E extracta mathematicae Vol. 32, Núm. 2, 209 – 211 (2017)

Corrigendum to
“Moore-Penrose Inverse and Operator Inequalities”

Extracta Mathematicae 30 (2015), 29 – 39

Ameur Seddik

Department of Mathematics, Faculty of Mathematics and Computer Science,
University of Batna 2, Batna, Algeria

seddikameur@hotmail.com

Presented by Manuel González Received January 18, 2017

Abstract: We correct a mistake which affect our main results, namely the proof of Lema 1.
The main results of the article remain unchanged.

Key words: Closed range operator, Moore-Penrose inverse, normal operator, operator in-
equality.

AMS Subject Class. (2010): 47A30, 47A03, 47B15.

The paper mentioned in the title includes the following result as Lemma 1:

Lemma 1. Let S ∈ B(H). If S is surjective or injective with closed range
and satisfies the following inequality

∀X ∈ B(H),
∥∥S2X∥∥ + ∥∥XS2∥∥ ≥ 2 ∥SXS∥ , (∗)

then S is normal.

In the proof of this lemma, the matrix representation of the operator R2

was computed incorrectly obtaining

[
S∗1S1 0
0 S∗2S2

]
, while the correct form of

this matrix is R2 =

[
S∗1S1 0
0 (S1S2)

∗
(S1S2)

]
. Since all the results of the paper

are based on this lemma, we shall give here a correct proof of it.

The original proof is given in two cases. The second case follows imme-
diately from the first one. The proof of the first case is divided in six steps.
The mistake is in the fourth step.

For S ∈ B(H) with closed range, S+ denotes the Moore-Penrose inverse
of S.

209



210 a. seddik

Proof of Lemma 1. Assume that S ̸= 0 and that all 2×2 matrices used in this
proof are given with respect to the orthogonal direct sum H = R(S) ⊕ ker S∗.

Then S =

[
S

1
S

2

0 0

]
.

We put P = |S| , Q = |S∗| , P1 = |S1| , P2 = |S2| , Q1 = (S1S∗1 + S2S
∗
2)

1
2 .

So we have S∗S = P 2 =

[
P 21 S

∗
1S2

S∗2S1 P
2
2

]
, SS∗ = Q2 =

[
Q2

1
0

0 0

]
. It is clear

that Q1 is invertible and Q
+ =

[
Q−11 0
0 0

]
.

Case 1. Assume that S is injective with closed range and satisfies (∗). Then
S+S = I, ker P = ker S = {0}, and R(P) = R(S∗S) is closed (since R(S∗) is
also closed). Thus ker P = {0} and R(P) = (ker P)⊥ = H. So, P is invertible.

Note that inequality (∗) implies the following inequality:

∀X ∈ B(H),
∥∥S2S+XS+∥∥ + ∥∥S+XS∥∥ ≥ 2 ∥∥SS+X∥∥ . (1)

The proof is given in four steps.
Step 1.

(
S2

)+
S = S+. See Step 2 of the original proof.

Step 2.
(
S2

)+
= (S+)2. See Step 3 of the original proof.

Step 3. ker S∗ = {0} . Since S is injective, then ker S∗ = {0} if and only if
S

2
= 0. Assume that S

2
̸= 0.

Since
(
S2

)+
= (S+)2, then the two operators S∗S and SS+ commute (see

[1, 2]). Thus P 2 =

[
P 2

1
0

0 P 22

]
, hence P =

[
P

1
0

0 P2

]
.

Since ker S∗ ̸= {0}, then σ(Q2) = σ(Q21)∪{0}. From the fact that σ(P
2) =

σ(Q2) − {0}, we have σ(P 2) = σ(Q21). Then σ(P
2
1
) ∪ σ(P 22 ) = σ(Q

2
1). Hence

σ(P 2
1
) ⊂ σ(Q21). Thus σ(P1) ⊂ σ(Q1).

Using the polar decomposition of S and S∗ in inequality (1), we obtain
the following inequality:

∀X ∈ B(H),
∥∥S2S+XP −1∥∥ + ∥∥Q+XQ∥∥ ≥ 2 ∥∥SS+X∥∥ .

By taking X =

[
X

1
0

0 0

]
(resp. X =

[
0 X2
0 0

]
), where X

1
∈ B(R(S))

(resp. X2 ∈ B(ker S∗, R(S))) in the last inequality, and since S2S+ =[
S

1
0

0 0

]
, we deduce the two following inequalities

∀X
1
∈ B(R(S)),

∥∥P1X1P −11 ∥∥ + ∥∥Q−11 X1Q1∥∥ ≥ 2 ∥X1∥ , (2)



corrigendum to “moore-penrose inverse and . . . ” 211

∀X2 ∈ B(ker S∗, R(S)),
∥∥P1X2P −12 ∥∥ ≥ 2 ∥X2∥ . (3)

By taking X2 = x ⊗ y (where x ∈ (R(S))1, y ∈ ker S∗) in (3), we obtain

∀x ∈ (R(S))1, ∀y ∈ ker S∗, ∥P1x∥
∥∥P −12 y∥∥ ≥ 2 ∥y∥ .

So we have

∀x ∈ (R(S))1, ∀y ∈ (ker S∗)1 , ∥P1x∥ ≥ 2 ∥P2y∥ .

Thus ∥P2y∥ ≤ k2, for every y ∈ (ker S
∗)1 (where k = inf∥x∥=1 ∥P1x∥ > 0),

and then
⟨
P 22 y, y

⟩
≤ k2/4, for every y ∈ (ker S∗)1. So we obtain σ(P

2
2 ) ⊂(

0, k
2

4

]
and σ(P 21 ) ⊂ [k

2, ∞).
Since σ(P

1
) ⊂ σ(Q

1
), and P

1
, Q

1
satisfy the inequality (2), then using

a variation of [3, Theorem 3.6] (in that paper Theorem 3.6 is stated with
equality between the spectra but the proof is the same for inclusion between
the spectra), we obtain P1 = Q1. Hence σ(Q

2
1
) = σ(P 21 ) = σ(P

2
1
) ∪ σ(P 22 ).

Then σ(P 22 ) ⊂ σ(P
2
1 ), that is impossible since

(
0, k

2

4

]
∩[k2, ∞) = ∅. Therefore

ker S∗ = {0}.
Step 4. S is normal. Since ker S∗ = {0}, we obtain R(S) = H. So that

S is invertible and satisfies the inequality (∗). Hence S satisfies the following
inequality

∀X ∈ B(H),
∥∥SXS−1∥∥ + ∥∥S−1XS∥∥ ≥ 2 ∥X∥ .

Therefore S is normal (using [4]).

Case 2. Assume that S is surjective and satisfies (∗). Then S∗ is injective
with closed range and satisfies inequality (∗). From Case 1, S∗ is normal.
Hence S is normal, and the proof is finished.

References

[1] R. Bouldin, The pseudo-inverse of a product, SIAM J. Appl. Math. 24 (1973),
489 – 495.

[2] S. Izumino, The product of operators with closed range and an extension of
the reverse order law, Tôhoku Math. J. (2) 34 (1) (1982), 43 – 52.

[3] A. Seddik, Some results related to the Corach-Porta-Recht inequality, Proc.
Amer. Math. Soc. 129 (2001), 3009 – 3015.

[4] A. Seddik, On the injective norm and characterization of some subclasses of
normal operators by inequalities or equalities, J. Math. Anal. Appl. 351 (1)
(2009), 277 – 284.