q = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 & 0 0 & 0 & 0 & - Y^{*} 0 & 0 & X & - Z^{*} 0 & Y & Z & W \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦,

(4.1)

where $Y \in C^{n_{1} \times n_{1}}$ is nonsingular and lower antitriangular, $W \in C^{n_{1} \times n_{1}}$ and $X \in C^{n_{2} \times n_{2}}$ are skew-Hermitian, and $Z \in C^{n_{1} \times n_{2}}$ . Then, either $inertia (X) = i (n_{2}, 0, 0)$ or $inertia (X) = i (0, 0, n_{2})$ .

Proof.

If the previous Theorem 4.1 is applied to $H = i A$ , it holds

˜ M = Q^{*} H Q = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 & 0 0 & 0 & 0 & {˜ Y}^{*} 0 & 0 & ˜ X & {˜ Z}^{*} 0 & ˜ Y & ˜ Z & ˜ W \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ .

(4.2)

If (4.2) is multiplied by $- i$ we obtain (4.1) by setting $M = - i ˜ M$ , $Y : = - i ˜ Y$ , $X : = - i ˜ X$ , $Z : = - i ˜ Z$ , $W : = - i ˜ W$ . It is easy to see that $X = - X^{*}$ and $W = - W^{*}$ . According to Theorem 4.1, matrix $˜ X$ is definite, therefore all eigenvalues of $X$ ,

λ_{k} (X) = λ_{k} (- i ˜ X) = - i λ_{k} (˜ X),

are situated at the same part of the imaginary axis. ∎

References

(1) Z. Bujanović, D. Kressner, A block algorithm for computing antitriangular factorizations of symmetric matrices, Numer. Algorithms 71 (1) (2016) 41–57.
(2) I. Gohberg, P. Lancaster, L. Rodman, Indefinite Linear Algebra and Applications, Birkhäuser, Basel, 2005.
(3) T. Laudadio, N. Mastronardi, P. Van Dooren, Numerical issues in computing the antitriangular factorization of symmetric indefinite matrices, Appl. Numer. Math. 116 (2016) 204–214.
(4) N. Mastronardi, P. Van Dooren, The antitriangular factorization of symmetric matrices, SIAM J. Matrix Anal. Appl. 34 (1) (2013) 173–196.
(5) J. Pestana, A. J. Wathen, The antitriangular factorization of saddle point matrices, SIAM J. Matrix Anal. Appl. 35 (2) (2014) 339–353.

The antitriangular factorization of skew-symmetric matrices

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1 Introduction

2 Factorization of a skew-symmetric matrix into antitriangular form

Theorem 2.1.

Proof.

3 Multi-arrowhead form of a skew-symmetric matrix

Theorem 3.1.

Proof.

4 Factorization of a skew-Hermitian matrix into block antitriangular form

Theorem 4.1.

Corollary 4.2.

Proof.

References

The antitriangular factorization of skew-symmetric matrices

Authors

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This week in AI

1 Introduction

2 Factorization of a skew-symmetric matrix into antitriangular form

Theorem 2.1.

Proof.

3 Multi-arrowhead form of a skew-symmetric matrix

Theorem 3.1.

Proof.

4 Factorization of a skew-Hermitian matrix into block antitriangular form

Theorem 4.1.

Corollary 4.2.

Proof.

References