EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 5844
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Singular Value Inequalities for Matrix Sums and
Products

Ahmad Al-Natoor1,∗, Rawan Al-smadi2, Aliaa Burqan2

1 Department of Mathematics, Faculty of Sciences, Isra University, Amman 11622, Jordan
2 Department of Mathematics, Faculty of Science, Zarqa University, Zarqa 13110, Jordan

Abstract. In this paper, we discuss some inequalities involving singular values for matrix sums
and products. In some of our results, we prove inequalities for functions of matrices and this
enables us to give a generalization of known recent result.
2020 Mathematics Subject Classifications: 15A18, 15A60 15A42, 15A16
Key Words and Phrases: Positive semidefinite matrix, singular value, spectral norm, inequality

1. Introduction

In this paper, the symbol Mn(C) denote the space of n × n complex matrices. The
numbers s1(A) ≥, ...,≥ sn(A) ≥ 0 are called the singular values of A ∈ Mn(C) which are

the eigenvalues of |A| = (A∗A)1/2 arranged in decreasing order and counted according to
multiplicity.

The spectral norm of A ∈ Mn (C), denoted by ∥A∥ , can be expressed as the largest
singular value of A, i.e, ∥A∥ = s1.

It is known (see [1] or [2]) that if X ∈ Mm and Y ∈ Mn are such that

[
X Z
Z∗ Y

]
≥ 0,

then

sj (Z) ≤ 1

2
sj

([
X Z
Z∗ Y

])
(1)

for j = 1, 2, ..., r, where r = min(n,m).
If A and B are positive semidefinite matrices, then by letting Z = AB, X = A2, and

Y = B2 in inequality (1), we have

sj (AB) ≤ 1

2
sj

([
A2 AB
BA B2

])
. (2)

Among other results in this paper, we give a general version of inequality (2). For
more singular value and norm inequalities for matrices, we refer the reader to [3], [4], [5],
[6], and [7].

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5844

Email addresses: ahmad.alnatoor@iu.edu.jo (A. Al-Natoor),
rawansmadi2@gmail.com (R. Al-smadi), aliaaburqan@zu.edu.jo (A. Burqan)

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



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 2 of 10

2. Main results

To start our analysis, we need the fact that if A,B ∈ Mn(C), then

sj(A) ≤ sj(B) iff sj(A⊕A) ≤ sj(B ⊕B) (3)

for j = 1, ..., 2n.

Lemma 1. Let A,B, Y ∈ Mn (C) be such that A and B are positive semidefinite. Then
for j = 1, ..., 2n, we have

sj ((AY − Y B)⊕ 0) ≤ max (∥A∥ , ∥B∥) sj (Y ⊕ Y ) .

In particular, if B = A, then

sj ((AY − Y A)⊕ 0) ≤ ∥A∥ sj (Y ⊕ Y ) .

Theorem 1. Let A,B ∈ Mn (C) be positive semidefinite and let g(t), h(t) be real polyno-
mials. Then for j = 1, ..., 2n, we have

sj ((ABh (B) + g (A)AB)⊕ 0) ≤
∥∥A2 +B2

∥∥ sj (g (A)⊕ h (B)) .

Proof. Let Q1 =

[
A 0
B 0

] [
A B
0 0

] [
g (A) 0
0 h (B)

]
=

[
A2g (A) ABh(B)
BAg (A) B2h (B)

]
,

and Q2 =

[
g (A) 0
0 h (B)

] [
A 0
−B 0

] [
A −B
0 0

]
=

[
g (A)A2 −g (A)AB

−h (B)BA h (B)B2

]
.

So, Q1 − Q2 =

[
A2g (A)− g (A)A2 ABh(B) + g (A)AB
BAg (A) + h (B)BA B2h (B)− h (B)B2

]
=

[
0 Q
Q∗ 0

]
, where

Q = ABh(B) + g (A)AB.
Now,

sj ((ABh(B) + g (A)AB)⊕ (ABh(B) + g (A)AB))

= sj((ABh(B) + g (A)AB)⊕ (BAg (A) + h (B)BA))

= sj (Q⊕Q∗)

= sj

([
0 Q
Q∗ 0

])
= sj (Q1 −Q2) . (4)

Using Lemma 1 and the fact that

[
A2 AB
BA B2

]
and

[
A2 −AB

−BA B2

]
are unitarily equiv-

alent via U=

[
I 0
0 −I

]
yield that



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 3 of 10

sj ((Q1 −Q2)⊕ 0) = sj


[

A 0
B 0

] [
A B
0 0

] [
g (A) 0
0 h (B)

]
−
[
g (A) 0
0 h (B)

] [
A 0
−B 0

] [
A −B
0 0

]


= sj




[
A2 AB
BA B2

] [
g (A) 0
0 h (B)

]
−
[
g (A) 0
0 h (B)

] [
A2 −AB

−BA B2

]
⊕ 0


≤ max

(∥∥∥∥[ A2 AB
BA B2

]∥∥∥∥ , ∥∥∥∥[ A2 −AB
−BA B2

]∥∥∥∥)
×sj

([
g (A) 0
0 h (B)

]
⊕
[
g (A) 0
0 h (B)

])
=

∥∥∥∥[ A2 AB
BA B2

]∥∥∥∥ sj ([ g (A) 0
0 h (B)

]
⊕
[
g (A) 0
0 h (B)

])
=

∥∥∥∥[ A2 AB
BA B2

]∥∥∥∥ sj ((g (A)⊕ h (B))⊕ (g (A)⊕ h (B))) .

Now, by using the fact that ∥T ∗T∥ = ∥TT ∗∥ for any T ∈ Mn (C) , we have∥∥∥∥[ A2 AB
BA B2

]∥∥∥∥ =

∥∥∥∥[ A 0
B 0

] [
A B
0 0

]∥∥∥∥
=

∥∥∥∥[ A B
0 0

] [
A 0
B 0

]∥∥∥∥
=

∥∥∥∥[ A2 +B2 0
0 0

]∥∥∥∥
=

∥∥A2 +B2
∥∥ .

So,

sj ((ABh(B) + g (A)AB)⊕ (ABh(B) + g (A)AB)⊕ 0)

≤
∥∥A2 +B2

∥∥ sj ((g (A)⊕ h (B))⊕ (g (A)⊕ h (B))) . (5)

Now, the desired result follows by inequalities (3), (4), and (5).

Lemma 2. [8]Let A,B, Y ∈ Mn (C) be such that A and B are positive semidefinite. Then
for j = 1, ..., 2n, we have

sj ((AY − Y B)⊕ 0) ≤ ∥Y ∥ sj (A⊕B) .



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 4 of 10

Theorem 2. Let A,B ∈ Mn (C) be positive semidefinite and let g(t), h(t) be real polyno-
mials. Then for j = 1, ..., 2n, we have

sj (ABh(B) + g (A)AB) ≤ max (∥g(A)∥ , ∥h(B)∥) sj
([

A2 AB
BA B2

])
. (6)

Proof. Let Q1, Q2, Q, and U be as in the proof of Theorem 1. Then by equation (4),
we have

sj ((ABh(B) + g (A)AB)⊕ (ABh(B) + g (A)AB))

= sj (Q1 −Q2)

≤
∥∥∥∥[ g(A) 0

0 h(B)

]∥∥∥∥ sj ([ A2 AB
BA B2

]
⊕
[

A2 −AB
−BA B2

])
(by Lemma 2)

= max (∥g(A)∥ , ∥h(B)∥) sj
([

A2 AB
BA B2

]
⊕
(
U

[
A2 −AB

−BA B2

]
U∗
))

= max (∥g(A)∥ , ∥h(B)∥) sj
([

A2 AB
BA B2

]
⊕
[

A2 AB
BA B2

])
. (7)

Now, the desired result follows by inequalities (3) and (7).

Inequality (6) represents a general version of inequality (2). In fact, letting h(t) =
g(t) = 1 in Corollary 2, we have

sj (AB) ≤ 1

2
sj

([
A2 AB
BA B2

])
,

which is inequality (2).
An application of Theorem 2 can be seen in the following corollary, which depends on

the following lemma .

Lemma 3. [9]Let A,B ∈ Mn (C). Then for j = 1, ..., 2n, we have

sj ((A±B)⊕ 0) ≤ sj (A⊕B) +
1

2
∥A±B∥ .

Corollary 1. Let A,B, Y ∈ Mn (C) be such that A and B are positive semidefinite. Then
for j = 1, ..., 2n, we have

sj ((ABh (B) + g (A)AB)⊕ 0)

≤ max (∥g(A)∥ , ∥h(B)∥) sj
(
A2 ⊕B2 ⊕AB ⊕AB

)
+
1

2

∥∥A2 +B2
∥∥ .



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 5 of 10

Proof. By inequality (6), we have

sj ((ABh (B) + g (A)AB)⊕ 0)

≤ max (∥g(A)∥ , ∥h(B)∥) sj
([

A2 AB
BA B2

])
= max (∥g(A)∥ , ∥h(B)∥) sj

([
A2 0
0 B2

]
+

[
0 AB

BA 0

])

≤ max (∥g(A)∥ , ∥h(B)∥)

 sj

([
A2 0
0 B2

]
⊕
[

0 AB
BA 0

])
+1

2

∥∥∥∥[ A2 0
0 B2

]
+

[
0 AB

BA 0

]∥∥∥∥


(by Lemma 3)

= max (∥g(A)∥ , ∥h(B)∥)

 sj

([
A2 0
0 B2

]
⊕
[

0 AB
BA 0

])
+1

2

∥∥∥∥[ A2 AB
BA B2

]∥∥∥∥


= max (∥g(A)∥ , ∥h(B)∥)
(
sj
(
A2 ⊕B2 ⊕AB ⊕AB

)
+

1

2

∥∥A2 +B2
∥∥) ,

as required.

Theorem 3. Let A,B ∈ Mn (C) be positive semidefinite and let g(t), h(t) be real polyno-
mials. Then for j = 1, 2, ..., 2n, we have

sj (ABh(B) + g(A)AB ⊕ 0)

≤ sj

([
A2g (A) ABh(B)
BAg(A) B2h (B)

])
+

1

2
∥ABh(B) + g(A)AB∥ . (8)

In particular, if j = 1, then

∥ABh(B) + g(A)AB∥ ≤ 2

∥∥∥∥[ A2g (A) ABh(B)
BAg(A) B2h (B)

]∥∥∥∥ .
Proof. Let Q1, Q2, Q, and U be as in the proof of Theorem 1. Then by equation (4),

we have

sj ((ABh(B) + g (A)AB)⊕ (ABh(B) + g (A)AB))

= sj (Q1 −Q2)

≤ sj (Q1 ⊕Q2) +
1

2
∥Q1 −Q2∥



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 6 of 10

(by Lemma 3)

≤ sj

([
A2g (A) ABh (B)
BAg (A) B2h (B)

]
⊕
[

g (A)A2 −g (A)AB
−h (B)BA h (B)B2

])
+
1

2

∥∥∥∥[ 0 ABhB + g (A)AB
BAg (A) + h (B)BA 0

]∥∥∥∥
= sj

([
A2g (A) ABh (B)
BAg (A) B2h (B)

]
⊕
(
U

[
g (A)A2 −g (A)AB

−h (B)BA h (B)B2

]
U∗
))

+
1

2
max (∥ABh (B) + g (A)AB∥ , ∥BAg (A) + h (B)BA∥)

= sj

([
A2g (A) ABh (B)
BAg (A) B2h (B)

]
⊕
[

g (A)A2 g (A)AB
h (B)BA h (B)B2

])
+
1

2
∥ABh (B) + g (A)AB∥ .

Consequently,

sj((ABh(B) + g(A)AB)⊕(ABh(B) + g(A)AB))

≤ sj

([
A2g (A) ABh (B)
BAg (A) B2h (B)

]
⊕
[

g (A)A2 g (A)AB
h (B)BA h (B)B2

])
+
1

2
∥ABh (B) + g (A)AB∥ . (9)

Now, the desired result follows from inequalities (3) and (9).

Lemma 4. [10]Let A,B ∈ Mn(C) be positive semidefinite. Then for r ≥ 0, we have∥∥∥A1/2(A+B)rB1/2
∥∥∥ ≤ 1

2

∥∥(A+B)r+1
∥∥ .

Corollary 2. Let A,B ∈ Mn(C) be positive semidefinite. Then for j = 1, ..., 2n, we have

sj

((
A1/2B3/2 +A3/2B12

)
⊕ 0
)
≤ sj

([
A2 A1/2B3/2

B1/2A3/2 B2

])
+

1

4

∥∥(A+B)2
∥∥

Proof. Let g(t) = h(t) = t2 in inequality (8). Then

sj
((
AB3 +A3B

)
⊕ 0
)

≤ sj

([
A4 AB3

BA3 B4

])
+

1

2

∥∥AB3 +A3B
∥∥

≤ sj

([
A4 AB3

BA3 B4

])
+

1

2

∥∥A(B2 +A2)B
∥∥ .

Replacing A by A1/2 and B by B1/2, we have

sj

((
A1/2B3/2 +A3/2B12

)
⊕ 0
)

≤ sj

([
A2 A1/2B3/2

B1/2A3/2 B2

])



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 7 of 10

+
1

2

∥∥∥A1/2(B +A)B1/2
∥∥∥

≤ sj

([
A2 A1/2B3/2

B1/2A3/2 B2

])
+

1

4

∥∥(A+B)2
∥∥

(by Lemma 4)

as required.

The author in [11] proved that if A,B, Y ∈ Mn(C) are such that Y is positive semidef-
inite, then

sj (AY B∗) ≤ 1

2
∥Y ∥ sj (A∗A+B∗B) (10)

for j = 1, 2, ..., n. Based on this inequality, we have the following lemma.

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

sj (AY B∗) ≤ 1

2
∥Y ∥ ∥A∥ ∥B∥ sj

(
A∗A

∥A∥2
+

B∗B

∥B∥2

)
(11)

for j = 1, ..., n.

Proof. In inequality (10), replacing A and B by
√

∥B∥
∥A∥A and

√
∥A∥
∥B∥B respectively, we

have

sj (AY B∗) ≤ 1

2
∥Y ∥ sj

(
∥B∥A∗A

∥A∥
+

∥A∥B∗B

∥B∥

)
=

1

2
∥Y ∥ ∥A∥ ∥B∥ sj

(
A∗A

∥A∥2
+

B∗B

∥B∥2

)
,

as required.

Theorem 4. Let A,B,C,D,E, F ∈ Mn(C) be such that C and D are positive semidefinite.
Then for j = 1, ..., 2n, we have

sj ((ACE∗ +BDF ∗)⊕ 0)

≤ max (∥C∥ , ∥D∥)
2
√
k1k2

sj

([
k2A

∗A+ k1E
∗E k2A

∗B + k1E
∗F

k2B
∗A+ k1F

∗E k2B
∗B + k1F

∗F

])
, (12)

where k1 = ∥AA∗ +BB∗∥ and k2 = ∥EE∗ + FF ∗∥ .

Proof. Let T1 =

[
A B
0 0

]
, T2 =

[
C 0
0 D

]
, and T3 =

[
E F
0 0

]
. Then

∥T1∥ =
√

∥T1T ∗
1 ∥ =

√
∥AA∗ +BB∗∥ =

√
k1,

∥T3∥ =
√

∥T3T ∗
3 ∥ =

√
∥EE∗ + FF ∗∥ =

√
k2



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 8 of 10

and
∥T2∥ = max (∥C∥ , ∥D∥) .

So,

sj (ACE∗ +BDF ∗ ⊕ 0) = sj (T1T2T
∗
3 )

≤ 1

2
∥T1∥ ∥T2∥ ∥T3∥ sj

(
T ∗
1 T1

∥T1∥2
+

T ∗
3 T3

∥T3∥2

)

=
1

2

√
k1k2max (∥C∥ , ∥D∥) sj


[

A∗A A∗B
B∗A BB

]
k1

+

[
E∗E E∗F
F ∗E F ∗F

]
k2



=
1

2

√
k1k2max (∥C∥ , ∥D∥) sj

k2

[
A∗A A∗B
B∗A B∗B

]
+ k1

[
E∗E E∗F
F ∗E F ∗F

]
k1k2


=

max (∥C∥ , ∥D∥)
2
√
k1k2

sj

([
k2A

∗A+ k1E
∗E k2A

∗B + k1E
∗F

k2B
∗A+ k1F

∗E k2B
∗B + k1F

∗F

])
,

as required.

Corollary 3. Let A,B,C,D ∈ Mn(C) be such that C and D are positive semidefinite.
Then for j = 1, ..., 2n, we have

sj ((ACB∗ +BDA∗)⊕ 0) ≤ max (∥C∥ , ∥D∥)
2

s2j

([
A B
B A

])
.

In particular, if C = D = I, we have

sj ((Re (AB
∗))⊕ 0) ≤ 1

4
s2j

([
A B
B A

])
where Re(T ) = T+T ∗

2 denotes the real part of T ∈ Mn(C).

Proof. Letting E = B and F = A in inequality (12), we have

sj ((ACB∗ +BDA∗)⊕ 0)

≤ max (∥C∥ , ∥D∥)
2k1

sj

([
k1A

∗A+ k1B
∗B k1A

∗B + k1B
∗A

k1B
∗A+ k1A

∗B k1B
∗B + k1A

∗A

])
=

max (∥C∥ , ∥D∥)
2

sj

([
A∗A+B∗B A∗B +B∗A
B∗A+A∗B B∗B +A∗A

])



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 9 of 10

=
max (∥C∥ , ∥D∥)

2
sj

([
A∗ B∗

B∗ A∗

] [
A B
B A

])
=

max (∥C∥ , ∥D∥)
2

sj

(∣∣∣∣[ A B
B A

]∣∣∣∣2
)

=
max (∥C∥ , ∥D∥)

2
s2j

([
A B
B A

])
=

max (∥C∥ , ∥D∥)
2

s2j

([
A B
B A

])
.

We end this paper by the following result, which gives a lower bound for singular values
of products and sums of matrices

Lemma 6. [12]Let A,B ∈ Mn(C). Then for j = 1, ..., n, we have

sj (AB) ≥ sn (A) sj (B) (13)

and
sj (AB) ≤ sj (A) s1 (B) . (14)

Theorem 5. Let A,B,C,D ∈ Mn(C). Then

sj ((AC +BD)⊕ 0) ≥ sn

((√
AA∗ +BB∗

)
⊕ 0
)
sj

((√
C∗C +D∗D

)
⊕ 0
)

(15)

and

sj ((AC +BD)⊕ 0) ≤ s1

((√
AA∗ +BB∗

)
⊕ 0
)
sj

((√
C∗C +D∗D

)
⊕ 0
)
. (16)

Proof. We have

sj ((AC +BD)⊕ 0)

= sj

([
A B
0 0

] [
C 0
D 0

])
≥ sn

([
A B
0 0

])
sj

([
C 0
D 0

])
(by inequality (13)

= sn

(∣∣∣∣[ A B
0 0

]∣∣∣∣) sj

(∣∣∣∣[ C 0
D 0

]∣∣∣∣)
= sn

(([
A∗ 0
B∗ 0

] [
A B
0 0

])1/2
)
sj

(([
C∗ D∗

0 0

] [
C 0
D 0

])1/2
)

= s1/2n

([
A∗ 0
B∗ 0

] [
A B
0 0

])
s
1/2
j

([
C∗ D∗

0 0

] [
C 0
D 0

])



A. Al-Natoor, R. Al-smadi, A. Burqan / Eur. J. Pure Appl. Math, 18 (2) (2025), 5844 10 of 10

= s1/2n

([
A B
0 0

] [
A∗ 0
B∗ 0

])
s
1/2
j

([
C∗ D∗

0 0

] [
C 0
D 0

])
= s1/2n

([
AA∗ +BB∗ 0

0 0

])
s
1/2
j

([
C∗C +D∗D 0

0 0

])
= sn

([ √
AA∗ +BB∗ 0

0 0

])
sj

([ √
C∗C +D∗D 0

0 0

])
= sn

((√
AA∗ +BB∗

)
⊕ 0
)
sj

((√
C∗C +D∗D

)
⊕ 0
)
.

which proves inequality (15). The inequality (16) follows by applying inequality (14) and
using the same argument that we use in proving inequality (15).

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