EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 1, Article Number 5689
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Singular Value Inequalities for Concave and Convex
Functions of Matrix Sums and Products

Ahmad Al-Natoor1,∗, Fadi Alrimawi2

1 Department of Mathematics, Isra University, Amman, Jordan
2 Department of Basic Sciences, Al-Ahliyya Amman University, Amman, Jordan

Abstract. In this paper, we present several singular value inequalities for special types of functions
of matrix sums and products. Some of special cases of our results give a generalization of some
recent inequalities.

2020 Mathematics Subject Classifications: 15A18, 15A42, 47A30, 93C05
Key Words and Phrases: Singular value, spectral norm, positive semidefinite matrix, concave
function, convex function, inequality, control theory

1. Introduction

Let Mn(C) be the C∗-algebra of all n × n complex matrices. A matrix X ∈ Mn(C)
is said to be positive semidefinite if x∗Ax ≥ 0 for all x ∈ Cn. The singular values of
X ∈ Mn(C), denoted by s1 (X) ≥ s2 (X) ≥ ... sn (X) ≥ 0 are the eigenvalues of |X|. In
this paper, when we write sj we mean s1 (X) , s2 (X) , ...,sn (X), i.e., j = 1, 2, ..., n.

The spectral norm of X ∈ Mn(C) is defined by ∥X∥ = max∥x∥=1 ∥Ax∥.
For X,Y ∈ Mn(C), let X ⊕ Y be the direct sum of X and Y , that is, the matrix given

by X ⊕ Y =

[
X 0
0 Y

]
. It is known that ∥X ⊕ Y ∥ = max (∥X∥ , ∥Y ∥) .

It is known [14] that if X,Y ∈ Mn(C), then

sj(X + Y ) ≤ 2sj(X ⊕ Y ), (1)

A generalization of inequality (1) has been given in [8] by

sj(XZ + ZY ) ≤ 2 ∥Z∥ sj(X ⊕ Y ). (2)

The authors in [7] have proved several singular value inequalities. One of these in-
equalities asserts that if X,Y ∈ Mn(C), then

sj (XY − Y X) ≤ ∥Y ∥ si (X ⊕X) +
1

2
sj−i+1 ((XY − Y X)⊕ (XY − Y X))

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5689

Email addresses: ahmad.alnatoor@iu.edu.jo (A. Al-Natoor), f.rimawi@ammanu.edu.jo (F. Alrimawi)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 2 of 11

for 1 ≤ i ≤ j ≤ n. In particular, if j = i, then

sj (XY − Y X) ≤ ∥Y ∥ sj (X ⊕X) +
1

2
∥XY − Y X∥ . (3)

Singular values of a matrix play a critical role in various applications, including data
compression, noise reduction, and the resolution of ill-posed problems (see, e.g., [1], [2],
and [12]), particularly in fields such as electrical and mechanical engineering, where these
concepts are used to optimize signal processing and system performance. The spectral
norm, defined as the largest singular value, is fundamental for assessing the stability
of the matrix and quantifying its maximum impact as a linear transformation, which
is crucial because it ensures that small perturbations in the input data do not lead to
disproportionately large errors in the output, making computations reliable and consistent
in practical applications.

In this paper, we give several singular value inequalities. Among other results, we give
a related inequality to inequality (2) and we give a generalization of inequality (3). For
recent articles related to matrix and singular value inequalities, we refer the reader to [4],
[3], [5], [10] and [11].

2. Main results

To start our analysis, we need the following lemmas. The first lemma is a consequence
of the spectral theorem for matrices (see, e.g., [13, p. 5]), the second lemma was given in
[7], while the third lemma can be found in [13, p. 75].

Lemma 1. Let X ∈ Mn(C) and let f be nonnegative increasing function on [0,∞). Then

f(sj(X)) = sj(f(|X|)),

where |X| is the absolute value of the matrix X.

Lemma 2. Let X,Y ∈ Mn(C). Then

sj (X + Y ) ≤ sj (X ⊕ Y ) +
1

2
∥X + Y ∥ .

Lemma 3. Let X,Y, Z ∈ Mn(C). Then

sj(XZY ) ≤ ∥X∥ ∥Y ∥ sj(Z).

Theorem 1. Let A,B,X, Y ∈ Mn(C). Then

(a)

sj (f (|XAY + Y BX|)) ≤ f (∥X∥) f (∥Y ∥) sj(f (|A|)⊕ f (|B|))

+f

(
1

2

)
f (∥XAY + Y BX∥) , (4)

where f is a nonnegative increasing submultiplicative concave function on [0,∞)
with f(0) = 0.



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 3 of 11

(b)

sj (f (|XAY + Y BX|)) ≤ f(2)

2
f (∥X∥) f (∥Y ∥) sj(f (|A|)⊕ f (|B|))

+
1

2
f (∥XAY + Y BX∥) , (5)

where f is a nonnegative increasing submultiplicative convex function on [0,∞) .

Proof. We have

sj (f (|XAY + Y BX|))

= f (sj (XAY + Y BX)) (by Lemma 1)

≤ f

(
sj (XAY ⊕ Y BX) +

1

2
∥XAY + Y BX∥

)
(6)

(by Lemma 2)

≤ f (sj (XAY ⊕ (Y BX)∗)) + f

(
1

2
∥XAY + Y BX∥

)
(since f is concave and f(0) = 0)

≤ f (sj (XAY ⊕ (Y BX)∗)) + f

(
1

2

)
f (∥XAY + Y BX∥)

(since f is submultiplicative)

= f

(
sj

([
X 0
0 X∗

] [
A 0
0 B∗

] [
Y 0
0 Y ∗

]))
+f

(
1

2

)
f (∥XAY + Y BX∥) (7)

≤ f

(∥∥∥∥[X 0
0 X

]∥∥∥∥ ∥∥∥∥[Y 0
0 Y

]∥∥∥∥ sj ([A 0
0 B

]))
+f

(
1

2

)
f (∥XAY + Y BX∥) (by Lemma 3)

= f

(
∥X∥ ∥Y ∥ sj

([
A 0
0 B

]))
+ f

(
1

2

)
f (∥XAY + Y BX∥)

≤ f(∥X∥)f(∥Y ∥)f(sj(A⊕B)) + f

(
1

2

)
f (∥XAY + Y BX∥)

= f(∥X∥)f(∥Y ∥)sj(f (|A|)⊕ f (|B|)) + f

(
1

2

)
f (∥XAY + Y BX∥) ,

which proves part (a). For part (b), we start from inequality (6), so we have

sj (f (|XAY + Y BX|))



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 4 of 11

≤ f

(
sj (XAY ⊕ Y BX) +

1

2
∥XAY + Y BX∥

)
≤ 1

2
f (2sj (XAY ⊕ (Y BX)∗)) +

1

2
f (∥XAY + Y BX∥) (since f is convex)

=
1

2
f (2sj (XAY ⊕ (Y BX)∗)) +

1

2
f (∥XAY + Y BX∥)

=
1

2
f

(
2sj

([
X 0
0 X∗

] [
A 0
0 B∗

] [
Y 0
0 Y ∗

]))
+
1

2
f (∥XAY + Y BX∥)

≤ 1

2
f

(
2

∥∥∥∥[X 0
0 X

]∥∥∥∥ ∥∥∥∥[Y 0
0 Y

]∥∥∥∥ sj ([A 0
0 B

]))

+
1

2
f (∥XAY + Y BX∥) (by Lemma 3)

=
1

2
f

(
2 ∥X∥ ∥Y ∥ sj

([
A 0
0 B

]))
+
1

2
f (∥XAY + Y BX∥)

≤ f(2)

2
f(∥X∥)f(∥Y ∥)f(sj(A⊕B)) +

1

2
f (∥XAY + Y BX∥)

=
f(2)

2
f(∥X∥)f(∥Y ∥)sj(f (|A|)⊕ f (|B|)) + 1

2
f (∥XAY + Y BX∥) ,

which completes the proof.

Taking f(t) = t, t ∈ [0,∞) in inequalities (4) and (5), we have

sj (XAY + Y BX) ≤ ∥X∥ ∥Y ∥ sj (A⊕B) +
1

2
∥XAY + Y BX∥ . (8)

Letting X = I in inequality (8), we obtain

sj (AY + Y B) ≤ ∥Y ∥ sj (A⊕B) +
1

2
∥AY + Y B∥ . (9)

Replacing A by X, Y by Z, and B by Y in inequality (9), we have

sj (XZ + ZY ) ≤ ∥Z∥ sj (X ⊕ Y ) +
1

2
∥XZ + ZY ∥ . (10)

Combining inequalities (2) and (10), we have

sj (XZ + ZY ) ≤ min

{
2 ∥Z∥ sj(X ⊕ Y ), ∥Z∥ sj (X ⊕ Y ) +

1

2
∥XZ + ZY ∥

}
,



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 5 of 11

which is a refinement of inequality (2).
Inequality (8) generalizes inequality (3). In fact, replacing B by −B in inequality (8),

we have

sj (XAY − Y BX) ≤ ∥X∥ ∥Y ∥ sj (A⊕B) +
1

2
∥XAY − Y BX∥ , (11)

which is a generalization of inequality (3). To see this, let X = I and then replace A and
B by X in inequality (11), we have

sj (XY − Y X) ≤ ∥Y ∥ sj (X ⊕X) +
1

2
∥XY − Y X∥ ,

which is inequality (3).
We need the following lemma [6] to give our second result.

Lemma 4. Let X,Y, Z ∈ Mn(C) be such that Z is positive semidefinite. Then

sj (XZY ∗) ≤ 1

2

∥∥∥∥X∗X

∥X∥2
+

Y ∗Y

∥Y ∥2

∥∥∥∥ ∥X∥ ∥Y ∥ sj (Z) .

Theorem 2. Let A,B,X, Y ∈ Mn(C) be such that A and B are positive semidefinite.
Then

(a)

sj (f (|XAY + Y BX|))

≤ f

(
1

2

)∥∥∥∥∥f
(
|X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

)∥∥∥∥∥ f (∥X∥) f (∥Y ∥) sj (f (A)⊕ f (B))

+f

(
1

2

)
f (∥XAY + Y BX∥) , (12)

where f is a nonnegative increasing submultiplicative concave function on [0,∞)
with f(0) = 0.

(b)

sj (f (|XAY + Y BX|))

≤ 1

2

∥∥∥∥∥f
(
|X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

)∥∥∥∥∥ f (∥X∥) f (∥Y ∥) sj (f (A)⊕ f (B))

+
1

2
f (∥XAY + Y BX∥) , (13)

where f is a nonnegative increasing submultiplicative convex function on [0,∞) .



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 6 of 11

Proof. By inequality (7), we have

sj (f (|XAY + Y BX|))

≤ f

(
sj

([
X 0
0 X∗

] [
A 0
0 B∗

] [
Y 0
0 Y ∗

]))
+f

(
1

2

)
f (∥XAY + Y BX∥)

≤ f


1
2

∥∥∥∥∥∥∥∥∥

X∗ 0
0 X

X 0
0 X∗


∥X∥2 +

Y ∗ 0
0 Y

Y 0
0 Y ∗


∥Y ∥2

∥∥∥∥∥∥∥∥∥
×∥X∥ ∥Y ∥ sj (A⊕B)


+f

(
1

2

)
f (∥XAY + Y BX∥)

(by Lemma 4)

= f

(
1

2

∥∥∥∥∥ |X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

∥∥∥∥∥ ∥X∥ ∥Y ∥ sj (A⊕B)

)

+f

(
1

2

)
f (∥XAY + Y BX∥)

≤ f

(
1

2

)
f

(∥∥∥∥∥ |X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

∥∥∥∥∥
)
f (∥X∥) f (∥Y ∥) f (sj (A⊕B))

+f

(
1

2

)
f (∥XAY + Y BX∥)

(since f is submultiplicative)

= f

(
1

2

)∥∥∥∥∥f
(
|X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

)∥∥∥∥∥ f (∥X∥) f (∥Y ∥) sj (f (A)⊕ f (B))

+f

(
1

2

)
f (∥XAY + Y BX∥) ,

which proves part (a). For part (b), we start from inequality (6), so we have

sj (f (|XAY + Y BX|))

≤ f

(
sj (XAY ⊕ Y BX) +

1

2
∥XAY + Y BX∥

)
= f

(
sj (XAY ⊕ (Y BX)∗) +

1

2
∥XAY + Y BX∥

)



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 7 of 11

≤ 1

2
f

(
2sj

([
X 0
0 X∗

] [
A 0
0 B

] [
Y 0
0 Y ∗

]))
+
1

2
f (∥XAY + Y BX∥)

≤ 1

2
f



∥∥∥∥∥∥∥∥∥

X∗ 0
0 X

X 0
0 X∗


∥X∥2 +

Y ∗ 0
0 Y

Y 0
0 Y ∗


∥Y ∥2

∥∥∥∥∥∥∥∥∥
×∥X∥ ∥Y ∥ sj (A⊕B)


+
1

2
f (∥XAY + Y BX∥)

(by Lemma 4)

=
1

2
f

(∥∥∥∥∥ |X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

∥∥∥∥∥ ∥X∥ ∥Y ∥ sj (A⊕B)

)
+
1

2
f (∥XAY + Y BX∥)

≤ 1

2
f

(∥∥∥∥∥ |X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

∥∥∥∥∥
)
f (∥X∥) f (∥Y ∥) f (sj (A⊕B))

+
1

2
f (∥XAY + Y BX∥)

(since f is submultiplicative)

=
1

2

∥∥∥∥∥f
(
|X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

)∥∥∥∥∥ f (∥X∥) f (∥Y ∥) sj (f (A)⊕ f (B))

+
1

2
f (∥XAY + Y BX∥) ,

this completes the proof.

Taking f(t) = t, t ∈ [0,∞) in inequalities (12) and (13), we have

sj (XAY + Y BX) ≤ 1

2

∥∥∥∥∥ |X|2 ⊕ |X∗|2

∥X∥2
+

|Y |2 ⊕ |Y ∗|2

∥Y ∥2

∥∥∥∥∥ ∥X∥ ∥Y ∥ sj (A⊕B)

+
1

2
∥XAY + Y BX∥ .

To state our next result, we need the following lemma [9].

Lemma 5. Let A,B,X ∈ Mn(C) be such that X is positive semidefinite. Then

sj (AXB∗) ≤ 1

2
∥X∥ sj (A∗A+B∗B) .



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 8 of 11

Theorem 3. Let A,B,X, Y ∈ Mn(C) be such that A and B are positive semidefinite.
Then

(a)

sj (f (|XAY + Y BX|))

≤ f

(
1

2

)
max (∥f(A)∥ , ∥f(B)∥) sj ((f (X∗X + Y Y ∗))⊕ (f (XX∗ + Y ∗Y )))

+f

(
1

2

)
f (∥XAY + Y BX∥) , (14)

where f is a nonnegative increasing submultiplicative concave function on [0,∞)
with f(0) = 0.

(b)

sj (f (|XAY + Y BX|))

≤ 1

2
max (∥f(A)∥ , ∥f(B)∥) sj ((f (X∗X + Y Y ∗))⊕ (f (XX∗ + Y ∗Y )))

+
1

2
f (∥XAY + Y BX∥) , (15)

where f is a nonnegative increasing submultiplicative convex function on [0,∞) .

Proof. By inequality (7), we have

sj (f (|XAY + Y BX|))

≤ f

(
sj

([
X 0
0 X∗

] [
A 0
0 B

] [
Y 0
0 Y ∗

]))
+f

(
1

2

)
f (∥XAY + Y BX∥)

≤ f

1

2

∥∥∥∥[A 0
0 B

]∥∥∥∥ sj


[
X∗ 0
0 X

] [
X 0
0 X∗

]
+

[
Y 0
0 Y ∗

] [
Y ∗ 0
0 Y

]



+f

(
1

2

)
f (∥XAY + Y BX∥) (by Lemma 5)

≤ f

(
1

2

)
f

(∥∥∥∥[A 0
0 B

]∥∥∥∥) f

(
sj

([
X∗X + Y Y ∗ 0

0 XX∗ + Y ∗Y

]))



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 9 of 11

+f

(
1

2

)
f (∥XAY + Y BX∥)

= f

(
1

2

)∥∥∥∥[f(A) 0
0 f(B)

]∥∥∥∥ sj (f ([X∗X + Y Y ∗ 0
0 XX∗ + Y ∗Y

]))
+f

(
1

2

)
f (∥XAY + Y BX∥)

= f

(
1

2

)
max (∥f(A)∥ , ∥f(B)∥) sj ((f (X∗X + Y Y ∗))⊕ (f (XX∗ + Y ∗Y )))

+f

(
1

2

)
f (∥XAY + Y BX∥) ,

which proves part (a). For part (b), we start from inequality (6), so we have

sj (f (|XAY + Y BX|))

≤ f

(
sj (XAY ⊕ Y BX) +

1

2
∥XAY + Y BX∥

)
= f

(
sj (XAY ⊕ (Y BX)∗) +

1

2
∥XAY + Y BX∥

)
≤ 1

2
f

(
2sj

([
X 0
0 X∗

] [
A 0
0 B

] [
Y 0
0 Y ∗

]))
+
1

2
f (∥XAY + Y BX∥)

≤ 1

2
f

∥∥∥∥[A 0
0 B

]∥∥∥∥ sj


[
X∗ 0
0 X

] [
X 0
0 X∗

]
+

[
Y 0
0 Y ∗

] [
Y ∗ 0
0 Y

]



+
1

2
f (∥XAY + Y BX∥) (by Lemma 5)

≤ 1

2
f

(∥∥∥∥[A 0
0 B

]∥∥∥∥) f

(
sj

([
X∗X + Y Y ∗ 0

0 XX∗ + Y ∗Y

]))
+
1

2
f (∥XAY + Y BX∥)

=
1

2

∥∥∥∥[f(A) 0
0 f(B)

]∥∥∥∥ sj ([f (X∗X + Y Y ∗) 0
0 f (XX∗ + Y ∗Y )

])
+
1

2
f (∥XAY + Y BX∥)

=
1

2
max (∥f(A)∥ , ∥f(B)∥) sj ((f (X∗X + Y Y ∗))⊕ (f (XX∗ + Y ∗Y )))

+
1

2
f (∥XAY + Y BX∥) ,



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 10 of 11

this completes the proof.

Taking f(t) = t, t ∈ [0,∞) in inequalities (14) and (15), we have

sj (XAY + Y BX) ≤ 1

2
max (∥A∥ , ∥B∥) sj ((X∗X + Y Y ∗)⊕ (XX∗ + Y ∗Y ))

+
1

2
∥XAY + Y BX∥ . (16)

Corollary 1. Let A,B,X, Y ∈ Mn(C) be such that A and B are positive semidefinite.
Then

sj (AY + Y B) ≤ max (∥A∥ , ∥B∥) sj (Y ⊕ Y ) +
1

2
∥AY + Y B∥ .

Proof. Letting X = I in inequality (16), we have

sj (AY + Y B) =
1

2
max (∥A∥ , ∥B∥) sj

([
I + Y Y ∗ 0

0 I + Y ∗Y

])
+
1

2
∥AY + Y B∥

=
1

2
max (∥A∥ , ∥B∥) sj

([
I 0
0 I

]
+

[
Y Y ∗ 0
0 Y ∗Y

])
+
1

2
∥AY + Y B∥

=
1

2
max (∥A∥ , ∥B∥)

(
1 + sj

([
Y Y ∗ 0
0 Y ∗Y

]))
+
1

2
∥AY + Y B∥

=
1

2
max (∥A∥ , ∥B∥) (1 + sj (Y Y ∗ ⊕ Y ∗Y )) +

1

2
∥AY + Y B∥

=
1

2
max (∥A∥ , ∥B∥)

(
1 + s2j (Y ⊕ Y )

)
+

1

2
∥AY + Y B∥ .

Replacing Y by tY and taking the min over t > 0, we have

sj (AY + Y B) ≤ max (∥A∥ , ∥B∥) sj (Y ⊕ Y ) +
1

2
∥AY + Y B∥ ,

as required.

Conclusion. In this paper, we present several inequalities related to singular values for
funcions of matrices and we give a general version of interesting recent results.

Acknowledgements

The authors are grateful to the reviewers for their careful reading and valuable sug-
gestions.



A. Al-Natoor, F. Alrimawi / Eur. J. Pure Appl. Math, 18 (1) (2025), 5689 11 of 11

References

[1] O. AbuAlghanam, O. Adwan, M. A. Al Shariah, and M. Qatawneh. Enhancing the
speed of the learning vector quantization (lvq) algorithm by adding partial distance
computation. Cybernetics and Information Technologies, 22:36–49, 2022.

[2] M. Abualhaj, A. A. Abu-Shareha, M. O. Hiari, Y. Alrabanah, M. Al-Zyoud, and M. A.
Alsharaiah. A paradigm for dos attack disclosure using machine learning techniques.
International Journal of Advanced Computer Science and Applications, 13:192–200,
2024.

[3] A. Al-Natoor. Norm inequalities for functions of matrices. Heliyon, 10, Issue 9:e30056.
[4] A. Al-Natoor. Norm inequalities for product of matrices. Acta Sci. Math. (Szeged),

https://doi.org/10.1007/s44146-024-00121-1, 2024.
[5] A. Al-Natoor and F. Alrimawi. Numerical radius inequalities involving 2 × 2 block

matrices. Eur. J. Pure Appl. Math., (Accepted).
[6] A. Al-Natoor, O. Hirzallah, and F. Kittaneh. Singular value inequalities for convex

functions of positive semidefinite matrices. Ann. Funct. Anal., 14:Papet No. 7:14 pp.,
2023.

[7] A. Al-Natoor, O. Hirzallah, and F. Kittaneh. Singular value and uni-
tarily invariant norm inequalities for matrices. Adv. Oper. Theory,
21:https://doi.org/10.1007/s43036–024–00319–8, 2024.

[8] A. Al-Natoor and F. Kittaneh. Singular value and norm inequalities for positive
semidefinite matrices. Linear Multilinear Algebra, 70:4498–4507, 2022.

[9] H. Albadawi. Singular values and arithmetic-geometric mean inequalities for opera-
tors. Ann. Funct. Anal., 3:10–18, 2024.

[10] F. Alrimawi, O. Hirzallah, and F. Kittaneh. Singular value inequalities involving
convex and concave functions of positive semidefinite matrices. Ann. Funct. Anal.,
11:1257–1273, 2020.

[11] F. Alrimawi, H. Kawariq, and F.A. Abushaheen. Some matrix and norm inequalities
for positive definite matrices. Int. J. Math. Comput. Sci., 15:845–855, 2020.

[12] M.A. Alsharaiah, L.H. Baniata, O. Adwan, O. AbuAlghanam, A. A. Abu-Shareha,
L. Alzboon, N. Mustafa, and M. Baniata. Neural network prediction model to explore
complex nonlinear behavior in dynamic biological network. Int. J. Interact. Mob.
Technol., 16:32–51, 2022.

[13] R. Bhatia. Matrix Analysis. Springer-Verlag, New York, 1997.
[14] O. Hirzalleh and F. Kittaneh. Inequalities for sums and direct sums of hilbert space

operator. Linear Algebra and its Applications, 424:71–82, 2007.