On Double Representation of q-k-normal Matrices

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ISSN(Online): 2319-8753 ISSN (Print): 2347-6710

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Vol. 8, Issue 1, January 2019

On Double Representation of q-

k

-normal

Matrices

R.Kavitha, K.Gunasekaran

Lecturer, Ramanujan Research Centre, PG and Research Department of Mathematics,

Government Arts College (Autonomous), Kumbakonam, Tamil Nadu, India.

Head of the Department, Ramanujan Research Centre, PG and Research Department of Mathematics,

Government Arts College (Autonomous), Kumbakonam, Tamil Nadu, India.

ABSTRACT: In this paper, the conjugate transpose of double representation of quaternion matrix is introduced. Also some basic results have been proved.

AMS Classifications : 15A09, 15A57, 15A24, 15A33, 15A15

KEYWORDS : q-k-normal matrices, double representation matrix, permutation matrix, Moore-Penrose inverse

I. INTRODUCTION

Wu.J.L and Zhang.P introduced bicomplex representation method for quaternion matrix is 2011[5]. The

multiplication

A B



was defined as

A B

₀ ₀



A B j

₁ ₁ where

A

₀



A j

₁



A

,

B

₀



B j

₁



B

,

A , B , A , B

₀ ₀ ₁ ₁ are

complex matrices with this concept, the conjugate transpose of bicomplex method developed and applied for the q-k

-normal matrices. Also elementary results are discussed.

II. SOME DEFINITIONS AND THEOREMS Definition 2.1

A quaternion matrix

A



H

n n is represented as

A

₀



A j

₁ , where

A

₀ and

A

₁ are in

C

n n , known as

double representation of quaternion matrix.

Definition 2.2

The product of double representation of A and B in

H

n n is defined as

AB



A B

₀ ₀



A B j

₁ ₁ with

n n

0 0 1 1

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Remark 2.3

If

A



H

n n then

* * * *

0 1 0 1

A



(A



A j)



A



A j

Remark 2.4

From the definition of q-k-Hermitian or q-k-normal matrix

KA K

*



KA K

₀*



KA Kj

₁* is easily understood

by one.

Theorem 2.5

If A is q-k-normal then

A

₀and

A

₁ are k-normal with K.

Proof

Since A is q-k-normal. So

(A

₀



A j)(KA K

₁ ₀*



KA Kj)

₁*



(KA K

₀*



KA Kj)(A

₁* ₀



A j).

₁

This implies that

A KA K

0 0*



A KA Kj

1 1*



KA KA

0* 0



KA KA j

1* 1

Equate the corresponding terms,

We have

A KA K

₀ ₀*



KA KA

₀* and

A KA K

₁ ₁*



KA KA

₁* . Thus

A

₀ and

A

₁ are q-k-normal with

respect to the permutation matrix K.

Hence proved.

Remark 2.6

The converse of the theorem (2.5) is also true. The proof is left to the reader.

Theorem 2.7

KA K

1



KA

₀1

K



KA Kj

₁1

Proof

If

A

1is q-k-normal. [by 3, theorem 2.4]

That is

A K(A ) K

1 1 *



K(A ) KA

1 * 1.

Since

A

1



A

₀1



A

₁1

j

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Therefore ,

KA K

1

KA

0 1

K

KA Kj

1 1

  





Hence proved.

Theorem 2.8

If

A

is q-k-normal then

A

₀Tand

A

₁T are k-normal.

Proof

If

A

† is q-k-normal [by 3, theorem 2.5]

That is

K(A ) KA

T *



AK(A ) K

T *

Since

A

T



A

₀T



A j

₁T

So,

KA K

T



KA K

0T



KA Kj

1T .

Thus,

T * T T * T * T T

0 1 0 1

K(A ) KA



(K(A ) K



K(A ) Kj)(A



A j)

T * T T * T

0 0 1 1

K(A ) KA

K(A ) KA j





(2.9)

A K(A ) K

T T *



(A

₀T



A j)(K(A ) K

₁T ₀T *



K(A ) Kj)

₁T *



A K(A ) K

₀T ₀T *



A K(A ) Kj

₁T ₁T *

(2.10) Since

A

Tis q-k-normal so equation (2.9) and equation (2.10) are equal.

Therefore ,

A K(A ) K

0T 0T *



K(A ) KA

0T * 0T and

T T * T * T

1 1 1 1

A K(A ) K



K(A ) KA

This gives as

A

₀T and

A

₁T are k-normal.

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Theorem 2.11

A

0* and * 1

A

are k-normal.

Proof

If

A

*is q-k-normal. [by 3, theorem 2.6]

Now,

A

*



A

₀*



A j

₁*

A K(A ) K

* * *



A KAK

*



(A

₀*



A j)(KA K

₁* ₀



KA K)

₁



A KA K

₀* ₀



A KA Kj

₁* ₁



A K(A ) K

₀* ₀* *



A K(A ) Kj

₁* ₁* *

(2.12)

K(A ) KA

* *



KAKA

*



(KA K

₀



KA Kj)(A

₁ ₀*



A j)

₁*



KA KA

₀ ₀*



KA KA j

₁ ₁*



K(A ) KA

₀* * ₀*



K(A ) KA j

₁* * ₁*

(2.13)

Since

A

*is q-k-normal. So equations (2.12) and (2.13) shows that

* * * * * * * * * * *

0 0 1 1 0 0 1 1

A K(A ) K



A K(A ) Kj



K(A ) KA



K(A ) KA j

Therefore ,

A K(A ) K

₀* ₀* *



K(A ) KA

₀* * ₀* and

A K(A ) K

₁* ₁* *



K(A ) KA

₁* * ₁*.

This implies that

A

₀* and

A

₁*are k-normal .

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Remark 2.14

If

A



H

n n then

m m

0 1

A



(A



A j)

,

m 1



,

m z



.

Since

A



A

₀



A j

₁ So

A

m



(A

₀



A j)(A

₁ ₀



A j)...m

₁ times.

Thus

A

m



A

₀m



A j

₁m because

AB



A B

₀ ₀



A B j

₁ ₁ where

A



A

₀



A j

₁ and

B



B

₀



B j

₁ for all

0 1 0 1

A , A , B , B

in

C

n n .

Theorem 2.15

Let

A



H

n n be a q-k-normal in the form of

A



A

0



A j

1 then

A

0 and

A

1 are k-normal.

Proof

Since A is q-k-normal then

A

₀ and

A

₁are k-normal (by theorem 2.5)

Now,

A [K(A ) K]

m m *



(A

₀



A j) [K((A

₁ m ₀



A j) ) Kj]

₁ m *

m m m * m *

0 1 0 1

(A

A j)[K(A

) K

K(A ) Kj]









A K(A

₀m ₀m *

) K



A K(A ) Kj

₁m ₁m *

(2.16)

m * m m * m

0 1 0 1

[K(A ) K]A



[K((A



A j) ) Kj](A



A j)

m * m * m m

0 1 0 1

[K(A

) K

K(A ) Kj](A

A j)









K(A

₀m *

) KA

₀m



K(A ) KA j

₁m * ₁m (2.17)

Since

A

0 and

A

1are k-normal. So equations (2.16) and (2.17) shows that

m m * m m * m * m m * m

0 0 1 1 0 0 1 1

A K(A

) K



A K(A ) Kj



K(A

) KA



K(A ) KA j

Therefore,

A K(A

₀m ₀m *

) K



K(A

₀m *

) KA

₀m and

A K(A ) Kj

₁m ₁m *



K(A ) KA j

₁m * ₁m

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Remark 2.18

In the case

A



H

n n is singular then we can find the Moore-Penrose inverse

A

†

satisfying the equation

AXA



A

,

XAX



X

,

(AX)

*



AX

and

(XA)

*



XA

. From the following theorem will be proved.

Theorem 2.19

A

†is also q-k-normal if

A

₀† and

A

₁†are existing and k-normal.

Proof

Since A is q-k-normal

Therefore,

A(KA K)

*



(KA K)A

*

Now,

A (K(A ) K)

† † *



(A

₀†

-

A j)(KA K

₁† ₀†



KA Kj)

₁†



(A

0

A j)(K(A ) K

1 0 *



K(A ) Kj)

1 *

† † † †



A K(A ) K

₀† ₀† *



A K(A ) Kj

₁† ₁† *



K(A K) A

₀† * ₀†



K(A ) KA Kj

₁† * ₁† [since

A

₀†and

A

₁†are q-k-normal]



(K(A ) K

₀† *



K(A ) Kj)(A

₁† * ₀†



A j)

₁†



K[(A )

₀† *



(A ) j]K(A

₁† * ₀



A j)

₁ †



K(A

₀†



A j) K(A

₁† * ₀



A j)

₁ †



K((A

₀



A j) ) KA

₁ † * †



K(A ) KA

† * †

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Theorem 2.20



A

₀ and



A

₁ are k-normal for any scalar



.

Proof

Since A is q-k-normal having the double representation

A



A

₀



A j

₁ , where

A

₀and

A

1



C

n n .

So,

A(KA K)

*



(KA K)A

* _,



A

_{is q-}k-normal

This implies that,



AK( A) K



*



K( A) KA



*

Now



A

 

A

₀

 

A j

₁ and

( A)



*

 

A

₀*

 

A j

₁*

This implies that,

* * *

0 1

K( A) K



 

(KA K)

 

(KA K) j

* * *

0 1 0 1

( A)K( A) K



 

( A

 

A j)( KA K



 

KA Kj)

 

A (KA K)

₀ ₀*

 

A (KA K) j

₁ ₁*

2 * *

0 0 1 1

[A (KA K) A (KA K) j]

 



2 *

A(KA K)

 

2 *

A(KA K)

 

This implies that,



2

A

₀ and



2

A

₁ are k-normal (by theorem 2.15)

Hence proved.

REFERENCES

1. Bhatia, Rajendara: Matrix Analysis; Springer Publications(1997) 159 – 164

2. Chen.L.X, “Inverse Matrix and Properties of Double Determinant over Quaternion TH Field,” Science in China (Series A), Vol. 34, No. 5, 1991, pp. 25-35.

3. Gunasekaran.K and Kavitha.R: On Quaternion-k-normal matrices; International Journal of Mathematical Archive-7(7),(2016) 93-101.

4. Li.T.S, “Properties of Double Determinant over Quaternion Field,” Journal of Central China Normal University, Vol. 1, 1995, 3-7.

5. Wu.J.L nd Zhang.P “ On Bicomplex Representation Methods and Application of Matrices over Quaternionic Division Algebra”, Ad.in Pure Maths. Vol.1 pp. 9-15 doi:10.4236/apm 2004(2011).

6. Zhang.Q.C, “Properties of Double Determinant over the Quaternion Field and Its Applications,” Acta MathematicaSinica, Vol. 38, No. 2, 1995, pp. 253-259.

7. Zhuang.W.J, “Inequalities of Eigenvalues and Singular Values for Quaternion Matrices,” Advances in Mathematics, Vol. 4, 1988, pp. 403-406.